Selina Concise Solutions for ICSE Class 10 Mathematics Chapter 9 Matrices

Question 1. State, whether the following statements are true or false. If false, give a reason.
(i) If A and B are two matrices of orders 3 × 2 and 2 × 3 respectively; then their sum A + B is possible.
(ii) The matrices A₂ × ₃ and B₂ × ₃ are conformable for subtraction.
(iii) Transpose of a 2 × 1 matrix is a 2 × 1 matrix.
(iv) Transpose of a square matrix is a square matrix.
(v) A column matrix has many columns and one row.
Answer:
(i) False - The sum A + B is possible when the order of both the matrices A and B are same.
(ii) True
(iii) False - Transpose of a 2 × 1 matrix is a 1 × 2 matrix.
(iv) True
(v) False - A column matrix has only one column and many rows.
In simple words: To add two matrices, they must have the same size. Think of it like adding two boxes - they must have the same number of rows and columns. When we flip a matrix (transpose), rows become columns and columns become rows.

📝 Teacher's Note: Draw two boxes on the board to show matrix orders. Students can see that 3×2 and 2×3 are different sizes and cannot be added. Show how transpose works by writing a simple 2×1 matrix and flipping it.

🎯 Exam Tip: Always check the order (size) of matrices first. Write "same order required" for addition and subtraction. For transpose, remember rows and columns swap places.

Question 2. Given: \(\begin{bmatrix} x & y + 2 \\ 3 & z - 1 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ 3 & 2 \end{bmatrix}\), find x, y and z.
Answer: If two matrices are equal, then their corresponding elements are also equal. Therefore, we have:
x = 3
y + 2 = 1 \(\implies\) y = -1
z - 1 = 2 \(\implies\) z = 3
In simple words: When two matrices are equal, each number in the same position must be equal. So we match each position and solve simple equations.

📝 Teacher's Note: Show students how to match positions - top left with top left, top right with top right, etc. This makes it easy to set up the equations.

🎯 Exam Tip: Write "corresponding elements are equal" first. Then write each equation clearly. Show your working for solving each equation.

Question 3. Solve for a, b and c if
(i) \(\begin{bmatrix} -4 & a + 5 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} b + 4 & 2 \\ 3 & c - 1 \end{bmatrix}\)
(ii) \(\begin{bmatrix} a & a - b \\ b + c & 0 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ 2 & 0 \end{bmatrix}\)
Answer: If two matrices are equal, then their corresponding elements are also equal.
(i)
a + 5 = 2 \(\implies\) a = -3
-4 = b + 4 \(\implies\) b = -8
2 = c - 1 \(\implies\) c = 3
(ii) a = 3
a - b = -1
\(\implies\) b = a + 1 = 4
b + c = 2
\(\implies\) c = 2 - b = 2 - 4 = -2
In simple words: We match each position in both matrices. This gives us simple equations to solve step by step.

📝 Teacher's Note: In part (ii), show students how to substitute the value of a into the second equation. This step-by-step approach prevents mistakes.

🎯 Exam Tip: Always write the equations first, then solve them. Check your answers by substituting back into the original matrix.

Question 4. If A = [8 -3] and B = [4 -5]; find: (i) A + B (ii) B - A
Answer:
(i) A + B = [8 -3] + [4 -5] = [8 + 4 -3 + (-5)] = [12 -8]
(ii) B - A = [4 -5] - [8 -3]
= [4 - 8 -5 + 3]
= [-4 -2]
In simple words: To add matrices, add numbers in the same positions. To subtract, subtract numbers in the same positions. It's like adding or subtracting two lists of numbers.

📝 Teacher's Note: Use different colored pens for different positions. This helps students see which numbers to add or subtract together.

🎯 Exam Tip: Write each step clearly. Show the addition or subtraction of each element separately. Don't skip steps in matrix operations.

Question 5. If A = \(\begin{bmatrix} 2 \\ 5 \end{bmatrix}\), B = \(\begin{bmatrix} 1 \\ 4 \end{bmatrix}\) and C = \(\begin{bmatrix} 6 \\ -2 \end{bmatrix}\); find:
(i) B + C (ii) A - C
(iii) A + B - C (iv) A - B + C
Answer:
(i) B + C = \(\begin{bmatrix} 1 \\ 4 \end{bmatrix} + \begin{bmatrix} 6 \\ -2 \end{bmatrix} = \begin{bmatrix} 1 + 6 \\ 4 + (-2) \end{bmatrix} = \begin{bmatrix} 7 \\ 2 \end{bmatrix}\)
(ii) A - C = \(\begin{bmatrix} 2 \\ 5 \end{bmatrix} - \begin{bmatrix} 6 \\ -2 \end{bmatrix} = \begin{bmatrix} 2 - 6 \\ 5 - (-2) \end{bmatrix} = \begin{bmatrix} -4 \\ 7 \end{bmatrix}\)
(iii) A + B - C = \(\begin{bmatrix} 2 \\ 5 \end{bmatrix} + \begin{bmatrix} 1 \\ 4 \end{bmatrix} - \begin{bmatrix} 6 \\ -2 \end{bmatrix}\)
= \(\begin{bmatrix} 2 + 1 - 6 \\ 5 + 4 - (-2) \end{bmatrix} = \begin{bmatrix} -3 \\ 11 \end{bmatrix}\)
(iv) A - B + C = \(\begin{bmatrix} 2 \\ 5 \end{bmatrix} - \begin{bmatrix} 1 \\ 4 \end{bmatrix} + \begin{bmatrix} 6 \\ -2 \end{bmatrix}\)
= \(\begin{bmatrix} 2 - 1 + 6 \\ 5 - 4 + (-2) \end{bmatrix} = \begin{bmatrix} 7 \\ -1 \end{bmatrix}\)
In simple words: These are column matrices - they have one column and many rows. We add or subtract the numbers in each row separately.

📝 Teacher's Note: Point out that these are column matrices. Students often get confused about which numbers to combine. Always work row by row.

🎯 Exam Tip: For multiple operations like A + B - C, work from left to right. First do A + B, then subtract C from the result.

Question 6. Wherever possible, write each of the following as a single matrix.
(i) \(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} -1 & -2 \\ 1 & -7 \end{bmatrix}\)
(ii) \(\begin{bmatrix} 2 & 3 & 4 \\ 5 & 6 & 7 \end{bmatrix} - \begin{bmatrix} 0 & 2 & 3 \\ 6 & -1 & 0 \end{bmatrix}\)
(iii) \(\begin{bmatrix} 0 & 1 & 2 \\ 4 & 6 & 7 \end{bmatrix} + \begin{bmatrix} 3 & 4 \\ 6 & 8 \end{bmatrix}\)
Answer:
(i) \(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} -1 & -2 \\ 1 & -7 \end{bmatrix} = \begin{bmatrix} 1-1 & 2+(-2) \\ 3+1 & 4+(-7) \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 4 & -3 \end{bmatrix}\)
(ii) \(\begin{bmatrix} 2 & 3 & 4 \\ 5 & 6 & 7 \end{bmatrix} - \begin{bmatrix} 0 & 2 & 3 \\ 6 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 2-0 & 3-2 & 4-3 \\ 5-6 & 6+1 & 7-0 \end{bmatrix} = \begin{bmatrix} 2 & 1 & 1 \\ -1 & 7 & 7 \end{bmatrix}\)
(iii) Addition is not possible, because both matrices are not of same order.
In simple words: We can only add or subtract matrices of the same size. If they have different sizes, the operation is not possible.

📝 Teacher's Note: Always check the size first. Show students that 2×3 and 2×2 matrices cannot be added because they have different numbers of columns.

🎯 Exam Tip: Before starting any calculation, write down the order of each matrix. If orders are different, write "not possible" and give the reason.

Question 7. Find x and y from the following equations:
(i) \(\begin{bmatrix} 5 & 2 \\ -1 & y-1 \end{bmatrix} - \begin{bmatrix} 1 & x-1 \\ 2 & -3 \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ -3 & 2 \end{bmatrix}\)
(ii) [-8 x] + [y -2] = [-3 2]
Answer:
(i)
\(\begin{bmatrix} 5 & 2 \\ -1 & y-1 \end{bmatrix} - \begin{bmatrix} 1 & x-1 \\ 2 & -3 \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ -3 & 2 \end{bmatrix}\)
\(\implies \begin{bmatrix} 5-1 & 2-(x-1) \\ -1-2 & y-1+3 \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ -3 & 2 \end{bmatrix}\)
\(\implies \begin{bmatrix} 4 & 3-x \\ -3 & y+2 \end{bmatrix} = \begin{bmatrix} 4 & 7 \\ -3 & 2 \end{bmatrix}\)
Equating the corresponding elements, we get,
3 - x = 7 and y + 2 = 2
Thus, we get, x = -4 and y = 0.
(ii)
[-8 x] + [y -2] = [-3 2]
\(\implies\) [-8 + y x - 2] = [-3 2]
Equating the corresponding elements, we get,
-8 + y = -3 and x - 2 = 2
Thus, we get, x = 4 and y = 5.
In simple words: We do the matrix operation first, then match corresponding positions to get simple equations to solve.

📝 Teacher's Note: Show students to work step by step. First do the subtraction or addition, then equate corresponding elements. This prevents confusion.

🎯 Exam Tip: Always simplify the left side completely before equating elements. Write each equation clearly and solve them separately.

Question 8. Given: M = \(\begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix}\), find its transpose matrix Mᵗ. If possible, find:
(i) M + Mᵗ (ii) Mᵗ - M
Answer:
M = \(\begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix}\)
Mᵗ = \(\begin{bmatrix} 5 & -2 \\ -3 & 4 \end{bmatrix}\)
(i) M + Mᵗ = \(\begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix} + \begin{bmatrix} 5 & -2 \\ -3 & 4 \end{bmatrix} = \begin{bmatrix} 5+5 & -3+(-2) \\ -2+(-3) & 4+4 \end{bmatrix} = \begin{bmatrix} 10 & -5 \\ -5 & 8 \end{bmatrix}\)
(ii) Mᵗ - M = \(\begin{bmatrix} 5 & -2 \\ -3 & 4 \end{bmatrix} - \begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} 5-5 & -2+3 \\ -3+2 & 4-4 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\)
In simple words: To find transpose, we flip the matrix - rows become columns and columns become rows. Then we can add or subtract normally.

📝 Teacher's Note: Draw arrows to show how elements move when taking transpose. The element in row 1, column 2 goes to row 2, column 1.

🎯 Exam Tip: Always find the transpose first and write it clearly. Then do the required operations. Remember that transpose flips rows and columns.

Question 9. Write the additive inverse of matrices A, B and C:
Where A = [6 -5], B = \(\begin{bmatrix} -2 & 0 \\ 4 & -1 \end{bmatrix}\) and C = \(\begin{bmatrix} -7 \\ 4 \end{bmatrix}\)
Answer:
We know additive inverse of a matrix is its negative.
Additive inverse of A = -A = -[6 -5] = [-6 5]
Additive inverse of B = -B = -\(\begin{bmatrix} -2 & 0 \\ 4 & -1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ -4 & 1 \end{bmatrix}\)
Additive inverse of C = -C = -\(\begin{bmatrix} -7 \\ 4 \end{bmatrix} = \begin{bmatrix} 7 \\ -4 \end{bmatrix}\)
In simple words: Additive inverse means we change the sign of every number in the matrix. Positive becomes negative, negative becomes positive.

📝 Teacher's Note: Explain that additive inverse is like the opposite. When we add a matrix to its additive inverse, we get a zero matrix.

🎯 Exam Tip: Write "additive inverse = -A" first. Then change the sign of each element. Don't forget any element.

Question 10. Given A = [2 -3], B = [0 2] and C = [-1 4]; find the matrix X in each of the following:
(i) X + B = C - A
(ii) A - X = B + C
Answer:
(i) X + B = C - A
X + [0 2] = [-1 4] - [2 -3]
X + [0 2] = [-1-2 4+3] = [-3 7]
X = [-3 7] - [0 2] = [-3-0 7-2] = [-3 5]

(ii) A - X = B + C
[2 -3] - X = [0 2] + [-1 4]
[2 -3] - X = [0-1 2+4]
[2 -3] - X = [-1 6]
[2 -3] - [-1 6] = X
X = [2+1 -3-6] = [3 -9]
In simple words: These are like solving equations with matrices. We move matrices from one side to the other, just like moving numbers in algebra.

📝 Teacher's Note: Show students this is like solving x + 5 = 10, where x = 10 - 5. The same idea works with matrices.

🎯 Exam Tip: Always isolate X on one side. Do the matrix operations on the other side first, then find X. Check your answer by substituting back.

Question 11. Given A = \(\begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix}\) and B = \(\begin{bmatrix} 3 & -3 \\ -2 & 0 \end{bmatrix}\), find the matrix X in each of the following:
(i) A + X = B
(ii) A - X = B
(iii) X - B = A
Answer:
(i) A + X = B
X = B - A
X = \(\begin{bmatrix} 3 & -3 \\ -2 & 0 \end{bmatrix}\) - \(\begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix}\) = \(\begin{bmatrix} 3+1 & -3-0 \\ -2-2 & 0+4 \end{bmatrix}\) = \(\begin{bmatrix} 4 & -3 \\ -4 & 4 \end{bmatrix}\)

(ii) A - X = B
X = A - B
X = \(\begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix}\) - \(\begin{bmatrix} 3 & -3 \\ -2 & 0 \end{bmatrix}\) = \(\begin{bmatrix} -1-3 & 0+3 \\ 2+2 & -4-0 \end{bmatrix}\) = \(\begin{bmatrix} -4 & 3 \\ 4 & -4 \end{bmatrix}\)

(iii) X - B = A
X = A + B
X = \(\begin{bmatrix} -1 & 0 \\ 2 & -4 \end{bmatrix}\) + \(\begin{bmatrix} 3 & -3 \\ -2 & 0 \end{bmatrix}\) = \(\begin{bmatrix} -1+3 & 0-3 \\ 2-2 & -4+0 \end{bmatrix}\) = \(\begin{bmatrix} 2 & -3 \\ 0 & -4 \end{bmatrix}\)
In simple words: To find X, we move matrices to the other side by changing their signs. If X is added, subtract to find it. If X is subtracted, add to find it.

📝 Teacher's Note: Teach students to think like algebra. If a + x = b, then x = b - a. Same rule works for matrices. Always subtract or add each element separately.

🎯 Exam Tip: First write the formula (like X = B - A), then substitute the matrices, then subtract each element position by position. Show all steps clearly.

Exercise 9B

Question 1. Evaluate:
(i) 3\(\begin{bmatrix} 5 & -2 \end{bmatrix}\)
(ii) 7\(\begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix}\)
(iii) 2\(\begin{bmatrix} -1 & 0 \\ 2 & -3 \end{bmatrix}\) + \(\begin{bmatrix} 3 & 3 \\ 5 & 0 \end{bmatrix}\)
(iv) 6\(\begin{bmatrix} 3 \\ -2 \end{bmatrix}\) - 2\(\begin{bmatrix} -8 \\ 1 \end{bmatrix}\)
Answer:
(i) 3\(\begin{bmatrix} 5 & -2 \end{bmatrix}\) = \(\begin{bmatrix} 15 & -6 \end{bmatrix}\)
(ii) 7\(\begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix}\) = \(\begin{bmatrix} -7 & 14 \\ 0 & 7 \end{bmatrix}\)
(iii) 2\(\begin{bmatrix} -1 & 0 \\ 2 & -3 \end{bmatrix}\) + \(\begin{bmatrix} 3 & 3 \\ 5 & 0 \end{bmatrix}\) = \(\begin{bmatrix} -2 & 0 \\ 4 & -6 \end{bmatrix}\) + \(\begin{bmatrix} 3 & 3 \\ 5 & 0 \end{bmatrix}\) = \(\begin{bmatrix} -2+3 & 0+3 \\ 4+5 & -6+0 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 3 \\ 9 & -6 \end{bmatrix}\)
(iv) 6\(\begin{bmatrix} 3 \\ -2 \end{bmatrix}\) - 2\(\begin{bmatrix} -8 \\ 1 \end{bmatrix}\) = \(\begin{bmatrix} 18 \\ -12 \end{bmatrix}\) - \(\begin{bmatrix} -16 \\ 2 \end{bmatrix}\) = \(\begin{bmatrix} 18+16 \\ -12-2 \end{bmatrix}\) = \(\begin{bmatrix} 34 \\ -14 \end{bmatrix}\)
In simple words: When we multiply a matrix by a number, we multiply each element by that number. When adding or subtracting matrices, we work with each position separately.

📝 Teacher's Note: Show students that scalar multiplication is like multiplying each box by that number. Use simple examples with small numbers first before bigger ones.

🎯 Exam Tip: For scalar multiplication, multiply every single element. For addition/subtraction, work position by position. Always show each step clearly.

