Frank Brothers Solutions for ICSE Class 10 Mathematics Chapter 11 Matrices

Exercise 11.1

Answer 2.
Answer:
\([2a+3b \quad a-b] = [19 \quad 2]\)

Since \([2a+3b \quad a-b]\) is \(1 \times 2\) matrix and \([19 \quad 2]\) is \(1 \times 2\) matrix

\(2a+3b = 19\) _____ (1)
\(a-b = 2\) _____ (2)

From equation (2): \(a = 2+b\)

Putting the value of \(a\) in equation (1):
\(2(2+b)+3b = 19\)
\(4+2b+3b = 19\)
\(5b = 15\)
\(b = 3\)

From equation (2):
\(a = 2+3 = 5\)

Therefore, \(a = 5\) and \(b = 3\)
In simple words: We matched the elements of two equal matrices. This gave us two simple equations. We solved them to find the values of a and b.

📝 Teacher's Note: When two matrices are equal, all their matching elements are equal. Teach students to compare position by position to make equations.

🎯 Exam Tip: Always write two separate equations when matrices are equal. Show clear substitution steps to get full marks.

Answer 3.
Answer:
\(\begin{bmatrix} 3x - y \\ 5 \end{bmatrix}_{2 \times 1} = \begin{bmatrix} 7 \\ x + y \end{bmatrix}_{2 \times 1}\)

Comparing corresponding elements:
\(3x - y = 7\) _____ (1)
\(x + y = 5\) _____ (2)

From equation (2): \(x = 5 - y\)

Putting the value of \(x\) in equation (1):
\(3(5 - y) - y = 7\)
\(15 - 3y - y = 7\)
\(-4y = -8\)
\(y = 2\)

From equation (2):
\(x + 2 = 5\)
\(x = 3\)

Therefore, \(x = 3\) and \(y = 2\)
In simple words: Two column matrices are equal, so their top elements are equal and bottom elements are equal. This gave us two equations to solve for x and y.

📝 Teacher's Note: Show students that matrices are like organized boxes. Elements in the same position must be equal when matrices are equal.

🎯 Exam Tip: Write "comparing corresponding elements" first. Then clearly show which elements match to form equations.

Answer 6.
Answer:
\(A = [4 \quad 7]_{1 \times 2}\), \(B = [3 \quad 1]_{1 \times 2}\)

(i) A + 2B
\(A + 2B = [4 \quad 7] + 2[3 \quad 1]\)
\(= [4 \quad 7] + [6 \quad 2]\)
\(= [4 + 6 \quad 7 + 2]\)
\(= [10 \quad 9]_{1 \times 2}\)

(ii) A - B
\(A - B = [4 \quad 7] - [3 \quad 1]\)
\(= [4-3 \quad 7-1]\)
\(= [1 \quad 6]_{1 \times 2}\)

(iii) 2A - 3B
\(2A - 3B = 2[4 \quad 7] - 3[3 \quad 1]\)
\(= [8 \quad 14] - [9 \quad 3]\)
\(= [8-9 \quad 14-3]\)
\(= [-1 \quad 11]_{1 \times 2}\)
In simple words: Matrix addition and subtraction is done element by element. First multiply matrices by numbers, then add or subtract matching positions.

📝 Teacher's Note: Emphasize that we can only add/subtract matrices of the same size. Show students to work position by position.

🎯 Exam Tip: Always check matrix sizes first. Show each step clearly - first multiply by scalars, then add/subtract elements in same positions.

Answer 7.
Answer:
\(P = \begin{bmatrix} 2 & 9 \\ 5 & 7 \end{bmatrix}_{2 \times 2}\), \(Q = \begin{bmatrix} 7 & 3 \\ 4 & 1 \end{bmatrix}_{2 \times 2}\)

(i) 2P + 3Q
\(2P + 3Q = 2\begin{bmatrix} 2 & 9 \\ 5 & 7 \end{bmatrix} + 3\begin{bmatrix} 7 & 3 \\ 4 & 1 \end{bmatrix}\)
\(= \begin{bmatrix} 4 & 18 \\ 10 & 14 \end{bmatrix} + \begin{bmatrix} 21 & 9 \\ 12 & 3 \end{bmatrix}\)
\(= \begin{bmatrix} 4+21 & 18+9 \\ 10+12 & 14+3 \end{bmatrix} = \begin{bmatrix} 25 & 27 \\ 22 & 17 \end{bmatrix}_{2 \times 2}\)

