OP Malhotra Class 12 Maths Solutions Chapter 6 Matrices Exercise 6 (B)

Question 1. Write each sum or difference as a single matrix.
(i) \( \left[\begin{array}{rr} -3 & -2 \\ 2 & 0 \end{array}\right]+\left[\begin{array}{rr} 3 & 2 \\ -2 & 0 \end{array}\right] \)
(ii) \( \left[\begin{array}{rrr} 1 & 2 & -1 \\ 3 & 1 & 4 \\ 3 & 1 & -6 \end{array}\right]+\left[\begin{array}{rrr} -1 & -2 & 1 \\ -3 & 0 & -4 \\ -2 & -1 & 6 \end{array}\right] \)
(iii) \( \left[\begin{array}{rr} \cos ^2 x & \sin ^2 x \\ \sin x & \cos ^2 x \end{array}\right]+\left[\begin{array}{rr} \sin ^2 x & \cos ^2 x \\ -\cos x & -\sin ^2 x \end{array}\right] \)
Answer:
(i) We add the corresponding elements of the matrices:
\( \left[\begin{array}{rr} -3 & -2 \\ 2 & 0 \end{array}\right]+\left[\begin{array}{rr} 3 & 2 \\ -2 & 0 \end{array}\right] = \left[\begin{array}{rr} -3+3 & -2+2 \\ 2-2 & 0+0 \end{array}\right] = \left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \)

(ii) We add the corresponding elements of the matrices:
\( \left[\begin{array}{rrr} 1 & 2 & -1 \\ 3 & 1 & 4 \\ 3 & 1 & -6 \end{array}\right]+\left[\begin{array}{rrr} -1 & -2 & 1 \\ -3 & 0 & -4 \\ -2 & -1 & 6 \end{array}\right] = \left[\begin{array}{ccc} 1+(-1) & 2+(-2) & -1+1 \\ 3+(-3) & 1+0 & 4+(-4) \\ 3+(-2) & 1+(-1) & -6+6 \end{array}\right] = \left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}\right] \)

(iii) We add the corresponding elements of the matrices:
\( \left[\begin{array}{rr} \cos ^2 x & \sin ^2 x \\ \sin x & \cos ^2 x \end{array}\right]+\left[\begin{array}{rr} \sin ^2 x & \cos ^2 x \\ -\cos x & -\sin ^2 x \end{array}\right] \)
\( = \left[\begin{array}{cc} \cos ^2 x+\sin ^2 x & \sin ^2 x+\cos ^2 x \\ \sin x-\cos x & \cos ^2 x-\sin ^2 x \end{array}\right] \)
Using the identity \( \sin^2 x + \cos^2 x = 1 \) and \( \cos^2 x - \sin^2 x = \cos 2x \):
\( = \left[\begin{array}{cc} 1 & 1 \\ \sin x-\cos x & \cos 2x \end{array}\right] \)
In simple words: To add or subtract matrices, you combine the numbers that are in the exact same spot in each matrix. This means the element in the first row, first column of the first matrix is added to the element in the first row, first column of the second matrix, and so on.

🎯 Exam Tip: Always make sure the matrices have the same number of rows and columns before you try to add or subtract them. Only corresponding elements can be combined.

Question 2. (i) Given \( A = \left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \\ 2 & 6 \end{array}\right], B = \left[\begin{array}{rr} 2 & -1 \\ 3 & -2 \\ 0 & 1 \end{array}\right] \) and \( C = \left[\begin{array}{rr} 4 & 2 \\ 1 & 0 \\ -2 & -4 \end{array}\right] \) compute the following
(b) (A + B) + C
(c) A + (B + C)
(d) A – B
(e) (A – B) + C
(f) B - A
(ii) Consider the answers to part (b) and (c), what law is illustrated?
(iii) Consider the parts (d) and (f), what conclusion can be drawn?
Answer:
(i)
First, let's calculate A + B:
\( A+B = \left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \\ 2 & 6 \end{array}\right]+\left[\begin{array}{rr} 2 & -1 \\ 3 & -2 \\ 0 & 1 \end{array}\right] = \left[\begin{array}{rr} 1+2 & 2-1 \\ 3+3 & 4-2 \\ 2+0 & 6+1 \end{array}\right] = \left[\begin{array}{ll} 3 & 1 \\ 6 & 2 \\ 2 & 7 \end{array}\right] \)