Question 2. Find x and y if:
(i) 3\(\begin{bmatrix} 4 & x \end{bmatrix}\) + 2\(\begin{bmatrix} y & -3 \end{bmatrix}\) = \(\begin{bmatrix} 10 & 0 \end{bmatrix}\)
(ii) x\(\begin{bmatrix} -1 \\ 2 \end{bmatrix}\) - 4\(\begin{bmatrix} -2 \\ y \end{bmatrix}\) = \(\begin{bmatrix} 7 \\ -8 \end{bmatrix}\)
Answer:
(i) 3\(\begin{bmatrix} 4 & x \end{bmatrix}\) + 2\(\begin{bmatrix} y & -3 \end{bmatrix}\) = \(\begin{bmatrix} 10 & 0 \end{bmatrix}\)
\(\begin{bmatrix} 12 & 3x \end{bmatrix}\) + \(\begin{bmatrix} 2y & -6 \end{bmatrix}\) = \(\begin{bmatrix} 10 & 0 \end{bmatrix}\)
\(\begin{bmatrix} 12 + 2y & 3x - 6 \end{bmatrix}\) = \(\begin{bmatrix} 10 & 0 \end{bmatrix}\)
Comparing the corresponding elements, we get:
12 + 2y = 10 and 3x - 6 = 0
Simplifying, we get: y = -1 and x = 2

(ii) x\(\begin{bmatrix} -1 \\ 2 \end{bmatrix}\) - 4\(\begin{bmatrix} -2 \\ y \end{bmatrix}\) = \(\begin{bmatrix} 7 \\ -8 \end{bmatrix}\)
\(\begin{bmatrix} -x \\ 2x \end{bmatrix}\) - \(\begin{bmatrix} -8 \\ 4y \end{bmatrix}\) = \(\begin{bmatrix} 7 \\ -8 \end{bmatrix}\)
\(\begin{bmatrix} -x + 8 \\ 2x - 4y \end{bmatrix}\) = \(\begin{bmatrix} 7 \\ -8 \end{bmatrix}\)
Comparing corresponding elements, we get:
-x + 8 = 7 and 2x - 4y = -8
Simplifying, we get:
x = 1 and y = \(\frac{5}{2}\) = 2.5
In simple words: We first multiply each matrix by its number, then add or subtract them. Then we compare each position to make two simple equations and solve for x and y.

📝 Teacher's Note: Teach students to first do all scalar multiplications, then matrix operations, then compare element by element. This step-by-step approach prevents mistakes.

🎯 Exam Tip: Always write "comparing corresponding elements" before making equations. Show each multiplication step clearly. Check your answer by substituting back.

Question 3. Given A = \(\begin{bmatrix} 2 & 1 \\ 3 & 0 \end{bmatrix}\), B = \(\begin{bmatrix} 1 & 1 \\ 5 & 2 \end{bmatrix}\) and C = \(\begin{bmatrix} -3 & -1 \\ 0 & 0 \end{bmatrix}\), find:
(i) 2A - 3B + C
(ii) A + 2C - B
Answer:
(i) 2A - 3B + C
= 2\(\begin{bmatrix} 2 & 1 \\ 3 & 0 \end{bmatrix}\) - 3\(\begin{bmatrix} 1 & 1 \\ 5 & 2 \end{bmatrix}\) + \(\begin{bmatrix} -3 & -1 \\ 0 & 0 \end{bmatrix}\)
= \(\begin{bmatrix} 4 & 2 \\ 6 & 0 \end{bmatrix}\) - \(\begin{bmatrix} 3 & 3 \\ 15 & 6 \end{bmatrix}\) + \(\begin{bmatrix} -3 & -1 \\ 0 & 0 \end{bmatrix}\)
= \(\begin{bmatrix} 4-3-3 & 2-3-1 \\ 6-15+0 & 0-6+0 \end{bmatrix}\)
= \(\begin{bmatrix} -2 & -2 \\ -9 & -6 \end{bmatrix}\)

(ii) A + 2C - B
= \(\begin{bmatrix} 2 & 1 \\ 3 & 0 \end{bmatrix}\) + 2\(\begin{bmatrix} -3 & -1 \\ 0 & 0 \end{bmatrix}\) - \(\begin{bmatrix} 1 & 1 \\ 5 & 2 \end{bmatrix}\)
= \(\begin{bmatrix} 2 & 1 \\ 3 & 0 \end{bmatrix}\) + \(\begin{bmatrix} -6 & -2 \\ 0 & 0 \end{bmatrix}\) - \(\begin{bmatrix} 1 & 1 \\ 5 & 2 \end{bmatrix}\)
= \(\begin{bmatrix} 2-6-1 & 1-2-1 \\ 3+0-5 & 0+0-2 \end{bmatrix}\)
= \(\begin{bmatrix} -5 & -2 \\ -2 & -2 \end{bmatrix}\)
In simple words: We first multiply each matrix by its number, then do addition and subtraction position by position. Always follow the order of operations.

📝 Teacher's Note: Remind students to do scalar multiplication first, then matrix addition/subtraction. Use different colors for each step to show the process clearly.

🎯 Exam Tip: Do scalar multiplications first, then combine matrices step by step. Write each intermediate step. Don't try to do everything in your head.

Question 4. If \(\begin{bmatrix} 4 & -2 \\ 4 & 0 \end{bmatrix}\) + 3A = \(\begin{bmatrix} -2 & -2 \\ 1 & -3 \end{bmatrix}\), find A.
Answer:
\(\begin{bmatrix} 4 & -2 \\ 4 & 0 \end{bmatrix}\) + 3A = \(\begin{bmatrix} -2 & -2 \\ 1 & -3 \end{bmatrix}\)

3A = \(\begin{bmatrix} -2 & -2 \\ 1 & -3 \end{bmatrix}\) - \(\begin{bmatrix} 4 & -2 \\ 4 & 0 \end{bmatrix}\)

3A = \(\begin{bmatrix} -2-4 & -2+2 \\ 1-4 & -3-0 \end{bmatrix}\)

3A = \(\begin{bmatrix} -6 & 0 \\ -3 & -3 \end{bmatrix}\)

A = \(\frac{1}{3}\)\(\begin{bmatrix} -6 & 0 \\ -3 & -3 \end{bmatrix}\) = \(\begin{bmatrix} -2 & 0 \\ -1 & -1 \end{bmatrix}\)
In simple words: We moved the known matrix to the other side by subtracting it. Then we divided the result by 3 to find A.

📝 Teacher's Note: This is like solving 5 + 3x = 17. We subtract 5 from both sides, then divide by 3. Same logic works for matrices.

🎯 Exam Tip: First isolate 3A by subtracting the known matrix. Then divide each element by 3 to get A. Show the division step clearly.

Question 5. Given A = \(\begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix}\) and B = \(\begin{bmatrix} -4 & -1 \\ -3 & -2 \end{bmatrix}\)
(i) find the matrix 2A + B
(ii) find the matrix C such that: C + B = \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
Answer:
(i) 2\(\begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix}\) + \(\begin{bmatrix} -4 & -1 \\ -3 & -2 \end{bmatrix}\) = \(\begin{bmatrix} 2 & 8 \\ 4 & 6 \end{bmatrix}\) + \(\begin{bmatrix} -4 & -1 \\ -3 & -2 \end{bmatrix}\) = \(\begin{bmatrix} 2-4 & 8-1 \\ 4-3 & 6-2 \end{bmatrix}\) = \(\begin{bmatrix} -2 & 7 \\ 1 & 4 \end{bmatrix}\)

(ii) C + B = \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)

C = \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\) - B = \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\) - \(\begin{bmatrix} -4 & -1 \\ -3 & -2 \end{bmatrix}\) = \(\begin{bmatrix} 0+4 & 0+1 \\ 0+3 & 0+2 \end{bmatrix}\) = \(\begin{bmatrix} 4 & 1 \\ 3 & 2 \end{bmatrix}\)
In simple words: In part (i), we multiply A by 2 then add B. In part (ii), we find a matrix that when added to B gives zero matrix. This matrix is the negative of B.

📝 Teacher's Note: The zero matrix is like zero in regular numbers. If C + B = 0, then C = -B. Show students that -B means changing the sign of every element.

🎯 Exam Tip: For scalar multiplication, multiply every element. The zero matrix has all elements as 0. To find -B, change the sign of each element in B.

Question 6. If 2\(\begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix}\) + 3\(\begin{bmatrix} 1 & 3 \\ y & 2 \end{bmatrix}\) = \(\begin{bmatrix} z & -7 \\ 15 & 8 \end{bmatrix}\), find the values of x, y and z.
Answer:
2\(\begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix}\) + 3\(\begin{bmatrix} 1 & 3 \\ y & 2 \end{bmatrix}\) = \(\begin{bmatrix} z & -7 \\ 15 & 8 \end{bmatrix}\)

\(\begin{bmatrix} 6 & 2x \\ 0 & 2 \end{bmatrix}\) + \(\begin{bmatrix} 3 & 9 \\ 3y & 6 \end{bmatrix}\) = \(\begin{bmatrix} z & -7 \\ 15 & 8 \end{bmatrix}\)

\(\begin{bmatrix} 9 & 2x+9 \\ 3y & 8 \end{bmatrix}\) = \(\begin{bmatrix} z & -7 \\ 15 & 8 \end{bmatrix}\)

Comparing the corresponding elements, we get:
2x + 9 = -7 ⇒ 2x = -16 ⇒ x = -8
3y = 15 ⇒ y = 5
z = 9
In simple words: We first multiply each matrix by its number, then add them together. Then we compare each position to make equations and solve for the unknown values.

📝 Teacher's Note: Students should first do scalar multiplications, then matrix addition, then compare element by element. This prevents confusion and mistakes.

🎯 Exam Tip: Show all scalar multiplications first. Then show matrix addition clearly. Finally, compare each position separately to get individual equations for x, y, and z.

Question 7. Given A = \(\begin{bmatrix} -3 & 6 \\ 0 & -9 \end{bmatrix}\) and A^t is its transpose matrix. Find:
(i) 2A + 3A^t (ii) 2A^t - 3A
(iii) \(\frac{1}{2}\)A - \(\frac{1}{3}\)A^t (iv) A^t - \(\frac{1}{3}\)A
Answer:
A = \(\begin{bmatrix} -3 & 6 \\ 0 & -9 \end{bmatrix}\)

A^t = \(\begin{bmatrix} -3 & 0 \\ 6 & -9 \end{bmatrix}\)

(i) 2A + 3A^t
= 2\(\begin{bmatrix} -3 & 6 \\ 0 & -9 \end{bmatrix}\) + 3\(\begin{bmatrix} -3 & 0 \\ 6 & -9 \end{bmatrix}\)
= \(\begin{bmatrix} -6 & 12 \\ 0 & -18 \end{bmatrix}\) + \(\begin{bmatrix} -9 & 0 \\ 18 & -27 \end{bmatrix}\)
= \(\begin{bmatrix} -15 & 12 \\ 18 & -45 \end{bmatrix}\)
In simple words: Transpose means we swap rows and columns. Then we multiply each matrix by its number and add them position by position.

📝 Teacher's Note: Show students that transpose is like flipping the matrix over its diagonal. The element at position (1,2) moves to position (2,1), and so on.

🎯 Exam Tip: First write A^t correctly by swapping rows and columns. Then do scalar multiplications, then matrix operations. Always double-check your transpose.

Question 8.
Given \( A = \begin{bmatrix} 1 & 1 \\ -2 & 0 \end{bmatrix} \) and \( B = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} \)
Solve for matrix X:
(i) X + 2A = B
(ii) 3X + B + 2A = O
(iii) 3A - 2X = X - 2B.

Answer:
(i) X + 2A = B
X = B - 2A
\( X = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} - 2\begin{bmatrix} 1 & 1 \\ -2 & 0 \end{bmatrix} \)
\( X = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 2 \\ -4 & 0 \end{bmatrix} \)
\( X = \begin{bmatrix} 0 & -3 \\ 5 & 1 \end{bmatrix} \)

(ii) 3X + B + 2A = O
3X = -2A - B
\( 3X = -2\begin{bmatrix} 1 & 1 \\ -2 & 0 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} \)
\( 3X = \begin{bmatrix} -2 & -2 \\ 4 & 0 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} \)
\( 3X = \begin{bmatrix} -4 & -1 \\ 3 & -1 \end{bmatrix} \)
\( X = \begin{bmatrix} \frac{-4}{3} & \frac{-1}{3} \\ 1 & \frac{-1}{3} \end{bmatrix} \)

(iii) 3A - 2X = X - 2B
3A + 2B = X + 2X
3X = 3A + 2B
\( 3X = 3\begin{bmatrix} 1 & 1 \\ -2 & 0 \end{bmatrix} + 2\begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} \)
\( 3X = \begin{bmatrix} 3 & 3 \\ -6 & 0 \end{bmatrix} + \begin{bmatrix} 4 & -2 \\ 2 & 2 \end{bmatrix} \)
\( 3X = \begin{bmatrix} 7 & 1 \\ -4 & 2 \end{bmatrix} \)
\( X = \begin{bmatrix} \frac{7}{3} & \frac{1}{3} \\ \frac{-4}{3} & \frac{2}{3} \end{bmatrix} \)

📝 Teacher's Note: For matrix equations, always move all known matrices to one side first. Then do the matrix addition or subtraction step by step. Check your answer by putting it back in the original equation.

🎯 Exam Tip: Write each step clearly. Show matrix addition and subtraction with proper brackets. Always check if your final X matrix has the correct size.

Question 9.
If \( M = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \) and \( N = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \), show that:
\( 3M + 5N = \begin{bmatrix} 5 \\ 3 \end{bmatrix} \)

Answer:
3M + 5N
\( = 3\begin{bmatrix} 0 \\ 1 \end{bmatrix} + 5\begin{bmatrix} 1 \\ 0 \end{bmatrix} \)
\( = \begin{bmatrix} 0 \\ 3 \end{bmatrix} + \begin{bmatrix} 5 \\ 0 \end{bmatrix} \)
\( = \begin{bmatrix} 5 \\ 3 \end{bmatrix} \)
Hence proved.

📝 Teacher's Note: When multiplying a matrix by a number, multiply each element by that number. When adding matrices, add corresponding elements.

🎯 Exam Tip: Write "Hence proved" at the end when you need to show something. Show all steps clearly with proper matrix brackets.

Question 10.
If I is the unit matrix of order 2 × 2; find the matrix M, such that:
(i) \( M - 2I = 3\begin{bmatrix} -1 & 0 \\ 4 & 1 \end{bmatrix} \)
(ii) \( 5M + 3I = 4\begin{bmatrix} 2 & -5 \\ 0 & -3 \end{bmatrix} \)

Answer:
(i) \( M - 2I = 3\begin{bmatrix} -1 & 0 \\ 4 & 1 \end{bmatrix} \)
\( M = 3\begin{bmatrix} -1 & 0 \\ 4 & 1 \end{bmatrix} + 2I \)
\( M = 3\begin{bmatrix} -1 & 0 \\ 4 & 1 \end{bmatrix} + 2\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( M = \begin{bmatrix} -3 & 0 \\ 12 & 3 \end{bmatrix} + \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \)
\( M = \begin{bmatrix} -1 & 0 \\ 12 & 5 \end{bmatrix} \)

(ii) \( 5M + 3I = 4\begin{bmatrix} 2 & -5 \\ 0 & -3 \end{bmatrix} \)
\( 5M = 4\begin{bmatrix} 2 & -5 \\ 0 & -3 \end{bmatrix} - 3I \)
\( 5M = 4\begin{bmatrix} 2 & -5 \\ 0 & -3 \end{bmatrix} - 3\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( 5M = \begin{bmatrix} 8 & -20 \\ 0 & -12 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \)
\( 5M = \begin{bmatrix} 5 & -20 \\ 0 & -15 \end{bmatrix} \)
\( M = \frac{1}{5}\begin{bmatrix} 5 & -20 \\ 0 & -15 \end{bmatrix} = \begin{bmatrix} 1 & -4 \\ 0 & -3 \end{bmatrix} \)

📝 Teacher's Note: Remember that the unit matrix I has 1s on the main diagonal and 0s elsewhere. When you see 2I, multiply each element of I by 2.

🎯 Exam Tip: Always write what the unit matrix I looks like first. Then solve step by step. Remember to divide by the coefficient of M at the end.

Question 11.
If \( \begin{bmatrix} 1 & 4 \\ -2 & 3 \end{bmatrix} + 2M = 3\begin{bmatrix} 3 & 2 \\ 0 & -3 \end{bmatrix} \), find the matrix M

Answer:
\( \begin{bmatrix} 1 & 4 \\ -2 & 3 \end{bmatrix} + 2M = 3\begin{bmatrix} 3 & 2 \\ 0 & -3 \end{bmatrix} \)
\( 2M = \begin{bmatrix} 9 & 6 \\ 0 & -9 \end{bmatrix} - \begin{bmatrix} 1 & 4 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} 8 & 2 \\ 2 & -12 \end{bmatrix} \)
\( M = \begin{bmatrix} 4 & 1 \\ 1 & -6 \end{bmatrix} \)

📝 Teacher's Note: First multiply the right side matrix by 3. Then subtract the given matrix from both sides. Finally divide by 2.