(ii) 2Q - P
\(2Q - P = 2\begin{bmatrix} 7 & 3 \\ 4 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 9 \\ 5 & 7 \end{bmatrix}\)
\(= \begin{bmatrix} 14 & 6 \\ 8 & 2 \end{bmatrix} - \begin{bmatrix} 2 & 9 \\ 5 & 7 \end{bmatrix}\)
\(= \begin{bmatrix} 14-2 & 6-9 \\ 8-5 & 2-7 \end{bmatrix} = \begin{bmatrix} 12 & -3 \\ 3 & -5 \end{bmatrix}_{2 \times 2}\)

(iii) 3P - 2Q
\(3P - 2Q = 3\begin{bmatrix} 2 & 9 \\ 5 & 7 \end{bmatrix} - 2\begin{bmatrix} 7 & 3 \\ 4 & 1 \end{bmatrix}\)
\(= \begin{bmatrix} 6 & 27 \\ 15 & 21 \end{bmatrix} - \begin{bmatrix} 14 & 6 \\ 8 & 2 \end{bmatrix}\)
\(= \begin{bmatrix} 6-14 & 27-6 \\ 15-8 & 21-2 \end{bmatrix} = \begin{bmatrix} -8 & 21 \\ 7 & 19 \end{bmatrix}_{2 \times 2}\)
In simple words: For 2×2 matrices, we multiply each number in the matrix by the given number, then add or subtract each position separately.

📝 Teacher's Note: Use grid paper to show students how each position in the matrix corresponds to a specific location. This helps avoid calculation errors.

🎯 Exam Tip: Write intermediate steps clearly. First show scalar multiplication, then the addition/subtraction. Always check your final matrix size.

Answer 8.
Answer:
\(A = \begin{bmatrix} 17 & 5 & 19 \\ 11 & 8 & 13 \end{bmatrix}_{2 \times 3}\), \(B = \begin{bmatrix} 9 & 3 & 7 \\ 1 & 6 & 5 \end{bmatrix}_{2 \times 3}\)

\(2A - 3B = 2\begin{bmatrix} 17 & 5 & 19 \\ 11 & 8 & 13 \end{bmatrix} - 3\begin{bmatrix} 9 & 3 & 7 \\ 1 & 6 & 5 \end{bmatrix}\)

\(= \begin{bmatrix} 34 & 10 & 38 \\ 22 & 16 & 26 \end{bmatrix} - \begin{bmatrix} 27 & 9 & 21 \\ 3 & 18 & 15 \end{bmatrix}\)

\(= \begin{bmatrix} 34-27 & 10-9 & 38-21 \\ 22-3 & 16-18 & 26-15 \end{bmatrix}\)

\(= \begin{bmatrix} 7 & 1 & 17 \\ 19 & -2 & 11 \end{bmatrix}_{2 \times 3}\)
In simple words: We multiplied each matrix by its number first, then subtracted each position. This works because both matrices have the same size (2 rows, 3 columns).

📝 Teacher's Note: Point out that matrix operations are only possible when matrices have the same order. Show the pattern: multiply first, then add/subtract.

🎯 Exam Tip: Always write the matrix dimensions. Show scalar multiplication step clearly before doing subtraction. This helps avoid arithmetic errors.

Answer 9.
Answer:
\(M = \begin{bmatrix} 8 & 3 \\ 9 & 7 \\ 4 & 3 \end{bmatrix}_{3 \times 2}\), \(N = \begin{bmatrix} 4 & 7 \\ 5 & 3 \\ 10 & 1 \end{bmatrix}_{3 \times 2}\)

(i) M + N
\(M + N = \begin{bmatrix} 8 & 3 \\ 9 & 7 \\ 4 & 3 \end{bmatrix} + \begin{bmatrix} 4 & 7 \\ 5 & 3 \\ 10 & 1 \end{bmatrix}\)

\(= \begin{bmatrix} 8+4 & 3+7 \\ 9+5 & 7+3 \\ 4+10 & 3+1 \end{bmatrix} = \begin{bmatrix} 12 & 10 \\ 14 & 10 \\ 14 & 4 \end{bmatrix}_{3 \times 2}\)

(ii) M - N
\(M - N = \begin{bmatrix} 8 & 3 \\ 9 & 7 \\ 4 & 3 \end{bmatrix} - \begin{bmatrix} 4 & 7 \\ 5 & 3 \\ 10 & 1 \end{bmatrix}\)

\(= \begin{bmatrix} 8-4 & 3-7 \\ 9-5 & 7-3 \\ 4-10 & 3-1 \end{bmatrix} = \begin{bmatrix} 4 & -4 \\ 4 & 4 \\ -6 & 2 \end{bmatrix}_{3 \times 2}\)
In simple words: For matrices with 3 rows and 2 columns, we add or subtract each position separately. The result matrix has the same size.

📝 Teacher's Note: Use colored pens to highlight corresponding positions when adding/subtracting. This visual aid helps students track which elements to combine.

🎯 Exam Tip: Be careful with negative numbers when subtracting. Double-check each calculation and keep the matrix structure neat.

Answer 10.
Answer:
\(A = \begin{bmatrix} 1 & 9 & 4 \\ 5 & 0 & 3 \end{bmatrix}_{2 \times 3}\)

(i) Negative A
\(-A = \begin{bmatrix} -1 & -9 & -4 \\ -5 & 0 & -3 \end{bmatrix}_{2 \times 3}\)

(ii) \(A^t\) (Transpose of A)
\(A^t = \begin{bmatrix} 1 & 5 \\ 9 & 0 \\ 4 & 3 \end{bmatrix}_{3 \times 2}\)
In simple words: Negative A means we change the sign of every number. Transpose means we change rows to columns and columns to rows.

📝 Teacher's Note: For transpose, teach students to "flip" the matrix. The first row becomes the first column. The second row becomes the second column.

🎯 Exam Tip: For negative matrix, put minus sign in front of every element. For transpose, notice how the matrix size changes from 2×3 to 3×2.

Answer 11.
Answer:
\(P = \begin{bmatrix} 8 & 5 \\ 7 & 2 \end{bmatrix}_{2 \times 2}\)

(i) \(P^t\)
\(P^t = \begin{bmatrix} 8 & 7 \\ 5 & 2 \end{bmatrix}_{2 \times 2}\)

(ii) \(P + P^t\)
\(P + P^t = \begin{bmatrix} 8 & 5 \\ 7 & 2 \end{bmatrix} + \begin{bmatrix} 8 & 7 \\ 5 & 2 \end{bmatrix}\)
\(= \begin{bmatrix} 16 & 12 \\ 12 & 4 \end{bmatrix}_{2 \times 2}\)

(iii) \(P - P^t\)
\(P - P^t = \begin{bmatrix} 8 & 5 \\ 7 & 2 \end{bmatrix} - \begin{bmatrix} 8 & 7 \\ 5 & 2 \end{bmatrix}\)
\(= \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}_{2 \times 2}\)
In simple words: We first found the transpose by swapping rows and columns. Then we added and subtracted the original matrix with its transpose.

📝 Teacher's Note: Point out that \(P + P^t\) gives a symmetric matrix (same when reflected across diagonal). \(P - P^t\) gives a skew-symmetric matrix.

🎯 Exam Tip: Find transpose first, then do the operations. Notice that diagonal elements in \(P - P^t\) are always zero.

Answer 12.
Answer:
\(B = \begin{bmatrix} 15 & 13 \\ 11 & 12 \\ 10 & 17 \end{bmatrix}_{3 \times 2}\)

\(B^t = \begin{bmatrix} 15 & 11 & 10 \\ 13 & 12 & 17 \end{bmatrix}_{2 \times 3}\)

To add two matrices, their order (number of rows and number of columns) should be same. In this case, order of \(B\) is \(3 \times 2\) and order of \(B^t\) is \(2 \times 3\). Since they are not same, we cannot add them.
In simple words: Matrix B is 3×2 size, but its transpose is 2×3 size. We can only add matrices of exactly the same size, so this addition is not possible.

📝 Teacher's Note: Always check matrix sizes before any operation. This is a fundamental rule that prevents many calculation errors.

🎯 Exam Tip: Write the matrix orders clearly and explain why addition is not possible. This shows the examiner you understand the basic rules.

Answer 13.
Answer:
\(A = \begin{bmatrix} 5 & r \\ p & 7 \end{bmatrix}\), \(B = \begin{bmatrix} q & 4 \\ 3 & s \end{bmatrix}\)

\(A + B = \begin{bmatrix} 5 & r \\ p & 7 \end{bmatrix} + \begin{bmatrix} q & 4 \\ 3 & s \end{bmatrix}\)

\(A + B = \begin{bmatrix} 5+q & r+4 \\ p+3 & 7+s \end{bmatrix}_{2 \times 2}\) _____ (1)

But, given
\(A + B = \begin{bmatrix} 9 & 7 \\ 5 & 8 \end{bmatrix}_{2 \times 2}\) _____ (2)