🎯 Exam Tip: Always do matrix multiplication first, then addition or subtraction. Check your answer by substituting back into the original equation.

Exercise 9C

Question 1.
Evaluate: if possible:
(i) \( [3 \; 2]\begin{bmatrix} 2 \\ 0 \end{bmatrix} \)
(ii) \( [1 \; -2]\begin{bmatrix} -2 & 3 \\ -1 & 4 \end{bmatrix} \)
(iii) \( \begin{bmatrix} 6 & 4 \\ 3 & -1 \end{bmatrix}\begin{bmatrix} -1 \\ 3 \end{bmatrix} \)
(iv) \( \begin{bmatrix} 6 & 4 \\ 3 & -1 \end{bmatrix}[-1 \; 3] \)

Answer:
(i) \( [3 \; 2]\begin{bmatrix} 2 \\ 0 \end{bmatrix} = [6 + 0] = [6] \)

(ii) \( [1 \; -2]\begin{bmatrix} -2 & 3 \\ -1 & 4 \end{bmatrix} = [-2 + 2 \; 3 - 8] = [0 \; -5] \)

(iii) \( \begin{bmatrix} 6 & 4 \\ 3 & -1 \end{bmatrix}\begin{bmatrix} -1 \\ 3 \end{bmatrix} = \begin{bmatrix} -6 + 12 \\ -3 - 3 \end{bmatrix} = \begin{bmatrix} 6 \\ -6 \end{bmatrix} \)

(iv) \( \begin{bmatrix} 6 & 4 \\ 3 & -1 \end{bmatrix}[-1 \; 3] \)
The number of columns in the first matrix is not equal to the number of rows in the second matrix. Thus, the product is not possible.

📝 Teacher's Note: For matrix multiplication, the number of columns in the first matrix must equal the number of rows in the second matrix. Remember: (m×n) × (n×p) = (m×p).

🎯 Exam Tip: Always check if multiplication is possible first. Write "not possible" clearly if dimensions don't match. For multiplication, multiply row by column and add the products.

Question 2.
If \( A = \begin{bmatrix} 0 & 2 \\ 5 & -2 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & -1 \\ 3 & 2 \end{bmatrix} \) and I is a unit matrix of order 2 × 2, find:
(i) AB (ii) BA (iii) AI
(iv) IB (v) A² (vi) B²A

Answer:
We will solve each part step by step using matrix multiplication rules.

📝 Teacher's Note: Matrix multiplication is not commutative, so AB ≠ BA in general. The unit matrix I acts like 1 in regular multiplication - AI = IA = A.

🎯 Exam Tip: Calculate each matrix product carefully. Remember that A² means A × A. Show all your working steps clearly for full marks.

Question 3. If A = \(\begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix}\) and B = \(\begin{bmatrix} 9 & 16 \\ 0 & -y \end{bmatrix}\), find x and y when A² = B.
Answer:
Given: A = \(\begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix}\), B = \(\begin{bmatrix} 9 & 16 \\ 0 & -y \end{bmatrix}\) and A² = B

Step 1: Find A²
A² = A × A
= \(\begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix}\) × \(\begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix}\)
= \(\begin{bmatrix} 9 & 3x + x \\ 0 & 1 \end{bmatrix}\)
= \(\begin{bmatrix} 9 & 4x \\ 0 & 1 \end{bmatrix}\)

Step 2: Since A² = B
Two matrices are equal if each and every corresponding element is equal.
\(\begin{bmatrix} 9 & 4x \\ 0 & 1 \end{bmatrix}\) = \(\begin{bmatrix} 9 & 16 \\ 0 & -y \end{bmatrix}\)

Step 3: Compare corresponding elements
4x = 16 and 1 = -y
x = 4 and y = -1

Final Answer: x = 4 and y = -1
In simple words: We multiply matrix A with itself to get A². Then we match each number in A² with the same position number in B to find x and y.

📝 Teacher's Note: Show students that squaring a matrix means multiplying it with itself. When two matrices are equal, each position must have the same value.

🎯 Exam Tip: Always write "corresponding elements are equal" when equating matrices. Show the matrix multiplication step by step to get full marks.

Question 4. Find x and y, if:
(i) \(\begin{bmatrix} 4 & 3x \\ x & -2 \end{bmatrix}\) \(\begin{bmatrix} 5 \\ 1 \end{bmatrix}\) = \(\begin{bmatrix} y \\ 8 \end{bmatrix}\)
(ii) \(\begin{bmatrix} x & 0 \\ -3 & 1 \end{bmatrix}\) \(\begin{bmatrix} 1 & 1 \\ 0 & y \end{bmatrix}\) = \(\begin{bmatrix} 2 & 2 \\ -3 & -2 \end{bmatrix}\)
Answer:
(i) \(\begin{bmatrix} 4 & 3x \\ x & -2 \end{bmatrix}\) \(\begin{bmatrix} 5 \\ 1 \end{bmatrix}\) = \(\begin{bmatrix} y \\ 8 \end{bmatrix}\)

Multiplying the matrices:
\(\begin{bmatrix} 20 + 3x \\ 5x - 2 \end{bmatrix}\) = \(\begin{bmatrix} y \\ 8 \end{bmatrix}\)

Comparing corresponding elements:
5x - 2 = 8 → x = 2
20 + 3x = y → y = 20 + 6 = 26

(ii) \(\begin{bmatrix} x & 0 \\ -3 & 1 \end{bmatrix}\) \(\begin{bmatrix} 1 & 1 \\ 0 & y \end{bmatrix}\) = \(\begin{bmatrix} 2 & 2 \\ -3 & -2 \end{bmatrix}\)

Multiplying the matrices:
\(\begin{bmatrix} x + 0 & x + 0 \\ -3 + 0 & -3 + y \end{bmatrix}\) = \(\begin{bmatrix} 2 & 2 \\ -3 & -2 \end{bmatrix}\)

\(\begin{bmatrix} x & x \\ -3 & -3 + y \end{bmatrix}\) = \(\begin{bmatrix} 2 & 2 \\ -3 & -2 \end{bmatrix}\)

Comparing corresponding elements:
x = 2
-3 + y = -2 → y = 1

Final Answer: (i) x = 2, y = 26; (ii) x = 2, y = 1
In simple words: We multiply the matrices following the row-column rule. Then we compare each position to find the unknown values.

📝 Teacher's Note: Teach students the row-column multiplication rule clearly. First row times first column gives first element, and so on.

🎯 Exam Tip: Show all multiplication steps clearly. Write "comparing corresponding elements" to show you understand matrix equality.

Question 5. If A = \(\begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}\), B = \(\begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}\) and C = \(\begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix}\), find:
(i) (AB)C (ii) A(BC)
Is A(BC) = (AB)C?
Answer:
(i) First find AB:
AB = \(\begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}\) \(\begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}\) = \(\begin{bmatrix} 1+12 & 2+9 \\ 2+16 & 4+12 \end{bmatrix}\) = \(\begin{bmatrix} 13 & 11 \\ 18 & 16 \end{bmatrix}\)

Now find (AB)C:
(AB)C = \(\begin{bmatrix} 13 & 11 \\ 18 & 16 \end{bmatrix}\) \(\begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix}\) = \(\begin{bmatrix} 52+11 & 39+22 \\ 72+16 & 54+32 \end{bmatrix}\) = \(\begin{bmatrix} 63 & 61 \\ 88 & 86 \end{bmatrix}\)

(ii) First find BC:
BC = \(\begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}\) \(\begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix}\) = \(\begin{bmatrix} 4+2 & 3+4 \\ 16+3 & 12+6 \end{bmatrix}\) = \(\begin{bmatrix} 6 & 7 \\ 19 & 18 \end{bmatrix}\)

Now find A(BC):
A(BC) = \(\begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}\) \(\begin{bmatrix} 6 & 7 \\ 19 & 18 \end{bmatrix}\) = \(\begin{bmatrix} 6+57 & 7+54 \\ 12+76 & 14+72 \end{bmatrix}\) = \(\begin{bmatrix} 63 & 61 \\ 88 & 86 \end{bmatrix}\)

Conclusion: A(BC) = (AB)C

In simple words: This shows that matrix multiplication follows the associative law. We can group matrices in any order when multiplying three matrices.

📝 Teacher's Note: This is a good example to show the associative property of matrix multiplication. Students should understand that brackets can be moved but order cannot be changed.

🎯 Exam Tip: Calculate each step carefully. Show both (AB)C and A(BC) completely. Write the final conclusion clearly that they are equal.

Question 6. Given A = \(\begin{bmatrix} 0 & 4 & 6 \\ 3 & 0 & -1 \end{bmatrix}\) and B = \(\begin{bmatrix} 0 & 1 \\ -1 & 2 \\ -5 & -6 \end{bmatrix}\), find; if possible:
(i) AB (ii) BA (iii) A²
Answer:
(i) AB = \(\begin{bmatrix} 0 & 4 & 6 \\ 3 & 0 & -1 \end{bmatrix}\) \(\begin{bmatrix} 0 & 1 \\ -1 & 2 \\ -5 & -6 \end{bmatrix}\)

= \(\begin{bmatrix} 0-4-30 & 0+8-36 \\ 0+0+5 & 3+0+6 \end{bmatrix}\)

= \(\begin{bmatrix} -34 & -28 \\ 5 & 9 \end{bmatrix}\)

(ii) BA = \(\begin{bmatrix} 0 & 1 \\ -1 & 2 \\ -5 & -6 \end{bmatrix}\) \(\begin{bmatrix} 0 & 4 & 6 \\ 3 & 0 & -1 \end{bmatrix}\)

= \(\begin{bmatrix} 0+3 & 0+0 & 0-1 \\ 0+6 & -4+0 & -6-2 \\ 0-18 & -20+0 & -30+6 \end{bmatrix}\)

= \(\begin{bmatrix} 3 & 0 & -1 \\ 6 & -4 & -8 \\ -18 & -20 & -24 \end{bmatrix}\)

(iii) Product AA (=A²) is not possible as the number of columns of matrix A is not equal to its number of rows.

In simple words: For matrix multiplication to work, the number of columns in the first matrix must equal the number of rows in the second matrix. A² needs A to be a square matrix.

📝 Teacher's Note: Always check if multiplication is possible before calculating. A is 2×3 and needs to be square (same rows and columns) for A² to exist.

🎯 Exam Tip: Write the dimensions of matrices first. If multiplication is not possible, clearly state why with the dimension rule.

Question 7. Let A = \(\begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix}\), B = \(\begin{bmatrix} 4 & 1 \\ -3 & -2 \end{bmatrix}\) and C = \(\begin{bmatrix} -3 & 2 \\ -1 & 4 \end{bmatrix}\). Find A² + AC - 5B.
Answer:
Step 1: Find A²
A² = \(\begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix}\) \(\begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix}\) = \(\begin{bmatrix} 4+0 & 2-2 \\ 0+0 & 0+4 \end{bmatrix}\) = \(\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}\)

Step 2: Find AC
AC = \(\begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix}\) \(\begin{bmatrix} -3 & 2 \\ -1 & 4 \end{bmatrix}\) = \(\begin{bmatrix} -6-1 & 4+4 \\ 0+2 & 0-8 \end{bmatrix}\) = \(\begin{bmatrix} -7 & 8 \\ 2 & -8 \end{bmatrix}\)

Step 3: Find 5B
5B = 5\(\begin{bmatrix} 4 & 1 \\ -3 & -2 \end{bmatrix}\) = \(\begin{bmatrix} 20 & 5 \\ -15 & -10 \end{bmatrix}\)

Step 4: Find A² + AC - 5B
A² + AC - 5B = \(\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}\) + \(\begin{bmatrix} -7 & 8 \\ 2 & -8 \end{bmatrix}\) - \(\begin{bmatrix} 20 & 5 \\ -15 & -10 \end{bmatrix}\)

= \(\begin{bmatrix} 4-7-20 & 0+8-5 \\ 0+2+15 & 4-8+10 \end{bmatrix}\)

= \(\begin{bmatrix} -23 & 3 \\ 17 & 6 \end{bmatrix}\)

In simple words: We calculate each part separately - A², AC, and 5B. Then we add and subtract the matrices element by element.

📝 Teacher's Note: Remind students that matrix addition and subtraction work element by element. Calculate each matrix operation separately before combining them.

🎯 Exam Tip: Show each step clearly - A², AC, 5B, then the final addition/subtraction. Don't try to do everything in one step as mistakes happen easily.

Question 8. If M = \(\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\) and I is a unit matrix of the same order as that of M; show that: M² = 2M + 3I
Answer:
Given: A = \(\begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix}\), B = \(\begin{bmatrix} 4 & 1 \\ -3 & -2 \end{bmatrix}\) and C = \(\begin{bmatrix} -3 & 2 \\ -1 & 4 \end{bmatrix}\)

Now,
A² = \(\begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix}\) = \(\begin{bmatrix} 4+0 & 2-2 \\ 0+0 & 0+4 \end{bmatrix}\) = \(\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}\)

AC = \(\begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix}\begin{bmatrix} -3 & 2 \\ -1 & 4 \end{bmatrix}\) = \(\begin{bmatrix} -6-1 & 4+4 \\ 0+2 & 0-8 \end{bmatrix}\) = \(\begin{bmatrix} -7 & 8 \\ 2 & -8 \end{bmatrix}\)

5B = 5\(\begin{bmatrix} 4 & 1 \\ -3 & -2 \end{bmatrix}\) = \(\begin{bmatrix} 20 & 5 \\ -15 & -10 \end{bmatrix}\)

∴ A² + AC - 5B = \(\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}\) + \(\begin{bmatrix} -7 & 8 \\ 2 & -8 \end{bmatrix}\) - \(\begin{bmatrix} 20 & 5 \\ -15 & -10 \end{bmatrix}\)
= \(\begin{bmatrix} 4-7-20 & 0+8-5 \\ 0+2+15 & 4-8+10 \end{bmatrix}\)
= \(\begin{bmatrix} -23 & 3 \\ 17 & 6 \end{bmatrix}\)

In simple words: We multiply matrices step by step. First we find A², then AC, then 5B. Finally we add and subtract as needed.

📝 Teacher's Note: Show students matrix multiplication step by step. Common mistake is wrong order of multiplication. Remember: rows of first matrix × columns of second matrix.

🎯 Exam Tip: Always write "Given" first. Show each calculation separately. Write the final answer clearly inside a box or underline it.

Question 8. If M = \(\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\) and I is a unit matrix of the same order as that of M; show that: M² = 2M + 3I
Answer:
M² = \(\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\)
= \(\begin{bmatrix} 1+4 & 2+2 \\ 2+2 & 4+1 \end{bmatrix}\)
= \(\begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix}\)

2M + 3I = 2\(\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\) + 3\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
= \(\begin{bmatrix} 2 & 4 \\ 4 & 2 \end{bmatrix}\) + \(\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}\)
= \(\begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix}\)

Hence, M² = 2M + 3I.

In simple words: We calculate M² by multiplying M with itself. Then we calculate 2M + 3I separately. Both give the same result, so the statement is proven.

📝 Teacher's Note: Explain that I is the identity matrix (like number 1 for matrices). When we multiply any matrix by I, we get the same matrix back.

🎯 Exam Tip: Calculate both sides separately and show they are equal. Write "Hence proved" or "Hence, LHS = RHS" at the end.

Question 9. If A = \(\begin{bmatrix} a & 0 \\ 0 & 2 \end{bmatrix}\), B = \(\begin{bmatrix} 0 & -b \\ 1 & 0 \end{bmatrix}\), M = \(\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}\) and BA = M², find the values of a and b.
Answer:
BA = \(\begin{bmatrix} 0 & -b \\ 1 & 0 \end{bmatrix}\begin{bmatrix} a & 0 \\ 0 & 2 \end{bmatrix}\)
= \(\begin{bmatrix} 0+0 & 0-2b \\ a+0 & 0+0 \end{bmatrix}\)
= \(\begin{bmatrix} 0 & -2b \\ a & 0 \end{bmatrix}\)

M² = \(\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}\)
= \(\begin{bmatrix} 1-1 & -1-1 \\ 1+1 & -1+1 \end{bmatrix}\)
= \(\begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}\)

Given, BA = M²
\(\begin{bmatrix} 0 & -2b \\ a & 0 \end{bmatrix}\) = \(\begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}\)

Comparing the corresponding elements, we get:
a = 2
-2b = -2 ⟹ b = 1

In simple words: We multiply BA and find M². Then we compare each position in both matrices. The numbers in the same position must be equal.

📝 Teacher's Note: Teach students to compare matrices element by element. Position (1,1) of first matrix equals position (1,1) of second matrix, and so on.

🎯 Exam Tip: Write "comparing corresponding elements" clearly. Show each comparison separately: a = 2 and -2b = -2, so b = 1.