From (1) and (2):
\(\begin{bmatrix} 5+q & r+4 \\ p+3 & 7+s \end{bmatrix} = \begin{bmatrix} 9 & 7 \\ 5 & 8 \end{bmatrix}\)

Comparing corresponding elements:
\(5 + q = 9\)
\(⇒ q = 4\)

\(r + 4 = 7\)
\(⇒ r = 3\)

\(p + 3 = 5\)
\(⇒ p = 2\)

\(7 + s = 8\)
\(⇒ s = 1\)

Therefore, \(p = 2\), \(q = 4\), \(r = 3\), \(s = 1\)
In simple words: We added matrices A and B with unknown values. Then we matched each position with the given result to find all the unknown values.

📝 Teacher's Note: This combines matrix addition with solving equations. Show students to work systematically through each position.

🎯 Exam Tip: Write the matrix addition first, then set it equal to the given result. Compare each position separately to get four simple equations.

Exercise 11.2

Answer 2.
Answer:
Given: \( A = \begin{bmatrix} 1 & 3 \\ 3 & 2 \end{bmatrix} \), \( B = \begin{bmatrix} -2 & 3 \\ -4 & 1 \end{bmatrix} \)

(a) \( AB = \begin{bmatrix} 1 & 3 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} -2 & 3 \\ -4 & 1 \end{bmatrix} \)

\( = \begin{bmatrix} -2-12 & 3+3 \\ -6-8 & 9+2 \end{bmatrix} = \begin{bmatrix} -14 & 6 \\ -14 & 11 \end{bmatrix} \)

(b) \( BA = \begin{bmatrix} -2 & 3 \\ -4 & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 3 & 2 \end{bmatrix} \)

\( = \begin{bmatrix} -2+9 & -6+6 \\ -4+3 & -12+2 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ -1 & -10 \end{bmatrix} \)

In simple words: Matrix multiplication means we multiply rows of the first matrix with columns of the second matrix. We add all the products to get each answer.

📝 Teacher's Note: Show students that AB is not equal to BA in matrices. Use simple 2×2 examples first. Matrix multiplication is not the same as number multiplication.

🎯 Exam Tip: Always write the matrices clearly. Show each step of multiplication. Check your addition carefully - most mistakes happen in adding the products.

Answer 3.
Answer:
Let \( A = \begin{bmatrix} p & q \\ r & s \end{bmatrix} \)

\( AI = \begin{bmatrix} p & q \\ r & s \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)

\( = \begin{bmatrix} p+0 & 0+q \\ r+0 & 0+s \end{bmatrix} = \begin{bmatrix} p & q \\ r & s \end{bmatrix} = A \)

\( IA = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} p & q \\ r & s \end{bmatrix} \)

\( = \begin{bmatrix} p+0 & q+0 \\ 0+r & 0+s \end{bmatrix} = \begin{bmatrix} p & q \\ r & s \end{bmatrix} = A \)

Hence proved: \( AI = IA = A \)

In simple words: The identity matrix I acts like the number 1 in multiplication. When you multiply any matrix by I, you get the same matrix back.

📝 Teacher's Note: Tell students that I is like the number 1. Just like 5×1 = 5, any matrix times I equals the same matrix. This is why I is called identity.

🎯 Exam Tip: Write "AI = IA = A" clearly at the end. Show both multiplications step by step. This proves that I is the identity matrix.

Answer 4.
Answer:
Given: \( P = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \), \( Q = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \)

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

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

\( P(QP) = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix} \)

\( = \begin{bmatrix} 4+10 & 5+8 \\ 8+5 & 10+4 \end{bmatrix} = \begin{bmatrix} 14 & 13 \\ 13 & 14 \end{bmatrix} \)

In simple words: We first find QP, then multiply P with that result. This shows how to do matrix multiplication when we have three matrices together.

📝 Teacher's Note: Explain that brackets matter in matrix multiplication. We must do QP first, then multiply by P. It's like doing (2×3)×4 in regular numbers.

🎯 Exam Tip: Always work from inside the brackets outward. Write "First find QP, then find P(QP)" to show you understand the order of operations.