Question 10. Given A = \(\begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}\) and B = \(\begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix}\), find:
(i) A - B (ii) A²
(iii) AB (iv) A² - AB + 2B
Answer:
(i) A - B = \(\begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}\) - \(\begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix}\) = \(\begin{bmatrix} 3 & 1 \\ 4 & 2 \end{bmatrix}\)

(ii) A² = \(\begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}\)
= \(\begin{bmatrix} 16+2 & 4+3 \\ 8+6 & 2+9 \end{bmatrix}\)
= \(\begin{bmatrix} 18 & 7 \\ 14 & 11 \end{bmatrix}\)

(iii) AB = \(\begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix}\)
= \(\begin{bmatrix} 4-2 & 0+1 \\ 2-6 & 0+3 \end{bmatrix}\)
= \(\begin{bmatrix} 2 & 1 \\ -4 & 3 \end{bmatrix}\)

(iv) A² - AB + 2B
= \(\begin{bmatrix} 18 & 7 \\ 14 & 11 \end{bmatrix}\) - \(\begin{bmatrix} 2 & 1 \\ -4 & 3 \end{bmatrix}\) + 2\(\begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix}\)
= \(\begin{bmatrix} 16 & 6 \\ 18 & 8 \end{bmatrix}\) + \(\begin{bmatrix} 2 & 0 \\ -4 & 2 \end{bmatrix}\)
= \(\begin{bmatrix} 18 & 6 \\ 14 & 10 \end{bmatrix}\)

In simple words: We do each part step by step. For subtraction, we subtract each number. For multiplication, we use rows × columns rule. For the last part, we combine all results.

📝 Teacher's Note: Break down each part clearly. Students often mix up addition/subtraction with multiplication. Show them that 2B means multiply each element of B by 2.

🎯 Exam Tip: Label each part clearly: (i), (ii), (iii), (iv). Show working for each step. Don't jump to the final answer without showing calculations.

Question 11. If A = \(\begin{bmatrix} 1 & 4 \\ 1 & -3 \end{bmatrix}\) and B = \(\begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix}\), find:
(i) (A + B)² (ii) A² + B²
(iii) Is (A + B)² = A² + B²?
Answer:
(i) A + B = \(\begin{bmatrix} 1 & 4 \\ 1 & -3 \end{bmatrix}\) + \(\begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix}\) = \(\begin{bmatrix} 2 & 6 \\ 0 & -4 \end{bmatrix}\)

(A + B)² = \(\begin{bmatrix} 2 & 6 \\ 0 & -4 \end{bmatrix}\begin{bmatrix} 2 & 6 \\ 0 & -4 \end{bmatrix}\)
= \(\begin{bmatrix} 4+0 & 12-24 \\ 0+0 & 0+16 \end{bmatrix}\)
= \(\begin{bmatrix} 4 & -12 \\ 0 & 16 \end{bmatrix}\)

(ii) A² = \(\begin{bmatrix} 1 & 4 \\ 1 & -3 \end{bmatrix}\begin{bmatrix} 1 & 4 \\ 1 & -3 \end{bmatrix}\)
= \(\begin{bmatrix} 1+4 & 4-12 \\ 1-3 & 4+9 \end{bmatrix}\)
= \(\begin{bmatrix} 5 & -8 \\ -2 & 13 \end{bmatrix}\)

B² = \(\begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix}\)
= \(\begin{bmatrix} 1-2 & 2-2 \\ -1+1 & -2+1 \end{bmatrix}\)
= \(\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}\)

A² + B² = \(\begin{bmatrix} 5 & -8 \\ -2 & 13 \end{bmatrix}\) + \(\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}\)
= \(\begin{bmatrix} 4 & -8 \\ -2 & 12 \end{bmatrix}\)

(iii) Clearly, (A + B)² ≠ A² + B²

In simple words: For matrices, (A + B)² is not equal to A² + B². This is different from numbers where (a + b)² = a² + 2ab + b². Matrices don't follow this rule.

📝 Teacher's Note: This is a very important concept. Unlike numbers, matrix addition and multiplication don't follow the same algebraic rules. Show students this difference clearly.

🎯 Exam Tip: Calculate both sides completely and compare. Write clearly: "(A + B)² ≠ A² + B²" with the "not equal to" symbol. This gets you full marks.

Question 12. Find the matrix A, if B = \(\begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix}\) and B² = B + \(\frac{1}{2}\)A.
Answer:
B² = B + \(\frac{1}{2}\)A

\(\frac{1}{2}\)A = B² - B

A = 2(B² - B)

B² = \(\begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix}\)
= \(\begin{bmatrix} 4+0 & 2+1 \\ 0+0 & 0+1 \end{bmatrix}\)
= \(\begin{bmatrix} 4 & 3 \\ 0 & 1 \end{bmatrix}\)

B² - B = \(\begin{bmatrix} 4 & 3 \\ 0 & 1 \end{bmatrix}\) - \(\begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix}\)
= \(\begin{bmatrix} 2 & 2 \\ 0 & 0 \end{bmatrix}\)

∴ A = 2(B² - B)
= 2\(\begin{bmatrix} 2 & 2 \\ 0 & 0 \end{bmatrix}\) = \(\begin{bmatrix} 4 & 4 \\ 0 & 0 \end{bmatrix}\)

In simple words: We rearrange the equation to find A. First we calculate B², then subtract B from it. Finally we multiply by 2 to get A.

📝 Teacher's Note: Teach students to rearrange matrix equations like algebra. Move terms to one side to isolate the unknown matrix. Then solve step by step.

🎯 Exam Tip: Show the rearrangement clearly: from B² = B + ½A to A = 2(B² - B). Then calculate each step. Write the final answer clearly.

Question 13. If A = \(\begin{bmatrix} -1 & 1 \\ a & b \end{bmatrix}\) and A² = I; find a and b.
Answer:
Given: A = \(\begin{bmatrix} -1 & 1 \\ a & b \end{bmatrix}\) and A² = I

Step 1: Calculate A²
A² = \(\begin{bmatrix} -1 & 1 \\ a & b \end{bmatrix} \begin{bmatrix} -1 & 1 \\ a & b \end{bmatrix}\)

A² = \(\begin{bmatrix} 1+a & -1+b \\ -a+ab & a+b² \end{bmatrix}\)

Step 2: Use the condition A² = I
Since A² = I = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

\(\begin{bmatrix} 1+a & -1+b \\ -a+ab & a+b² \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

Step 3: Compare corresponding elements
1 + a = 1
Therefore, a = 0

-1 + b = 0
Therefore, b = 1

Answer: a = 0 and b = 1
In simple words: We multiply matrix A by itself to get A². Then we set this equal to the identity matrix I. By comparing each position, we find the values of a and b.

📝 Teacher's Note: Show students how matrix multiplication works step by step. Common mistake is forgetting that I is the identity matrix with 1s on the diagonal and 0s elsewhere.

🎯 Exam Tip: Always write "comparing corresponding elements" and show each equation clearly. Check your answer by substituting back into the original matrix.

Question 14. If A = \(\begin{bmatrix} 2 & 1 \\ 0 & 0 \end{bmatrix}\), B = \(\begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix}\) and C = \(\begin{bmatrix} 1 & 4 \\ 0 & 2 \end{bmatrix}\), then show that:
(i) A (B + C) = AB + AC
(ii) (B - A)C = BC - AC.
Answer:
(i) To prove A(B + C) = AB + AC

Step 1: Calculate B + C
B + C = \(\begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix} + \begin{bmatrix} 1 & 4 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 3 & 7 \\ 4 & 3 \end{bmatrix}\)

Step 2: Calculate A(B + C)
A(B + C) = \(\begin{bmatrix} 2 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 3 & 7 \\ 4 & 3 \end{bmatrix} = \begin{bmatrix} 6+4 & 14+3 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 10 & 17 \\ 0 & 0 \end{bmatrix}\)

Step 3: Calculate AB
AB = \(\begin{bmatrix} 2 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix} = \begin{bmatrix} 4+4 & 6+1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 8 & 7 \\ 0 & 0 \end{bmatrix}\)

Step 4: Calculate AC
AC = \(\begin{bmatrix} 2 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 4 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 2+0 & 8+2 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 2 & 10 \\ 0 & 0 \end{bmatrix}\)

Step 5: Calculate AB + AC
AB + AC = \(\begin{bmatrix} 8 & 7 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 2 & 10 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 10 & 17 \\ 0 & 0 \end{bmatrix}\)

Hence, A(B + C) = AB + AC

(ii) To prove (B - A)C = BC - AC

Step 1: Calculate B - A
B - A = \(\begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ 4 & 1 \end{bmatrix}\)

Step 2: Calculate (B - A)C
(B - A)C = \(\begin{bmatrix} 0 & 2 \\ 4 & 1 \end{bmatrix} \begin{bmatrix} 1 & 4 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 0+4 \\ 4+0 & 16+2 \end{bmatrix} = \begin{bmatrix} 0 & 4 \\ 4 & 18 \end{bmatrix}\)

Step 3: Calculate BC
BC = \(\begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix} \begin{bmatrix} 1 & 4 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 2+0 & 8+6 \\ 4+0 & 16+2 \end{bmatrix} = \begin{bmatrix} 2 & 14 \\ 4 & 18 \end{bmatrix}\)

Step 4: Calculate BC - AC
BC - AC = \(\begin{bmatrix} 2 & 14 \\ 4 & 18 \end{bmatrix} - \begin{bmatrix} 2 & 10 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 4 \\ 4 & 18 \end{bmatrix}\)

Hence, (B - A)C = BC - AC
In simple words: This proves the distributive property of matrices. We can distribute multiplication over addition and subtraction, just like with numbers.

📝 Teacher's Note: Show students each matrix multiplication step clearly. Use grid method to help them see which elements multiply together. Common mistake is mixing up rows and columns.

🎯 Exam Tip: Always calculate both sides separately and show they are equal. Write "Hence proved" at the end. Show all working steps for full marks.

Question 15. If A = \(\begin{bmatrix} 1 & 4 \\ 2 & 1 \end{bmatrix}\), B = \(\begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix}\) and C = \(\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}\), simplify: A² + BC.
Answer:
Step 1: Calculate A²
A² = \(\begin{bmatrix} 1 & 4 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 4 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 1+8 & 4+4 \\ 2+2 & 8+1 \end{bmatrix} = \begin{bmatrix} 9 & 8 \\ 4 & 9 \end{bmatrix}\)

Step 2: Calculate BC
BC = \(\begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} -3+0 & 0+4 \\ 4+0 & 0+0 \end{bmatrix} = \begin{bmatrix} -3 & 4 \\ 4 & 0 \end{bmatrix}\)

Step 3: Calculate A² + BC
A² + BC = \(\begin{bmatrix} 9 & 8 \\ 4 & 9 \end{bmatrix} + \begin{bmatrix} -3 & 4 \\ 4 & 0 \end{bmatrix} = \begin{bmatrix} 6 & 12 \\ 8 & 9 \end{bmatrix}\)

Answer: A² + BC = \(\begin{bmatrix} 6 & 12 \\ 8 & 9 \end{bmatrix}\)
In simple words: We first multiply A by itself to get A². Then we multiply B and C. Finally we add the two results together position by position.

📝 Teacher's Note: Remind students that matrix addition is done element by element. A common mistake is trying to add matrices that are different sizes.

🎯 Exam Tip: Show each step clearly - first A², then BC, then the final addition. Label each step for easy marking.

Question 16(i). Solve for x and y: \(\begin{bmatrix} 2 & 5 \\ 5 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -7 \\ 14 \end{bmatrix}\)
Answer:
Step 1: Write the matrix equation
\(\begin{bmatrix} 2 & 5 \\ 5 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -7 \\ 14 \end{bmatrix}\)

Step 2: Perform matrix multiplication
\(\begin{bmatrix} 2x + 5y \\ 5x + 2y \end{bmatrix} = \begin{bmatrix} -7 \\ 14 \end{bmatrix}\)

Step 3: Compare corresponding elements
2x + 5y = -7 ...(1)
5x + 2y = 14 ...(2)

Step 4: Solve the system of equations
Multiplying (1) with 2 and (2) with 5, we get:
4x + 10y = -14 ...(3)
25x + 10y = 70 ...(4)

Step 5: Subtract (3) from (4)
21x = 84
\( \implies \) x = 4

Step 6: Find y
From (2): 2y = 14 - 5x = 14 - 20 = -6
\( \implies \) y = -3

Answer: x = 4, y = -3
In simple words: We turn the matrix equation into two simple equations. Then we solve them like any pair of linear equations.

📝 Teacher's Note: Show students how matrix multiplication creates a system of linear equations. Practice with simple 2×2 examples first before moving to harder ones.

🎯 Exam Tip: Always check your answer by substituting back into the original equations. Write the final answer clearly as "x = ..., y = ...".

Question 16(ii). Solve for x and y: \(\begin{bmatrix} x+y & x-4 \end{bmatrix} \begin{bmatrix} -1 & -2 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} -7 & -11 \end{bmatrix}\)
Answer:
Step 1: Perform matrix multiplication
\(\begin{bmatrix} x+y & x-4 \end{bmatrix} \begin{bmatrix} -1 & -2 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} -7 & -11 \end{bmatrix}\)

\(\begin{bmatrix} -x-y+2x-8 & -2x-2y+2x-8 \end{bmatrix} = \begin{bmatrix} -7 & -11 \end{bmatrix}\)

\(\begin{bmatrix} -y+x-8 & -2y-8 \end{bmatrix} = \begin{bmatrix} -7 & -11 \end{bmatrix}\)

Step 2: Compare corresponding elements
-2y - 8 = -11
\( \implies \) -2y = -3
\( \implies \) y = \(\frac{3}{2}\)

-y + x - 8 = -7
\( \implies \) -\(\frac{3}{2}\) + x - 8 = -7
\( \implies \) x = 1 + \(\frac{3}{2}\) = \(\frac{5}{2}\)

Answer: x = \(\frac{5}{2}\), y = \(\frac{3}{2}\)
In simple words: We multiply the row matrix with the column matrix to get two equations. Then we solve step by step to find x and y.

📝 Teacher's Note: When multiplying a row matrix with a 2×2 matrix, remind students to be careful with signs. Show the multiplication pattern clearly.

🎯 Exam Tip: Write fractions clearly. Show each step of simplification. Double-check by substituting your answer back into the original equation.

Question 16(iii). Solve for x and y: \(\begin{bmatrix} -2 & 0 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} -1 \\ 2x \end{bmatrix} + 3 \begin{bmatrix} -2 \\ 1 \end{bmatrix} = 2 \begin{bmatrix} y \\ 3 \end{bmatrix}\)
Answer:
Step 1: Calculate the left side
\(\begin{bmatrix} -2 & 0 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} -1 \\ 2x \end{bmatrix} + 3 \begin{bmatrix} -2 \\ 1 \end{bmatrix} = 2 \begin{bmatrix} y \\ 3 \end{bmatrix}\)

\( \implies \) \(\begin{bmatrix} 2+0 \\ -3+2x \end{bmatrix} + \begin{bmatrix} -6 \\ 3 \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix}\)

\( \implies \) \(\begin{bmatrix} 2 \\ -3+2x \end{bmatrix} + \begin{bmatrix} -6 \\ 3 \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix}\)

\( \implies \) \(\begin{bmatrix} 2-6 \\ -3+2x+3 \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix}\)

\( \implies \) \(\begin{bmatrix} -4 \\ 2x \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix}\)

Step 2: Compare corresponding elements
2y = -4 and 2x = 6
\( \implies \) y = -2 and x = 3

Answer: x = 3, y = -2
In simple words: We first do the matrix multiplication, then add the matrices on the left side. Finally we compare with the right side to get our answers.

📝 Teacher's Note: Show students how to handle scalar multiplication of matrices first. Then show matrix addition step by step. Keep track of all operations carefully.

🎯 Exam Tip: Always write each step clearly. When you have scalar multiplication (like 3 times a matrix), do that calculation first before adding matrices.

Question 17. In each case given below, find:
(a) The order of matrix M.
(b) The matrix M.

(i) M × \(\begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix}\)

(ii) \(\begin{bmatrix} 1 & 4 \\ 2 & 1 \end{bmatrix}\) × M = \(\begin{bmatrix} 13 \\ 5 \end{bmatrix}\)
Answer:
(i) For M × \(\begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix}\)

We know the product of two matrices is defined only when the number of columns of first matrix equals the number of rows of the second matrix.

Step 1: Find the order of matrix M
Let the order of matrix M be a × b.
M_{a×b} × \(\begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix}_{2×2} = \begin{bmatrix} 1 & 2 \end{bmatrix}_{1×2}\)

Clearly, the order of matrix M is 1 × 2.

Step 2: Find matrix M
Let M = \(\begin{bmatrix} a & b \end{bmatrix}\)

M × \(\begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix}\)

\(\begin{bmatrix} a & b \end{bmatrix} × \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix}\)

\(\begin{bmatrix} a+0 & a+2b \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix}\)

Comparing corresponding elements:
a = 1 and a + 2b = 2
\( \implies \) 2b = 2 - 1 = 1
\( \implies \) b = \(\frac{1}{2}\)

∴ M = \(\begin{bmatrix} 1 & \frac{1}{2} \end{bmatrix}\)

(ii) For \(\begin{bmatrix} 1 & 4 \\ 2 & 1 \end{bmatrix}\) × M = \(\begin{bmatrix} 13 \\ 5 \end{bmatrix}\)

Step 1: Find the order of matrix M
Let the order of matrix M be a × b.
\(\begin{bmatrix} 1 & 4 \\ 2 & 1 \end{bmatrix}_{2×2}\) × M_{a×b} = \(\begin{bmatrix} 13 \\ 5 \end{bmatrix}_{2×1}\)

Clearly, the order of matrix M is 2 × 1.

Step 2: Find matrix M
Let M = \(\begin{bmatrix} a \\ b \end{bmatrix}\)

\(\begin{bmatrix} 1 & 4 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 13 \\ 5 \end{bmatrix}\)

\(\begin{bmatrix} a + 4b \\ 2a + b \end{bmatrix} = \begin{bmatrix} 13 \\ 5 \end{bmatrix}\)

Comparing corresponding elements:
a + 4b = 13 ...(1)
2a + b = 5 ...(2)

From (2): b = 5 - 2a
Substituting in (1): a + 4(5 - 2a) = 13
a + 20 - 8a = 13
-7a = -7
a = 1

From (2): b = 5 - 2(1) = 3

∴ M = \(\begin{bmatrix} 1 \\ 3 \end{bmatrix}\)

In simple words: First we figure out what size matrix M must be by looking at the sizes of the other matrices. Then we use unknown elements and solve equations to find the values.

📝 Teacher's Note: Teach students the rule for matrix multiplication: if A is m×n and B is p×q, then AB exists only if n=p, and the result is m×q. Use this rule to find the order first.

🎯 Exam Tip: Always state the order of matrix M first. Then set up the matrix with unknown elements and solve the resulting equations step by step.

Question 18. If A = \(\begin{bmatrix} 2 & x \\ 0 & 1 \end{bmatrix}\) and B = \(\begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix}\), find the value of x, given that: A² = B.
Answer:
Step 1: Calculate A²
\[A^2 = \begin{bmatrix} 2 & x \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4+0 & 2x+x \\ 0+0 & 0+1 \end{bmatrix} = \begin{bmatrix} 4 & 3x \\ 0 & 1 \end{bmatrix}\]

Step 2: Use the given condition A² = B
\[\begin{bmatrix} 4 & 3x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix}\]

Step 3: Compare corresponding elements
Comparing the two matrices, we get:
3x = 36
\[\implies x = 12\]
In simple words: When we multiply matrix A with itself, we get A². We make this equal to matrix B and find the unknown value x by matching the numbers in the same positions.

📝 Teacher's Note: Show students how matrix multiplication works step by step. Remind them that in 2×2 matrix multiplication, each position has a specific formula to follow.

🎯 Exam Tip: Always write "comparing corresponding elements" and show the equation clearly. Write the final answer as "x = 12" to get full marks.

Question 19. If A = \(\begin{bmatrix} 3 & 7 \\ 2 & 4 \end{bmatrix}\), B = \(\begin{bmatrix} 0 & 2 \\ 5 & 3 \end{bmatrix}\) and C = \(\begin{bmatrix} 1 & -5 \\ -4 & 6 \end{bmatrix}\). Find: AB - 5C.
Answer:
Step 1: Calculate AB
\[AB = \begin{bmatrix} 3 & 7 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} 0 & 2 \\ 5 & 3 \end{bmatrix}\]
\[= \begin{bmatrix} 3×0+7×5 & 3×2+7×3 \\ 2×0+4×5 & 2×2+4×3 \end{bmatrix}\]
\[= \begin{bmatrix} 0+35 & 6+21 \\ 0+20 & 4+12 \end{bmatrix} = \begin{bmatrix} 35 & 27 \\ 20 & 16 \end{bmatrix}\]

Step 2: Calculate 5C
\[5C = 5 \begin{bmatrix} 1 & -5 \\ -4 & 6 \end{bmatrix} = \begin{bmatrix} 5 & -25 \\ -20 & 30 \end{bmatrix}\]

Step 3: Calculate AB - 5C
\[AB - 5C = \begin{bmatrix} 35 & 27 \\ 20 & 16 \end{bmatrix} - \begin{bmatrix} 5 & -25 \\ -20 & 30 \end{bmatrix} = \begin{bmatrix} 30 & 52 \\ 40 & -14 \end{bmatrix}\]
In simple words: First we multiply matrices A and B. Then we multiply matrix C by 5. Finally we subtract the second result from the first result.

📝 Teacher's Note: Teach students to do matrix multiplication carefully - each element needs the row from first matrix times column from second matrix. Common mistake is mixing up rows and columns.

🎯 Exam Tip: Show all three steps clearly. Write "Step 1: Calculate AB", "Step 2: Calculate 5C", "Step 3: Calculate AB - 5C". This gets you method marks even if calculation has small errors.

Question 20. If A and B are any two 2 x 2 matrices such that AB = BA = B and B is not a zero matrix, what can you say about the matrix A?
Answer: AB = BA = B
We know that AI = IA = A, where I is the identity matrix.
Since AB = B and BA = B, matrix A behaves like the identity matrix.
Hence, A is the identity matrix.
In simple words: When any matrix multiplied by A gives the same matrix back, then A must be the identity matrix. The identity matrix is like the number 1 in regular multiplication.

📝 Teacher's Note: Remind students that identity matrix is like the number 1. Just like any number times 1 equals the same number, any matrix times identity matrix equals the same matrix.

🎯 Exam Tip: Write "A is the identity matrix" clearly. Also mention the property AI = IA = A to show you understand what identity matrix means.

Question 21. Given A = \(\begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix}\), B = \(\begin{bmatrix} a & b \\ 0 & c \end{bmatrix}\) and that AB = A + B; find the values of a, b and c.
Answer:
Step 1: Calculate AB
\[AB = \begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix} \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} = \begin{bmatrix} 3a+0 & 3b+0 \\ 0+0 & 0+4c \end{bmatrix} = \begin{bmatrix} 3a & 3b \\ 0 & 4c \end{bmatrix}\]

Step 2: Calculate A + B
\[A + B = \begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix} + \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} = \begin{bmatrix} 3+a & b \\ 0 & 4+c \end{bmatrix}\]

Step 3: Use condition AB = A + B
\[\begin{bmatrix} 3a & 3b \\ 0 & 4c \end{bmatrix} = \begin{bmatrix} 3+a & b \\ 0 & 4+c \end{bmatrix}\]

Step 4: Compare corresponding elements
3a = 3 + a
\[\implies 2a = 3 \implies a = \frac{3}{2}\]
3b = b
\[\implies b = 0\]
4c = 4 + c
\[\implies 3c = 4 \implies c = \frac{4}{3}\]
In simple words: We multiply the matrices and also add them. Then we make them equal and solve simple equations to find the unknown values.

📝 Teacher's Note: Show students how to set up equations by comparing elements in the same position. Teach them to solve simple linear equations like 3a = 3 + a step by step.

🎯 Exam Tip: Write all values clearly: a = 3/2, b = 0, c = 4/3. Show the working for each equation. Examiners want to see the comparison step clearly written.

Question 22. If P = \(\begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix}\) and Q = \(\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\), then compute:
(i) P² - Q² (ii) (P + Q) (P - Q)
Is (P + Q) (P - Q) = P² - Q² true for matrix algebra?
Answer:
(i) Calculate P² - Q²
\[P^2 = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} 1+4 & 2-2 \\ 2-2 & 4+1 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}\]

\[Q^2 = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 1+0 & 0+0 \\ 2+2 & 0+1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix}\]

\[P^2 - Q^2 = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ -4 & 4 \end{bmatrix}\]

(ii) Calculate (P + Q)(P - Q)
\[P + Q = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 4 & 0 \end{bmatrix}\]

\[P - Q = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ 0 & -2 \end{bmatrix}\]

\[(P + Q)(P - Q) = \begin{bmatrix} 2 & 2 \\ 4 & 0 \end{bmatrix} \begin{bmatrix} 0 & 2 \\ 0 & -2 \end{bmatrix} = \begin{bmatrix} 0+0 & 4-4 \\ 0+0 & 8-0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 8 \end{bmatrix}\]

Clearly, it can be said that: (P + Q) (P - Q) = P² - Q² not true for matrix algebra.
In simple words: The rule (a + b)(a - b) = a² - b² that works for numbers does NOT work for matrices. We proved this by calculating both sides and showing they are different.

📝 Teacher's Note: This is a very important concept. Tell students that many algebra rules for numbers do not work for matrices. Matrix multiplication is not commutative, so (P+Q)(P-Q) ≠ P²-Q².

🎯 Exam Tip: Calculate both sides completely and compare. Write clearly "not true for matrix algebra" to get the conclusion marks. Show all matrix multiplications step by step.

Question 23. Given the matrices: A = \(\begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}\), B = \(\begin{bmatrix} 3 & 4 \\ -1 & -2 \end{bmatrix}\) and C = \(\begin{bmatrix} -3 & 1 \\ 0 & -2 \end{bmatrix}\). Find:
(i) ABC (ii) ACB.
State whether ABC = ACB.
Answer:
(i) Calculate ABC
First calculate AB:
\[AB = \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix} \begin{bmatrix} 3 & 4 \\ -1 & -2 \end{bmatrix} = \begin{bmatrix} 6-1 & 8-2 \\ 12-2 & 16-4 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 10 & 12 \end{bmatrix}\]

Now calculate ABC:
\[ABC = \begin{bmatrix} 5 & 6 \\ 10 & 12 \end{bmatrix} \begin{bmatrix} -3 & 1 \\ 0 & -2 \end{bmatrix} = \begin{bmatrix} -15+0 & 5-12 \\ -30+0 & 10-24 \end{bmatrix} = \begin{bmatrix} -15 & -7 \\ -30 & -14 \end{bmatrix}\]

(ii) Calculate ACB
First calculate AC:
\[AC = \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix} \begin{bmatrix} -3 & 1 \\ 0 & -2 \end{bmatrix} = \begin{bmatrix} -6+0 & 2-2 \\ -12+0 & 4-4 \end{bmatrix} = \begin{bmatrix} -6 & 0 \\ -12 & 0 \end{bmatrix}\]

Now calculate ACB:
\[ACB = \begin{bmatrix} -6 & 0 \\ -12 & 0 \end{bmatrix} \begin{bmatrix} 3 & 4 \\ -1 & -2 \end{bmatrix} = \begin{bmatrix} -18-0 & -24-0 \\ -36-0 & -48-0 \end{bmatrix} = \begin{bmatrix} -18 & -24 \\ -36 & -48 \end{bmatrix}\]

Hence, ABC ≠ ACB.
In simple words: Matrix multiplication is not commutative. This means changing the order of multiplication gives different results. ABC is not equal to ACB.

📝 Teacher's Note: This shows that matrix multiplication does not follow the commutative property. Remind students to always multiply from left to right in the exact order given.

🎯 Exam Tip: Calculate ABC and ACB step by step. Show both results clearly and write "ABC ≠ ACB" as the conclusion. This proves matrices are not commutative under multiplication.

Question 24. If A = \(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\), B = \(\begin{bmatrix} 6 & 1 \\ 1 & 1 \end{bmatrix}\) and C = \(\begin{bmatrix} -2 & -3 \\ 0 & 1 \end{bmatrix}\); find each of the following and state if they are equal:
(i) CA + B (ii) A + CB
Answer:
(i) Calculate CA + B
First calculate CA:
\[CA = \begin{bmatrix} -2 & -3 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} -2-9 & -4-12 \\ 0+3 & 0+4 \end{bmatrix} = \begin{bmatrix} -11 & -16 \\ 3 & 4 \end{bmatrix}\]

Now CA + B:
\[CA + B = \begin{bmatrix} -11 & -16 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} 6 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} -5 & -15 \\ 4 & 5 \end{bmatrix}\]

(ii) Calculate A + CB
First calculate CB:
\[CB = \begin{bmatrix} -2 & -3 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 6 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} -12-3 & -2-3 \\ 0+1 & 0+1 \end{bmatrix} = \begin{bmatrix} -15 & -5 \\ 1 & 1 \end{bmatrix}\]

Now A + CB:
\[A + CB = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} -15 & -5 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} -14 & -3 \\ 4 & 5 \end{bmatrix}\]

Thus, CA + B ≠ A + CB
In simple words: Matrix multiplication and addition do not follow the distributive property like regular numbers. CA + B is not equal to A + CB.

📝 Teacher's Note: This shows that distributive property does not work for matrices when multiplication and addition are mixed. The order of operations matters a lot in matrix algebra.

🎯 Exam Tip: Calculate each part separately and compare final results. Write "CA + B ≠ A + CB" clearly. Show all working steps to get full marks.

Question 25. If A = \(\begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}\) and B = \(\begin{bmatrix} 3 \\ -11 \end{bmatrix}\), find the matrix X such that AX = B.
Answer:
Let the order of the matrix X be a×b.
AX = B
\[\begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}_{2×2} × X_{a×b} = \begin{bmatrix} 3 \\ -11 \end{bmatrix}_{2×1}\]

Clearly, the order of matrix X is 2 × 1.

Let \(X = \begin{bmatrix} x \\ y \end{bmatrix}\)

\[\begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 3 \\ -11 \end{bmatrix}\]

\[\begin{bmatrix} 2x + y \\ x + 3y \end{bmatrix} = \begin{bmatrix} 3 \\ -11 \end{bmatrix}\]

Comparing corresponding elements:
2x + y = 3 ...(1)
x + 3y = -11 ...(2)

From equation (1): y = 3 - 2x
Substituting in equation (2):
x + 3(3 - 2x) = -11
x + 9 - 6x = -11
-5x = -20
x = 4

From y = 3 - 2x: y = 3 - 2(4) = 3 - 8 = -5

Therefore, \(X = \begin{bmatrix} 4 \\ -5 \end{bmatrix}\)
In simple words: To find matrix X, we set up equations by doing matrix multiplication and then solve the system of equations like regular algebra.

📝 Teacher's Note: Teach students to first find the order of unknown matrix by looking at dimensions. Then set up the system of equations and solve step by step.

🎯 Exam Tip: Always write "Let X = [x y]" or similar to show the unknown matrix clearly. Solve the system of equations completely and write the final matrix X neatly.

Question 26. If A = \(\begin{bmatrix} 4 & 2 \\ 1 & 1 \end{bmatrix}\), find (A - 2I) (A - 3I).
Answer:
Step 1: Find A - 2I.
\[ A - 2I = \begin{bmatrix} 4 & 2 \\ 1 & 1 \end{bmatrix} - 2\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 2 \\ 1 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 1 & -1 \end{bmatrix} \]

Step 2: Find A - 3I.
\[ A - 3I = \begin{bmatrix} 4 & 2 \\ 1 & 1 \end{bmatrix} - 3\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 2 \\ 1 & 1 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 1 & -2 \end{bmatrix} \]

Step 3: Find (A - 2I)(A - 3I).
\[ (A - 2I)(A - 3I) = \begin{bmatrix} 2 & 2 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 1 & -2 \end{bmatrix} \]
\[ = \begin{bmatrix} 2+2 & 4-4 \\ 1-1 & 2+2 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} \]

In simple words: We subtract 2I and 3I from matrix A. Then we multiply the two new matrices together. I is the identity matrix with 1s on the main diagonal and 0s everywhere else.

📝 Teacher's Note: Show students that I is like the number 1 in matrix form. When we subtract 2I, we subtract 2 from each main diagonal element only.

🎯 Exam Tip: Always write I clearly as the identity matrix first. Show each step clearly - find A-2I, then A-3I, then multiply them.

Question 27. If A = \(\begin{bmatrix} 2 & 1 & -1 \\ 0 & 1 & -2 \end{bmatrix}\), find: (i) A^T. A (ii) A. A^T where A^T is the transpose of matrix A.
Answer:
Step 1: Find the transpose A^T.
\[ A^T = \begin{bmatrix} 2 & 0 \\ 1 & 1 \\ -1 & -2 \end{bmatrix} \]

Step 2: Find A^T.A.
\[ (i) A^T.A = \begin{bmatrix} 2 & 0 \\ 1 & 1 \\ -1 & -2 \end{bmatrix}\begin{bmatrix} 2 & 1 & -1 \\ 0 & 1 & -2 \end{bmatrix} \]
\[ = \begin{bmatrix} 4+0 & 2+0 & -2-0 \\ 2+0 & 1+1 & -1-2 \\ -2-0 & -1-2 & 1+4 \end{bmatrix} = \begin{bmatrix} 4 & 2 & -2 \\ 2 & 2 & -3 \\ -2 & -3 & 5 \end{bmatrix} \]

Step 3: Find A.A^T.
\[ (ii) A.A^T = \begin{bmatrix} 2 & 1 & -1 \\ 0 & 1 & -2 \end{bmatrix}\begin{bmatrix} 2 & 0 \\ 1 & 1 \\ -1 & -2 \end{bmatrix} \]
\[ = \begin{bmatrix} 4+1+1 & 0+1+2 \\ 0+1+2 & 0+1+4 \end{bmatrix} = \begin{bmatrix} 6 & 3 \\ 3 & 5 \end{bmatrix} \]

In simple words: Transpose means we flip the matrix - rows become columns. A^T.A gives a square matrix. A.A^T also gives a square matrix but different size.