Answer 14.
Answer:
Given: \( A = \begin{bmatrix} p & q \\ 8 & r \end{bmatrix} \), \( B = \begin{bmatrix} 3p & 5q \\ 2q & 7 \end{bmatrix} \)

\( A + B = \begin{bmatrix} p+3q & q+5q \\ 8+2q & 5+7 \end{bmatrix} = \begin{bmatrix} 4p & 6q \\ 8+2q & 12 \end{bmatrix} \) ...(1)

But given: \( A + B = \begin{bmatrix} 12 & 6 \\ 2r & 3s \end{bmatrix} \) ...(2)

From (1) and (2):
\( \begin{bmatrix} 4p & 6q \\ 8+2q & 12 \end{bmatrix} = \begin{bmatrix} 12 & 6 \\ 2r & 3s \end{bmatrix} \)

\( 4p = 12 \)
\( \Rightarrow p = 3 \)
\( 6q = 6 \)
\( \Rightarrow q = 1 \)
\( 8 + 2q = 2r \)
\( 8 + 2 = 2r \)
\( \Rightarrow r = 5 \)
\( 12 = 3s \)
\( \Rightarrow s = 4 \)

In simple words: We add the two matrices and compare each position with the given result. This gives us four equations to find p, q, r, and s.

📝 Teacher's Note: Show students that when two matrices are equal, each corresponding element must be equal. This gives us simple equations to solve.

🎯 Exam Tip: Write "comparing corresponding elements" to show you understand how matrix equality works. Find one variable at a time clearly.

Answer 15.
Answer:
Given: \( \begin{bmatrix} 2a+b & c \\ d & 3a-b \end{bmatrix} = \begin{bmatrix} 4 & 3a \\ 7 & 6 \end{bmatrix} \)

\( 2a + b = 4 \) ...(1)
\( 3a - b = 6 \) ...(2)

Adding (1) and (2), we get:
\( 5a = 10 \)
\( \Rightarrow a = 2 \)

From (1): \( 2(2) + b = 4 \)
\( \Rightarrow b = 0 \)

\( c = 3a \)
\( \Rightarrow c = 3 \times 2 = 6 \)
\( \Rightarrow d = 7 \)

In simple words: We compare each position in the matrices to get equations. Then we solve these equations to find a, b, c, and d.

📝 Teacher's Note: Show students how to solve simultaneous equations by addition method. Adding equations eliminates one variable and makes solving easier.

🎯 Exam Tip: Always check your answers by substituting back into the original equations. Write all values clearly: a=2, b=0, c=6, d=7.

Answer 17.
Answer:
Given: \( A = \begin{bmatrix} 15 & 7 \\ 13 & 8 \end{bmatrix} \), \( B = \begin{bmatrix} 16 & 12 \\ 27 & 11 \end{bmatrix} \)

(i) \( A + X = B \)
\( X = B - A \)
\( X = \begin{bmatrix} 16 & 12 \\ 27 & 11 \end{bmatrix} - \begin{bmatrix} 15 & 7 \\ 13 & 8 \end{bmatrix} = \begin{bmatrix} 1 & 5 \\ 14 & 3 \end{bmatrix} \)

(ii) \( 2A - X = B \)
\( X = 2A - B \)
\( X = \begin{bmatrix} 30 & 14 \\ 26 & 16 \end{bmatrix} - \begin{bmatrix} 16 & 12 \\ 27 & 11 \end{bmatrix} = \begin{bmatrix} 14 & 2 \\ -1 & 5 \end{bmatrix} \)

In simple words: We solve for matrix X like solving for x in regular algebra. We move terms to one side to isolate X.

📝 Teacher's Note: Explain that matrix equations work like regular equations. If A + X = B, then X = B - A. The same algebra rules apply to matrices.

🎯 Exam Tip: Always rearrange the equation to get X alone on one side. Show the subtraction step by step for each element of the matrices.

Answer 18.
Answer:
Given: \( P = \begin{bmatrix} 14 & 17 \\ 13 & 1 \end{bmatrix} \), \( Q = \begin{bmatrix} 2 & 1 \\ 3 & -3 \end{bmatrix} \)

\( P - M = 3Q \)
\( M = P - 3Q \)

\( M = \begin{bmatrix} 14 & 17 \\ 13 & 1 \end{bmatrix} - \begin{bmatrix} 6 & 3 \\ 9 & -9 \end{bmatrix} \)

\( = \begin{bmatrix} 14-6 & 17-3 \\ 13-9 & 1+9 \end{bmatrix} = \begin{bmatrix} 8 & 14 \\ 4 & 10 \end{bmatrix} \)

In simple words: We first multiply Q by 3 to get 3Q. Then we subtract this from P to find M.

📝 Teacher's Note: Remind students that 3Q means multiply every element of Q by 3. Then do the subtraction element by element.

🎯 Exam Tip: Show the step "3Q = ..." before doing the subtraction. This shows you understand scalar multiplication of matrices.