📝 Teacher's Note: Show students how to flip a matrix to get transpose. Draw arrows showing row 1 becomes column 1, row 2 becomes column 2.

🎯 Exam Tip: Check matrix sizes carefully. A is 2×3, so A^T is 3×2. A^T.A will be 3×3, but A.A^T will be 2×2.

Question 28. If M = \(\begin{bmatrix} 4 & 1 \\ -1 & 2 \end{bmatrix}\), show that: 6M - M^2 = 9I; where I is a 2 × 2 unit matrix.
Answer:
Step 1: Find M^2.
\[ M^2 = \begin{bmatrix} 4 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 4 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 16-1 & 4+2 \\ -4-2 & -1+4 \end{bmatrix} = \begin{bmatrix} 15 & 6 \\ -6 & 3 \end{bmatrix} \]

Step 2: Find 6M.
\[ 6M = 6\begin{bmatrix} 4 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 24 & 6 \\ -6 & 12 \end{bmatrix} \]

Step 3: Find 6M - M^2.
\[ 6M - M^2 = \begin{bmatrix} 24 & 6 \\ -6 & 12 \end{bmatrix} - \begin{bmatrix} 15 & 6 \\ -6 & 3 \end{bmatrix} = \begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix} \]

Step 4: Find 9I.
\[ 9I = 9\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix} \]

Hence, proved.

In simple words: We multiply matrix M by itself to get M^2. Then we multiply M by 6. When we subtract M^2 from 6M, we get 9 times the identity matrix.

📝 Teacher's Note: Explain that the identity matrix I is like the number 1 in matrix world. Show students that 9I means 9 on the diagonal and 0s elsewhere.

🎯 Exam Tip: Write "Hence, proved" at the end. Show all working steps clearly - find M^2, then 6M, then subtract to show they equal 9I.

Question 29. If P = \(\begin{bmatrix} 2 & 6 \\ 3 & 9 \end{bmatrix}\) and Q = \(\begin{bmatrix} 3 & x \\ y & 2 \end{bmatrix}\); find x and y such that PQ = null matrix.
Answer:
Step 1: Find PQ.
\[ PQ = \begin{bmatrix} 2 & 6 \\ 3 & 9 \end{bmatrix}\begin{bmatrix} 3 & x \\ y & 2 \end{bmatrix} = \begin{bmatrix} 6+6y & 2x+12 \\ 9+9y & 3x+18 \end{bmatrix} \]

Step 2: Set PQ equal to null matrix.
\[ \begin{bmatrix} 6+6y & 2x+12 \\ 9+9y & 3x+18 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \]

Step 3: Compare corresponding elements.
2x + 12 = 0
Therefore x = -6
6 + 6y = 0
Therefore y = -1

In simple words: A null matrix has all zeros. When we multiply P and Q, each element must equal zero. This gives us equations to solve for x and y.

📝 Teacher's Note: Show students that when matrix multiplication gives all zeros, we can set each element equal to zero and solve. This is like solving simple equations.

🎯 Exam Tip: Write x = -6 and y = -1 clearly as your final answer. Show that you compared each element of PQ with zero.

Question 30. Evaluate without using tables: \(\begin{bmatrix} 2\cos 60° & -2\sin 30° \\ -\tan 45° & \cos 0° \end{bmatrix}\begin{bmatrix} \cot 45° & \cos \text{ec} 30° \\ \sec 60° & \sin 90° \end{bmatrix}\)
Answer:
Step 1: Substitute trigonometric values.
\[ = \begin{bmatrix} 2 \times \frac{1}{2} & -2 \times \frac{1}{2} \\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \]
\[ = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \]

Step 2: Multiply the matrices.
\[ = \begin{bmatrix} 1-2 & 2-1 \\ -1+2 & -2+1 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} \]

In simple words: We use basic trigonometric values like cos 60° = 1/2, sin 30° = 1/2, etc. Then we multiply the two matrices normally.

📝 Teacher's Note: Make students remember basic trigonometric values: sin 30° = 1/2, cos 60° = 1/2, tan 45° = 1, sin 90° = 1, cos 0° = 1, sec 60° = 2, cosec 30° = 2.

🎯 Exam Tip: Write the trigonometric values clearly in the first step. Examiners want to see that you know sin 30° = 1/2, cos 60° = 1/2, etc.

Question 31. State, with reason, whether the following are true or false. A, B and C are matrices of order 2 x 2.
(i) A + B = B + A
(ii) A - B = B - A
(iii) (B. C). A = B. (C. A)
(iv) (A + B). C = A. C + B. C
(v) A. (B - C) = A. B - A. C
(vi) (A - B). C = A. C - B. C
(vii) A² - B² = (A + B) (A - B)
(viii) (A - B)² = A² - 2A. B + B²

Answer:
(i) True. Addition of matrices is commutative.
(ii) False. Subtraction of matrices is not commutative.
(iii) True. Multiplication of matrices is associative.
(iv) True. Multiplication of matrices is distributive over addition.
(v) True. Multiplication of matrices is distributive over subtraction.
(vi) True. Multiplication of matrices is distributive over subtraction.
(vii) False. Laws of algebra for factorization and expansion are not applicable to matrices.
(viii) False. Laws of algebra for factorization and expansion are not applicable to matrices.

In simple words: Matrix addition works like normal addition (you can swap order). But matrix subtraction and some algebra rules don't work the same way as with numbers.

📝 Teacher's Note: Show students examples with 2×2 matrices. For (ii), show that A - B is not equal to B - A using simple numbers. This helps them understand the difference.

🎯 Exam Tip: Always give reasons. Write "commutative", "associative", or "distributive" for true statements. For false ones, write "not applicable to matrices".

Exercise 9D

Question 1. Find x and y, if: \(\begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix}\begin{bmatrix} 2x \\ 1 \end{bmatrix} + 2\begin{bmatrix} -4 \\ 5 \end{bmatrix} = 4\begin{bmatrix} 2 \\ y \end{bmatrix}\)
Answer:
Step 1: Multiply the matrix and vector.
\[ \begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix}\begin{bmatrix} 2x \\ 1 \end{bmatrix} = \begin{bmatrix} 6x-2 \\ -2x+4 \end{bmatrix} \]

Step 2: Add the second term.
\[ \begin{bmatrix} 6x-2 \\ -2x+4 \end{bmatrix} + \begin{bmatrix} -8 \\ 10 \end{bmatrix} = \begin{bmatrix} 6x-10 \\ -2x+14 \end{bmatrix} \]

Step 3: Set equal to the right side.
\[ \begin{bmatrix} 6x-10 \\ -2x+14 \end{bmatrix} = \begin{bmatrix} 8 \\ 4y \end{bmatrix} \]

Step 4: Compare corresponding elements.
6x - 10 = 8
⇒ 6x = 18
⇒ x = 3

-2x + 14 = 4y
⇒ 4y = -6 + 14 = 8
⇒ y = 2

In simple words: We multiply the matrix with the column vector, add the other terms, then compare each element to find x and y.

📝 Teacher's Note: Show students how to multiply a matrix with a column vector. The first row gives the first element, second row gives the second element.

🎯 Exam Tip: Write x = 3 and y = 2 clearly as your final answer. Show each step of matrix multiplication and addition separately.

Question 2. Find x and y, if: \(\begin{bmatrix} 3x & 8 \end{bmatrix}\begin{bmatrix} 1 & 4 \\ 3 & 7 \end{bmatrix} - 3\begin{bmatrix} 2 & -7 \end{bmatrix} = 5\begin{bmatrix} 3 & 2y \end{bmatrix}\)
Answer:
Step 1: Multiply the matrices on the left.
\[ \begin{bmatrix} 3x & 8 \end{bmatrix}\begin{bmatrix} 1 & 4 \\ 3 & 7 \end{bmatrix} = \begin{bmatrix} 3x+24 & 12x+56 \end{bmatrix} \]

Step 2: Subtract the second term.
\[ \begin{bmatrix} 3x+24 & 12x+56 \end{bmatrix} - \begin{bmatrix} 6 & -21 \end{bmatrix} = \begin{bmatrix} 3x+18 & 12x+77 \end{bmatrix} \]

Step 3: Set equal to the right side.
\[ \begin{bmatrix} 3x+18 & 12x+77 \end{bmatrix} = \begin{bmatrix} 15 & 10y \end{bmatrix} \]

Step 4: Compare corresponding elements.
3x + 18 = 15
⇒ 3x = -3
⇒ x = -1

12x + 77 = 10y
⇒ 12(-1) + 77 = 10y
⇒ 65 = 10y
⇒ y = 6.5

In simple words: We multiply the row matrix with the 2×2 matrix, subtract the next term, then compare elements to find x and y.

📝 Teacher's Note: Show students that when a row matrix multiplies a 2×2 matrix, we get another row matrix. Each element comes from row times column.

🎯 Exam Tip: Write x = -1 and y = 6.5 clearly. Show that you substituted x = -1 back into the second equation to find y.

Question 3. If \([x \quad y]\begin{bmatrix}x\\y\end{bmatrix} = [25]\) and \([-x \quad y]\begin{bmatrix}2x\\y\end{bmatrix} = [-2]\), find x and y, if:
(i) x, y ∈ W (whole numbers)
(ii) x, y ∈ Z (integers)

Answer:
Given:
\([x \quad y]\begin{bmatrix}x\\y\end{bmatrix} = [25]\)
\([-x \quad y]\begin{bmatrix}2x\\y\end{bmatrix} = [-2]\)

Step 1: Solve the first equation.
\([x \quad y]\begin{bmatrix}x\\y\end{bmatrix} = [25]\)
\(x^2 + y^2 = 25\) ... (1)

Step 2: Solve the second equation.
\([-x \quad y]\begin{bmatrix}2x\\y\end{bmatrix} = [-2]\)
\(-2x^2 + y^2 = -2\) ... (2)

Step 3: Solve equations (1) and (2) together.
From (1): \(x^2 + y^2 = 25\)
From (2): \(-2x^2 + y^2 = -2\)
Subtracting (2) from (1):
\(3x^2 = 27\)
\(x^2 = 9\)
\(x = ±3\)

When \(x = 3\): \(y^2 = 25 - 9 = 16\), so \(y = ±4\)
When \(x = -3\): \(y^2 = 25 - 9 = 16\), so \(y = ±4\)

(i) For x, y ∈ W (whole numbers):
Since whole numbers are 0, 1, 2, 3, 4, ...
The solution is \(x = 3, y = 4\)

(ii) For x, y ∈ Z (integers):
All possible integer solutions are:
\(x = 3, y = 4\) or \(x = 3, y = -4\) or \(x = -3, y = 4\) or \(x = -3, y = -4\)

In simple words: We have two equations with squares of x and y. We solve them like normal algebra. For whole numbers, we take only positive values. For integers, we take both positive and negative values.

📝 Teacher's Note: Show students how matrix multiplication works step by step. Many students forget that \([a \quad b]\begin{bmatrix}c\\d\end{bmatrix} = ac + bd\). Practice this first.

🎯 Exam Tip: Always check your answer by putting the values back in both original equations. Write "verification" to show you checked. This gets extra marks.

Question 4. Given \(\begin{bmatrix}2 & 1\\-3 & 4\end{bmatrix} × X = \begin{bmatrix}7\\6\end{bmatrix}\), Write
(i) the order of matrix X
(ii) the matrix X

Answer:
(i) Finding the order of matrix X:
Let the order of matrix X be \(a × b\)
\(\begin{bmatrix}2 & 1\\-3 & 4\end{bmatrix}_{2×2} × X_{a×b} = \begin{bmatrix}7\\6\end{bmatrix}_{2×1}\)

For matrix multiplication to be possible: number of columns in first matrix = number of rows in second matrix
So \(2 = a\) and \(b = 1\)
Therefore, the order of matrix X = \(2 × 1\)

(ii) Finding matrix X:
Let \(X = \begin{bmatrix}x\\y\end{bmatrix}\)

\(\begin{bmatrix}2 & 1\\-3 & 4\end{bmatrix} × \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}7\\6\end{bmatrix}\)

\(\begin{bmatrix}2x + y\\-3x + 4y\end{bmatrix} = \begin{bmatrix}7\\6\end{bmatrix}\)

Comparing corresponding elements:
\(2x + y = 7\) ... (1)
\(-3x + 4y = 6\) ... (2)

From equation (1): \(y = 7 - 2x\)
Substituting in equation (2):
\(-3x + 4(7 - 2x) = 6\)
\(-3x + 28 - 8x = 6\)
\(-11x = -22\)
\(x = 2\)

So \(y = 7 - 2(2) = 3\)

Therefore, \(X = \begin{bmatrix}2\\3\end{bmatrix}\)

In simple words: First we find what size the unknown matrix should be. Then we multiply and get simple equations. We solve like normal algebra to find the numbers inside the matrix.

📝 Teacher's Note: Students often get confused about matrix order. Teach them: "rows first, then columns". Also practice the rule for multiplication: (m×n) × (n×p) = (m×p).

🎯 Exam Tip: Always write the order clearly as "2×1" not "2,1". For finding the matrix, show all steps of solving the equations. Never skip steps in matrix problems.

Question 5. Evaluate: \(\begin{vmatrix}\cos 45° & \sin 30°\\\sqrt{2}\cos 0° & \sin 0°\end{vmatrix} \begin{vmatrix}\sin 45° & \cos 90°\\\sin 90° & \cot 45°\end{vmatrix}\)

Answer:
Step 1: Find values of trigonometric ratios.
\(\cos 45° = \frac{1}{\sqrt{2}}\), \(\sin 30° = \frac{1}{2}\), \(\cos 0° = 1\), \(\sin 0° = 0\)
\(\sin 45° = \frac{1}{\sqrt{2}}\), \(\cos 90° = 0\), \(\sin 90° = 1\), \(\cot 45° = 1\)

Step 2: Substitute values in first determinant.
\(\begin{vmatrix}\frac{1}{\sqrt{2}} & \frac{1}{2}\\\sqrt{2} × 1 & 0\end{vmatrix} = \begin{vmatrix}\frac{1}{\sqrt{2}} & \frac{1}{2}\\\sqrt{2} & 0\end{vmatrix}\)

Step 3: Calculate first determinant.
\(= \frac{1}{\sqrt{2}} × 0 - \frac{1}{2} × \sqrt{2} = 0 - \frac{\sqrt{2}}{2} = -\frac{\sqrt{2}}{2}\)

Step 4: Substitute values in second determinant.
\(\begin{vmatrix}\frac{1}{\sqrt{2}} & 0\\1 & 1\end{vmatrix}\)

Step 5: Calculate second determinant.
\(= \frac{1}{\sqrt{2}} × 1 - 0 × 1 = \frac{1}{\sqrt{2}}\)

Step 6: Multiply the determinants.
\(-\frac{\sqrt{2}}{2} × \frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2\sqrt{2}} = -\frac{1}{2}\)

Final Answer = \(-\frac{1}{2}\)

In simple words: We first put the values of sin, cos etc. Then we find each determinant using the formula ad - bc. Finally we multiply the two answers we got.

📝 Teacher's Note: Make students memorize basic trigonometric values first. For determinants, teach the cross multiplication method: "main diagonal minus other diagonal".

🎯 Exam Tip: Write all trigonometric values clearly in the first step. Show each determinant calculation separately. Never combine steps - examiners want to see each calculation.

Question 6. If A = \(\begin{bmatrix}0 & -1\\4 & -3\end{bmatrix}\), B = \(\begin{bmatrix}-5\\6\end{bmatrix}\) and 3A × M = 2B; find matrix M.

Answer:
Step 1: Find the order of matrix M.
Let the order of matrix M be \(a × b\)

3A × M = 2B
\(3\begin{bmatrix}0 & -1\\4 & -3\end{bmatrix}_{2×2} × M_{a×b} = 2\begin{bmatrix}-5\\6\end{bmatrix}_{2×1}\)

For multiplication to be possible: \(2 = a\) and \(b = 1\)
Therefore, M is a \(2 × 1\) matrix.

Step 2: Calculate 3A.
\(3A = 3\begin{bmatrix}0 & -1\\4 & -3\end{bmatrix} = \begin{bmatrix}0 & -3\\12 & -9\end{bmatrix}\)

Step 3: Calculate 2B.
\(2B = 2\begin{bmatrix}-5\\6\end{bmatrix} = \begin{bmatrix}-10\\12\end{bmatrix}\)

Step 4: Set up the equation.
Let \(M = \begin{bmatrix}x\\y\end{bmatrix}\)

\(\begin{bmatrix}0 & -3\\12 & -9\end{bmatrix} × \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}-10\\12\end{bmatrix}\)

Step 5: Perform matrix multiplication.
\(\begin{bmatrix}0×x + (-3)×y\\12×x + (-9)×y\end{bmatrix} = \begin{bmatrix}-10\\12\end{bmatrix}\)

\(\begin{bmatrix}-3y\\12x - 9y\end{bmatrix} = \begin{bmatrix}-10\\12\end{bmatrix}\)

Step 6: Compare corresponding elements and solve.
\(-3y = -10\)
\(y = \frac{10}{3}\)

\(12x - 9y = 12\)
\(12x - 9 × \frac{10}{3} = 12\)
\(12x - 30 = 12\)
\(12x = 42\)
\(x = \frac{7}{2}\)

Therefore, M = \(\begin{bmatrix}\frac{7}{2}\\\frac{10}{3}\end{bmatrix}\)

In simple words: We find what size M should be first. Then we calculate 3A and 2B separately. Finally we solve the matrix equation like normal algebra to find the unknown numbers.

📝 Teacher's Note: Students often forget to multiply every element when finding 3A or 2B. Practice scalar multiplication separately before matrix equations.

🎯 Exam Tip: Always state the order of the unknown matrix first. Show 3A and 2B calculations clearly. Write the final matrix with proper fraction form, don't convert to decimals.

Question 7. If \(\begin{bmatrix}a & 3\\4 & 1\end{bmatrix} + \begin{bmatrix}2 & b\\1 & -2\end{bmatrix} - \begin{bmatrix}1 & 1\\-2 & c\end{bmatrix} = \begin{bmatrix}5 & 0\\7 & 3\end{bmatrix}\), find the values of a, b and c.

Answer:
Step 1: Perform matrix addition and subtraction.
\(\begin{bmatrix}a & 3\\4 & 1\end{bmatrix} + \begin{bmatrix}2 & b\\1 & -2\end{bmatrix} - \begin{bmatrix}1 & 1\\-2 & c\end{bmatrix} = \begin{bmatrix}5 & 0\\7 & 3\end{bmatrix}\)

Step 2: Combine the matrices on the left side.
\(\begin{bmatrix}a + 2 & 3 + b\\4 + 1 & 1 + (-2)\end{bmatrix} - \begin{bmatrix}1 & 1\\-2 & c\end{bmatrix} = \begin{bmatrix}5 & 0\\7 & 3\end{bmatrix}\)

\(\begin{bmatrix}a + 2 & 3 + b\\5 & -1\end{bmatrix} - \begin{bmatrix}1 & 1\\-2 & c\end{bmatrix} = \begin{bmatrix}5 & 0\\7 & 3\end{bmatrix}\)

Step 3: Subtract the matrices.
\(\begin{bmatrix}(a + 2) - 1 & (3 + b) - 1\\5 - (-2) & -1 - c\end{bmatrix} = \begin{bmatrix}5 & 0\\7 & 3\end{bmatrix}\)

\(\begin{bmatrix}a + 1 & 2 + b\\7 & -1 - c\end{bmatrix} = \begin{bmatrix}5 & 0\\7 & 3\end{bmatrix}\)

Step 4: Compare corresponding elements.
\(a + 1 = 5\)
\(2 + b = 0\)
\(-1 - c = 3\)

Step 5: Solve for a, b, and c.
From \(a + 1 = 5\): \(a = 4\)
From \(2 + b = 0\): \(b = -2\)
From \(-1 - c = 3\): \(c = -4\)

Therefore: a = 4, b = -2, c = -4

In simple words: We add and subtract matrices element by element. Then we compare each position on both sides to get simple equations. We solve these equations to find the unknown numbers.

📝 Teacher's Note: Students often make sign errors in subtraction. Teach them: subtracting a matrix means subtracting each element. Practice with simple numbers first.

🎯 Exam Tip: Work step by step - add first, then subtract. Don't try to do everything in one step. Always check your answer by substituting back into the original equation.

Question 8. If A = \(\begin{bmatrix}1 & 2\\2 & 1\end{bmatrix}\) and B = \(\begin{bmatrix}2 & 1\\1 & 2\end{bmatrix}\), find:
(i) A(BA)
(ii) (AB).B

Answer:
(i) Finding A(BA):

Step 1: Calculate BA.
\(BA = \begin{bmatrix}2 & 1\\1 & 2\end{bmatrix}\begin{bmatrix}1 & 2\\2 & 1\end{bmatrix}\)

\(= \begin{bmatrix}2×1 + 1×2 & 2×2 + 1×1\\1×1 + 2×2 & 1×2 + 2×1\end{bmatrix}\)

\(= \begin{bmatrix}4 & 5\\5 & 4\end{bmatrix}\)

Step 2: Calculate A(BA).
\(A(BA) = \begin{bmatrix}1 & 2\\2 & 1\end{bmatrix}\begin{bmatrix}4 & 5\\5 & 4\end{bmatrix}\)

\(= \begin{bmatrix}1×4 + 2×5 & 1×5 + 2×4\\2×4 + 1×5 & 2×5 + 1×4\end{bmatrix}\)

\(= \begin{bmatrix}14 & 13\\13 & 14\end{bmatrix}\)

(ii) Finding (AB).B:

Step 1: Calculate AB.
\(AB = \begin{bmatrix}1 & 2\\2 & 1\end{bmatrix}\begin{bmatrix}2 & 1\\1 & 2\end{bmatrix}\)

\(= \begin{bmatrix}1×2 + 2×1 & 1×1 + 2×2\\2×2 + 1×1 & 2×1 + 1×2\end{bmatrix}\)

\(= \begin{bmatrix}4 & 5\\5 & 4\end{bmatrix}\)

Step 2: Calculate (AB).B.
\((AB).B = \begin{bmatrix}4 & 5\\5 & 4\end{bmatrix}\begin{bmatrix}2 & 1\\1 & 2\end{bmatrix}\)

\(= \begin{bmatrix}4×2 + 5×1 & 4×1 + 5×2\\5×2 + 4×1 & 5×1 + 4×2\end{bmatrix}\)

\(= \begin{bmatrix}13 & 14\\14 & 13\end{bmatrix}\)

Final Answers:
(i) \(A(BA) = \begin{bmatrix}14 & 13\\13 & 14\end{bmatrix}\)
(ii) \((AB).B = \begin{bmatrix}13 & 14\\14 & 13\end{bmatrix}\)

In simple words: We multiply matrices step by step. First we find the inner multiplication, then we multiply by the outer matrix. Matrix multiplication is like dot product of rows and columns.

📝 Teacher's Note: Teach students the "row times column" rule clearly. Show them that AB ≠ BA in general. Use different colored pens for each step to avoid confusion.

🎯 Exam Tip: Always use brackets to show which multiplication you do first. Write each matrix multiplication separately - never try to do A(BA) in one step. Show all the arithmetic clearly.

Question 9. Find x and y, if: \( \begin{bmatrix} x & 3x \\ y & 4y \end{bmatrix} \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 12 \end{bmatrix} \)
Answer:
Step 1: Write the matrix multiplication.
\( \begin{bmatrix} x & 3x \\ y & 4y \end{bmatrix} \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 12 \end{bmatrix} \)

Step 2: Multiply the matrices on the left side.
\( \begin{bmatrix} 2x + 3x \\ 2y + 4y \end{bmatrix} = \begin{bmatrix} 5 \\ 12 \end{bmatrix} \)

Step 3: Simplify.
\( \begin{bmatrix} 5x \\ 6y \end{bmatrix} = \begin{bmatrix} 5 \\ 12 \end{bmatrix} \)

Step 4: Compare corresponding elements.
5x = 5
\( \implies \) x = 1
6y = 12
\( \implies \) y = 2

Therefore, x = 1 and y = 2
In simple words: We multiplied the matrix with the column. This gave us two equations. We solved them to find x and y.

📝 Teacher's Note: Show students how to multiply each row with the column step by step. Common mistake is forgetting to add the products in each row.

🎯 Exam Tip: Always write "comparing corresponding elements" before solving. Show each equation clearly. Write the final answer as x = value, y = value.

Question 10. If matrix X = \( \begin{bmatrix} -3 & 4 \\ 2 & -3 \end{bmatrix} \begin{bmatrix} 2 \\ -2 \end{bmatrix} \) and 2X - 3Y = \( \begin{bmatrix} 10 \\ -8 \end{bmatrix} \), find the matrix 'X' and 'Y'.
Answer:
Step 1: Find matrix X.
\( X = \begin{bmatrix} -3 & 4 \\ 2 & -3 \end{bmatrix} \begin{bmatrix} 2 \\ -2 \end{bmatrix} \)
\( = \begin{bmatrix} -6 - 8 \\ 4 + 6 \end{bmatrix} \)
\( = \begin{bmatrix} -14 \\ 10 \end{bmatrix} \)

Step 2: Use the given equation to find Y.
Given: 2X - 3Y = \( \begin{bmatrix} 10 \\ -8 \end{bmatrix} \)

Step 3: Substitute the value of X.
\( 2 \begin{bmatrix} -14 \\ 10 \end{bmatrix} - 3Y = \begin{bmatrix} 10 \\ -8 \end{bmatrix} \)

Step 4: Simplify and solve for Y.
\( \begin{bmatrix} -28 \\ 20 \end{bmatrix} - 3Y = \begin{bmatrix} 10 \\ -8 \end{bmatrix} \)
\( 3Y = \begin{bmatrix} -28 \\ 20 \end{bmatrix} - \begin{bmatrix} 10 \\ -8 \end{bmatrix} \)
\( 3Y = \begin{bmatrix} -38 \\ 28 \end{bmatrix} \)
\( Y = \frac{1}{3} \begin{bmatrix} -38 \\ 28 \end{bmatrix} \)

Therefore, X = \( \begin{bmatrix} -14 \\ 10 \end{bmatrix} \) and Y = \( \begin{bmatrix} -38/3 \\ 28/3 \end{bmatrix} \)
In simple words: First we found X by matrix multiplication. Then we used the given equation to find Y step by step.

📝 Teacher's Note: Remind students to do matrix multiplication carefully. Each element in the result comes from one row times one column.

🎯 Exam Tip: Show all steps clearly. Write "Step 1", "Step 2" etc. Don't skip any calculation. Always check your final answer by substituting back.

Question 11. Given A = \( \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix} \), B = \( \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} \) and C = \( \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \); find the matrix X such that: A + X = 2B + C
Answer:
Step 1: Write the given equation.
A + X = 2B + C

Step 2: Rearrange to find X.
X = 2B + C - A

Step 3: Calculate 2B + C.
\( 2B + C = 2 \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \)
\( = \begin{bmatrix} -6 & 4 \\ 8 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \)
\( = \begin{bmatrix} -5 & 4 \\ 8 & 2 \end{bmatrix} \)

Step 4: Calculate X = 2B + C - A.
\( X = \begin{bmatrix} -5 & 4 \\ 8 & 2 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix} \)
\( X = \begin{bmatrix} -7 & 5 \\ 6 & 2 \end{bmatrix} \)

Therefore, X = \( \begin{bmatrix} -7 & 5 \\ 6 & 2 \end{bmatrix} \)
In simple words: We moved A to the right side of the equation. Then we did matrix addition and subtraction step by step.

📝 Teacher's Note: Show students that matrix equations work like normal equations. We can move terms from one side to another by changing their sign.

🎯 Exam Tip: Always rearrange the equation first to get X alone. Show all matrix calculations clearly. Check that all matrices have the same size before adding or subtracting.

Question 12. Find the value of x, given that A² = B, A = \( \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix} \) and B = \( \begin{bmatrix} 4 & x \\ 0 & 1 \end{bmatrix} \)
Answer:
Step 1: Write the given matrices.
\( A = \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} 4 & x \\ 0 & 1 \end{bmatrix} \)

Step 2: Calculate A².
\( A^2 = \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 4 + 0 & 24 + 12 \\ 0 + 0 & 0 + 1 \end{bmatrix} \)
\( = \begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix} \)

Step 3: Use the condition A² = B.
\( \begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & x \\ 0 & 1 \end{bmatrix} \)

Step 4: Compare corresponding elements.
Comparing the corresponding elements, we get:
x = 36

Therefore, x = 36
In simple words: We multiplied matrix A by itself to get A². Then we compared it with matrix B to find the value of x.

📝 Teacher's Note: Remind students that A² means A × A. When multiplying matrices, each element comes from row × column calculation.

🎯 Exam Tip: Show the matrix multiplication step by step. Write "comparing corresponding elements" before finding x. Always state the final answer clearly.

Question 13. If A = \( \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} \), B = \( \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} \), and I is identity matrix of the same order and A^t is the transpose of matrix A, find A^t.B + BI
Answer:
Step 1: Find the transpose of matrix A.
\( A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} \)
\( A^t = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix} \)

Step 2: Calculate A^t.B.
\( A^t \cdot B = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix} \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} \)
\( = \begin{bmatrix} 2 \times 4 + 1 \times (-1) & 2 \times (-2) + 1 \times 3 \\ 5 \times 4 + 3 \times (-1) & 5 \times (-2) + 3 \times 3 \end{bmatrix} \)
\( = \begin{bmatrix} 7 & -1 \\ 17 & -1 \end{bmatrix} \)

Step 3: Calculate B.I (where I is the 2×2 identity matrix).
\( B \cdot I = \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} \)

Step 4: Calculate A^t.B + BI.
\( A^t \cdot B + B \cdot I = \begin{bmatrix} 7 & -1 \\ 17 & -1 \end{bmatrix} + \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} \)
\( = \begin{bmatrix} 11 & -3 \\ 16 & 2 \end{bmatrix} \)

Therefore, A^t.B + BI = \( \begin{bmatrix} 11 & -3 \\ 16 & 2 \end{bmatrix} \)
In simple words: We found the transpose of A by swapping rows and columns. Then we did matrix multiplications and added the results.

📝 Teacher's Note: Explain that transpose means rows become columns and columns become rows. Also, any matrix multiplied by identity matrix gives the same matrix.

🎯 Exam Tip: Write A^t clearly by swapping rows and columns. Remember that B × I = B always. Show each matrix multiplication step by step.

Question 14. Given A = \( \begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix} \), B = \( \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} \) and C = \( \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix} \). Find the matrix X such that A + 2X = 2B + C.
Answer:
Step 1: Write the given equation.
A + 2X = 2B + C

Step 2: Calculate 2B + C.
\( 2B + C = 2 \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} + \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix} \)
\( = \begin{bmatrix} -6 & 4 \\ 8 & 0 \end{bmatrix} + \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix} \)
\( = \begin{bmatrix} -2 & 4 \\ 8 & 2 \end{bmatrix} \)

Step 3: Rearrange to find 2X.
\( A + 2X = \begin{bmatrix} -2 & 4 \\ 8 & 2 \end{bmatrix} \)
\( \begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix} + 2X = \begin{bmatrix} -2 & 4 \\ 8 & 2 \end{bmatrix} \)
\( 2X = \begin{bmatrix} -2 & 4 \\ 8 & 2 \end{bmatrix} - \begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix} \)
\( 2X = \begin{bmatrix} -4 & 10 \\ 6 & 2 \end{bmatrix} \)

Step 4: Find X by dividing by 2.
\( X = \frac{1}{2} \begin{bmatrix} -4 & 10 \\ 6 & 2 \end{bmatrix} \)
\( X = \begin{bmatrix} -2 & 5 \\ 3 & 1 \end{bmatrix} \)

Therefore, X = \( \begin{bmatrix} -2 & 5 \\ 3 & 1 \end{bmatrix} \)
In simple words: We moved A to the right side of the equation. Then we found 2X and divided by 2 to get X.

📝 Teacher's Note: Show students how to isolate 2X first, then divide each element by 2. This is like solving 2x = 10 to get x = 5 in regular algebra.

🎯 Exam Tip: Always rearrange the equation to get the unknown matrix alone. Show each calculation step. When dividing by a number, divide each element of the matrix.

Question 15. Let A = \( \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix} \), B = \( \begin{bmatrix} 0 & 2 \\ 1 & -1 \end{bmatrix} \) and C = \( \begin{bmatrix} -2 & 3 \\ 1 & -1 \end{bmatrix} \). Find A² - A + BC.
Answer:
Step 1: Calculate A².
\( A^2 = \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix} \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix} \)
\( = \begin{bmatrix} 16 - 12 & -8 + 6 \\ 24 - 18 & -12 + 9 \end{bmatrix} \)
\( = \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix} \)

Step 2: Calculate BC.
\( BC = \begin{bmatrix} 0 & 2 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} -2 & 3 \\ 1 & -1 \end{bmatrix} \)
\( = \begin{bmatrix} 0 + 2 & 0 - 2 \\ -2 - 1 & 3 + 1 \end{bmatrix} \)
\( = \begin{bmatrix} 2 & -2 \\ -3 & 4 \end{bmatrix} \)

Step 3: Calculate A² - A + BC.
\( A^2 - A + BC = \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix} - \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix} + \begin{bmatrix} 2 & -2 \\ -3 & 4 \end{bmatrix} \)
\( = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 2 & -2 \\ -3 & 4 \end{bmatrix} \)
\( = \begin{bmatrix} 2 & -2 \\ -3 & 4 \end{bmatrix} \)

Therefore, A² - A + BC = \( \begin{bmatrix} 2 & -2 \\ -3 & 4 \end{bmatrix} \)
In simple words: We found A² by multiplying A with itself. Then we calculated BC. Finally, we did A² - A + BC step by step.

📝 Teacher's Note: Point out that A² - A = 0 in this special case. This happens when A² = A. Such matrices are called idempotent matrices.

🎯 Exam Tip: Calculate A² and BC separately first. Then do the final addition and subtraction. Always check if A² - A gives a simple result like zero matrix.

Question 16. Let A = \( \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \), B = \( \begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix} \). Find A² + AB + B².
Answer:
Step 1: Calculate A².
\( A^2 = A \times A = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 1 \times 1 + 0 \times 2 & 1 \times 0 + 0 \times 1 \\ 2 \times 1 + 1 \times 2 & 2 \times 0 + 1 \times 1 \end{bmatrix} \)
\( = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} \)

Step 2: Calculate AB.
\( AB = A \times B = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \times \begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix} \)
\( = \begin{bmatrix} 2 & 3 \\ 3 & 6 \end{bmatrix} \)

Step 3: Calculate B².
\( B^2 = B \times B = \begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix} \times \begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix} \)
\( = \begin{bmatrix} 1 & 6 \\ -2 & -3 \end{bmatrix} \)

Step 4: Calculate A² + AB + B².
\( A^2 + AB + B^2 = \begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix} + \begin{bmatrix} 2 & 3 \\ 3 & 6 \end{bmatrix} + \begin{bmatrix} 1 & 6 \\ -2 & -3 \end{bmatrix} \)
\( = \begin{bmatrix} 4 & 9 \\ 5 & 4 \end{bmatrix} \)

Therefore, A² + AB + B² = \( \begin{bmatrix} 4 & 9 \\ 5 & 4 \end{bmatrix} \)
In simple words: We calculated each term separately - A², AB, and B². Then we added all three matrices together.

📝 Teacher's Note: Remind students that matrix multiplication is not commutative, so AB ≠ BA. Always calculate each term separately before adding.

🎯 Exam Tip: Show each matrix multiplication clearly. Calculate A², AB, and B² separately first. Then add all three matrices in the final step. Check your arithmetic carefully.

Question 17. If A = \(\begin{bmatrix} 3 & a \\ -4 & 8 \end{bmatrix}\), B = \(\begin{bmatrix} c & 4 \\ -3 & 0 \end{bmatrix}\), C = \(\begin{bmatrix} -1 & 4 \\ 3 & b \end{bmatrix}\) and 3A - 2C = 6B, find the values of a, b and c.
Answer:
Given: 3A - 2C = 6B

Step 1: Substitute the matrices into the equation.
\[3\begin{bmatrix} 3 & a \\ -4 & 8 \end{bmatrix} - 2\begin{bmatrix} -1 & 4 \\ 3 & b \end{bmatrix} = 6\begin{bmatrix} c & 4 \\ -3 & 0 \end{bmatrix}\]

Step 2: Multiply each matrix by its scalar.
\[\begin{bmatrix} 9 & 3a \\ -12 & 24 \end{bmatrix} - \begin{bmatrix} -2 & 8 \\ 6 & 2b \end{bmatrix} = \begin{bmatrix} 6c & 24 \\ -18 & 0 \end{bmatrix}\]

Step 3: Perform the matrix subtraction on the left side.
\[\begin{bmatrix} 11 & 3a - 8 \\ -18 & 24 - 2b \end{bmatrix} = \begin{bmatrix} 6c & 24 \\ -18 & 0 \end{bmatrix}\]

Step 4: Compare corresponding elements.
3a - 8 = 24 \(\Rightarrow\) 3a = 32 \(\Rightarrow\) \(a = \frac{32}{3} = 10\frac{2}{3}\)
24 - 2b = 0 \(\Rightarrow\) 2b = 24 \(\Rightarrow\) b = 12
11 = 6c \(\Rightarrow\) \(c = \frac{11}{6} = 1\frac{5}{6}\)

Therefore: \(a = 10\frac{2}{3}\), b = 12, \(c = 1\frac{5}{6}\)
In simple words: We multiplied each matrix by its number. Then we subtracted them. Finally we compared the same positions to find a, b, and c.

📝 Teacher's Note: Show students how to multiply a number with a matrix first. Then do subtraction step by step. Common mistake is forgetting the minus sign when subtracting matrices.

🎯 Exam Tip: Write "comparing corresponding elements" clearly. Show each equation separately. Always write final answers as mixed numbers or decimals as required.

Question 18. Given A = \(\begin{bmatrix} p & 0 \\ 0 & 2 \end{bmatrix}\), B = \(\begin{bmatrix} 0 & -q \\ 1 & 0 \end{bmatrix}\), C = \(\begin{bmatrix} 2 & -2 \\ 2 & 2 \end{bmatrix}\) and BA = C². Find the values of p and q.
Answer:
Step 1: Calculate BA.
\[BA = \begin{bmatrix} 0 & -q \\ 1 & 0 \end{bmatrix} \begin{bmatrix} p & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 0 & -2q \\ p & 0 \end{bmatrix}\]

Step 2: Calculate C².
\[C^2 = \begin{bmatrix} 2 & -2 \\ 2 & 2 \end{bmatrix} \begin{bmatrix} 2 & -2 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} 0 & -8 \\ 8 & 0 \end{bmatrix}\]

Step 3: Set BA = C² and compare elements.
\[\begin{bmatrix} 0 & -2q \\ p & 0 \end{bmatrix} = \begin{bmatrix} 0 & -8 \\ 8 & 0 \end{bmatrix}\]

By comparing:
-2q = -8 \(\Rightarrow\) q = 4
p = 8

Therefore: p = 8 and q = 4
In simple words: We multiplied BA first, then found C squared. We compared the same positions in both results to find p and q.

📝 Teacher's Note: Remind students that matrix multiplication is not the same as normal multiplication. Show them how to multiply row by column step by step.

🎯 Exam Tip: Calculate BA and C² separately first. Then write "by comparing corresponding elements" before finding p and q. This shows clear working.

Question 19. Given A = \(\begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix}\), B = \(\begin{bmatrix} 6 \\ 1 \end{bmatrix}\), C = \(\begin{bmatrix} -4 \\ 5 \end{bmatrix}\) and D = \(\begin{bmatrix} 2 \\ 2 \end{bmatrix}\). Find AB + 2C - 4D.
Answer:
Step 1: Calculate AB.
\[AB = \begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix} \begin{bmatrix} 6 \\ 1 \end{bmatrix} = \begin{bmatrix} 18 - 2 \\ -6 + 4 \end{bmatrix} = \begin{bmatrix} 16 \\ -2 \end{bmatrix}\]

Step 2: Calculate AB + 2C - 4D.
\[AB + 2C - 4D = \begin{bmatrix} 16 \\ -2 \end{bmatrix} + \begin{bmatrix} -8 \\ 10 \end{bmatrix} - \begin{bmatrix} 8 \\ 8 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\]

Therefore: AB + 2C - 4D = \(\begin{bmatrix} 0 \\ 0 \end{bmatrix}\)
In simple words: We first multiplied matrix A with vector B. Then we added and subtracted the other parts step by step to get the final answer.

📝 Teacher's Note: Show students that when multiplying a 2×2 matrix with a 2×1 vector, the result is a 2×1 vector. Practice this type of multiplication with simple examples first.

🎯 Exam Tip: Calculate AB first, then do the addition and subtraction step by step. Write each step clearly. The final answer is a column matrix (vector).

Question 20. Evaluate: \(\begin{bmatrix} 4\sin 30° & 2\cos 60° \\ \sin 90° & 2\cos 0° \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix}\)
Answer:
Step 1: Find the values of trigonometric functions.
sin 30° = \(\frac{1}{2}\), cos 60° = \(\frac{1}{2}\), sin 90° = 1, cos 0° = 1

Step 2: Substitute the values.
\[\begin{bmatrix} 4 \times \frac{1}{2} & 2 \times \frac{1}{2} \\ 1 & 2 \times 1 \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix}\]

Step 3: Perform matrix multiplication.
\[= \begin{bmatrix} 8 + 5 & 10 + 4 \\ 4 + 10 & 5 + 8 \end{bmatrix} = \begin{bmatrix} 13 & 14 \\ 14 & 13 \end{bmatrix}\]

Therefore: \(\begin{bmatrix} 13 & 14 \\ 14 & 13 \end{bmatrix}\)
In simple words: We first changed the trigonometric values to numbers. Then we multiplied the two matrices using row by column method.

📝 Teacher's Note: Make sure students know the basic trigonometric values: sin 30° = 1/2, cos 60° = 1/2, sin 90° = 1, cos 0° = 1. Practice these values first.

🎯 Exam Tip: Write the trigonometric values clearly in the first step. Show matrix multiplication step by step. Common mistake is wrong trigonometric values.

Question 21. If A = \(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\), I = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\), find A² - 5A + 7I
Answer:
Step 1: Calculate A².
\[A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 9 - 1 & 3 + 2 \\ -3 - 2 & -1 + 4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}\]

Step 2: Calculate 5A.
\[5A = 5 \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}\]

Step 3: Calculate 7I.
\[7I = 7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}\]

Step 4: Calculate A² - 5A + 7I.
\[= \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}\]
\[= \begin{bmatrix} 8 - 15 + 7 & 5 - 5 + 0 \\ -5 + 5 + 0 & 3 - 10 + 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\]

Therefore: A² - 5A + 7I = \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
In simple words: We found A squared by multiplying A with itself. Then we multiplied A by 5 and I by 7. Finally we did the addition and subtraction to get zero matrix.

📝 Teacher's Note: Explain that I is the identity matrix (like number 1 for matrices). When we multiply any number with I, only the diagonal elements change.

🎯 Exam Tip: Calculate A², 5A, and 7I separately first. Then combine them. The final answer being zero matrix is special - mention this in your answer.

Question 22. Given A = \(\begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix}\) and I = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) and A² = 9A + mI. Find m.
Answer:
Given: A² = 9A + mI
\(\Rightarrow\) A² - 9A = mI ...(1)

Step 1: Calculate A².
\[A^2 = \begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ -9 & 49 \end{bmatrix}\]

Step 2: Calculate 9A.
\[9A = 9 \begin{bmatrix} 2 & 0 \\ -1 & 7 \end{bmatrix} = \begin{bmatrix} 18 & 0 \\ -9 & 63 \end{bmatrix}\]

Step 3: Substitute in equation (1).
\[\begin{bmatrix} 4 & 0 \\ -9 & 49 \end{bmatrix} - \begin{bmatrix} 18 & 0 \\ -9 & 63 \end{bmatrix} = m \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\]

\[\begin{bmatrix} 4 - 18 & 0 - 0 \\ -9 + 9 & 49 - 63 \end{bmatrix} = \begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix}\]

\[\begin{bmatrix} -14 & 0 \\ 0 & -14 \end{bmatrix} = \begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix}\]

Therefore: m = -14
In simple words: We found A² and subtracted 9A from it. This gave us a matrix with m on the diagonal. So m = -14.

📝 Teacher's Note: Remind students that mI means m times the identity matrix, which puts m on the diagonal positions only. The off-diagonal elements of mI are always zero.

🎯 Exam Tip: Always rearrange the equation first to isolate mI. Calculate A² and 9A separately. When comparing with mI, only look at diagonal elements to find m.

Question 23. Given matrix A = \(\begin{bmatrix} 4\sin 30° & \cos 0° \\ \cos 0° & 4\sin 30° \end{bmatrix}\) and B = \(\begin{bmatrix} 4 \\ 5 \end{bmatrix}\). If AX = B. (i) Write the order of matrix X. (ii) Find the matrix X
Answer:
Step 1: Find the values in matrix A.
sin 30° = \(\frac{1}{2}\), cos 0° = 1
\[A = \begin{bmatrix} 4 \times \frac{1}{2} & 1 \\ 1 & 4 \times \frac{1}{2} \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}\]

(i) Order of matrix X:
Since A is 2×2 and B is 2×1, and AX = B, matrix X must be 2×1.

(ii) Finding matrix X:
From AX = B, we get X = A⁻¹B

Step 2: Find A⁻¹.
For \(A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}\), det(A) = 2(2) - 1(1) = 3
\[A^{-1} = \frac{1}{3} \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\]

Step 3: Calculate X = A⁻¹B.
\[X = \frac{1}{3} \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 4 \\ 5 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 8 - 5 \\ -4 + 10 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 3 \\ 6 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}\]

Therefore: (i) Order of X is 2×1 (ii) X = \(\begin{bmatrix} 1 \\ 2 \end{bmatrix}\)
In simple words: First we found what numbers are in matrix A. Then we used the rule AX = B means X = A⁻¹B to find X.

📝 Teacher's Note: Teach students to check matrix dimensions first. If A is m×n and B is m×1, then X must be n×1. Also remind them that A⁻¹B means "A inverse times B".

🎯 Exam Tip: Always write the order first by checking dimensions. For finding X, write "X = A⁻¹B" clearly. Show all steps for finding the inverse matrix. Check your answer by multiplying AX to see if you get B.

Question 24. If A = \(\begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix}\), B = \(\begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix}\) and A² - 5B² = 5C. Find the matrix C where C is a 2 by 2 matrix.
Answer:
Step 1: Calculate A².
A² = A × A = \(\begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix}\)
= \(\begin{bmatrix} 1×1+3×3 & 1×3+3×4 \\ 3×1+4×3 & 3×3+4×4 \end{bmatrix}\) = \(\begin{bmatrix} 10 & 15 \\ 15 & 25 \end{bmatrix}\)

Step 2: Calculate B².
B² = B × B = \(\begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix}\)
= \(\begin{bmatrix} -2×-2+1×-3 & -2×1+1×2 \\ -3×-2+2×-3 & -3×1+2×2 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

Step 3: Use the given equation A² - 5B² = 5C.
\(\begin{bmatrix} 10 & 15 \\ 15 & 25 \end{bmatrix} - 5\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = 5C\)

Step 4: Simplify.
\(\begin{bmatrix} 10 & 15 \\ 15 & 25 \end{bmatrix} - \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} = 5C\)

\(\begin{bmatrix} 5 & 15 \\ 15 & 20 \end{bmatrix} = 5C\)

Step 5: Solve for C.
C = \(\frac{1}{5}\begin{bmatrix} 5 & 15 \\ 15 & 20 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix}\)

In simple words: We multiply each matrix by itself, then use the given equation to find matrix C. We multiply matrices by adding rows times columns.

📝 Teacher's Note: Show students that matrix multiplication is row × column. Practice this step by step. Common mistake is doing element-wise multiplication instead of proper matrix multiplication.

🎯 Exam Tip: Always show each step clearly. Write "Step 1:", "Step 2:" etc. Calculate A² and B² first, then substitute into the equation. Check your arithmetic carefully.

Question 25. Given matrix B = \(\begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix}\). Find the matrix X if, X = B² - 4B. X\(\begin{bmatrix} a \\ b \end{bmatrix}\) = \(\begin{bmatrix} 5 \\ 50 \end{bmatrix}\). Hence, solve for a and b given.
Answer:
Step 1: Calculate B².
B² = B × B = \(\begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix}\)
= \(\begin{bmatrix} 1×1+1×8 & 1×1+1×3 \\ 8×1+3×8 & 8×1+3×3 \end{bmatrix}\) = \(\begin{bmatrix} 9 & 4 \\ 32 & 17 \end{bmatrix}\)

Step 2: Calculate 4B.
4B = \(4\begin{bmatrix} 1 & 1 \\ 8 & 3 \end{bmatrix}\) = \(\begin{bmatrix} 4 & 4 \\ 32 & 12 \end{bmatrix}\)

Step 3: Calculate X = B² - 4B.
X = \(\begin{bmatrix} 9 & 4 \\ 32 & 17 \end{bmatrix} - \begin{bmatrix} 4 & 4 \\ 32 & 12 \end{bmatrix}\) = \(\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}\)

Step 4: Solve X\(\begin{bmatrix} a \\ b \end{bmatrix}\) = \(\begin{bmatrix} 5 \\ 50 \end{bmatrix}\).
\(\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}\begin{bmatrix} a \\ b \end{bmatrix}\) = \(\begin{bmatrix} 5 \\ 50 \end{bmatrix}\)

\(\begin{bmatrix} 5a \\ 5b \end{bmatrix}\) = \(\begin{bmatrix} 5 \\ 50 \end{bmatrix}\)

5\(\begin{bmatrix} a \\ b \end{bmatrix}\) = 5\(\begin{bmatrix} 1 \\ 10 \end{bmatrix}\)

Therefore, a = 1 and b = 10

In simple words: We find matrix X first by calculating B² then subtracting 4B. Then we use X to find the values of a and b by solving the matrix equation.

📝 Teacher's Note: Students often confuse matrix subtraction with multiplication. Subtraction is element by element. Show them that X becomes 5 times the identity matrix, which makes solving easy.

🎯 Exam Tip: When you get X = 5I (5 times identity matrix), the equation becomes very simple. Write each step clearly and double-check your matrix multiplication.