(b) Now, calculate (A + B) + C:
\( (A+B)+C = \left[\begin{array}{ll} 3 & 1 \\ 6 & 2 \\ 2 & 7 \end{array}\right]+\left[\begin{array}{rr} 4 & 2 \\ 1 & 0 \\ -2 & -4 \end{array}\right] = \left[\begin{array}{rr} 3+4 & 1+2 \\ 6+1 & 2+0 \\ 2-2 & 7-4 \end{array}\right] = \left[\begin{array}{ll} 7 & 3 \\ 7 & 2 \\ 0 & 3 \end{array}\right] \)

Next, let's calculate B + C:
\( B+C = \left[\begin{array}{rr} 2 & -1 \\ 3 & -2 \\ 0 & 1 \end{array}\right]+\left[\begin{array}{rr} 4 & 2 \\ 1 & 0 \\ -2 & -4 \end{array}\right] = \left[\begin{array}{rr} 2+4 & -1+2 \\ 3+1 & -2+0 \\ 0-2 & 1-4 \end{array}\right] = \left[\begin{array}{rr} 6 & 1 \\ 4 & -2 \\ -2 & -3 \end{array}\right] \)

(c) Now, calculate A + (B + C):
\( A+(B+C) = \left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \\ 2 & 6 \end{array}\right]+\left[\begin{array}{rr} 6 & 1 \\ 4 & -2 \\ -2 & -3 \end{array}\right] = \left[\begin{array}{rr} 1+6 & 2+1 \\ 3+4 & 4-2 \\ 2-2 & 6-3 \end{array}\right] = \left[\begin{array}{ll} 7 & 3 \\ 7 & 2 \\ 0 & 3 \end{array}\right] \)

(d) Calculate A - B:
\( A-B = \left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \\ 2 & 6 \end{array}\right]-\left[\begin{array}{rr} 2 & -1 \\ 3 & -2 \\ 0 & 1 \end{array}\right] = \left[\begin{array}{rr} 1-2 & 2-(-1) \\ 3-3 & 4-(-2) \\ 2-0 & 6-1 \end{array}\right] = \left[\begin{array}{rr} -1 & 3 \\ 0 & 6 \\ 2 & 5 \end{array}\right] \)

(e) Now, calculate (A - B) + C:
\( (A-B)+C = \left[\begin{array}{rr} -1 & 3 \\ 0 & 6 \\ 2 & 5 \end{array}\right]+\left[\begin{array}{rr} 4 & 2 \\ 1 & 0 \\ -2 & -4 \end{array}\right] = \left[\begin{array}{rr} -1+4 & 3+2 \\ 0+1 & 6+0 \\ 2-2 & 5-4 \end{array}\right] = \left[\begin{array}{ll} 3 & 5 \\ 1 & 6 \\ 0 & 1 \end{array}\right] \)

(f) Calculate B - A:
\( B-A = \left[\begin{array}{rr} 2 & -1 \\ 3 & -2 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \\ 2 & 6 \end{array}\right] = \left[\begin{array}{rr} 2-1 & -1-2 \\ 3-3 & -2-4 \\ 0-2 & 1-6 \end{array}\right] = \left[\begin{array}{rr} 1 & -3 \\ 0 & -6 \\ -2 & -5 \end{array}\right] \)

(ii) From parts (b) and (c), we see that \( (A+B)+C = A+(B+C) \). This shows that the **associative law** holds true for matrix addition. This means when you add three matrices, the way you group them does not change the final sum. The order of operations does not matter here.

(iii) From parts (d) and (f), we can see that \( A-B \neq B-A \). Instead, \( A-B = -(B-A) \). This means that the **commutative law** does not hold for matrix subtraction. The order in which you subtract matrices matters; switching the order gives you the negative of the original result.
In simple words: For part (i), we added and subtracted matrices by combining the numbers in the same positions. For parts (ii) and (iii), we checked some rules. We found that the grouping rule (associative law) works for adding matrices, but the order rule (commutative law) does not work for subtracting matrices.

🎯 Exam Tip: Remember that matrix addition is associative and commutative, but matrix subtraction is neither. Pay close attention to the order of matrices when subtracting.

Question 3. Solve the equation \( X + \left[\begin{array}{III} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}\right]=\left[\begin{array}{III} 2 & 1 & 2 \\ 3 & 2 & 3 \\ 4 & 3 & 4 \end{array}\right] \) for the \( 3 \times 3 \) matrix X.
Answer:
Given the equation:
\( X + \left[\begin{array}{III} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}\right]=\left[\begin{array}{III} 2 & 1 & 2 \\ 3 & 2 & 3 \\ 4 & 3 & 4 \end{array}\right] \)

To find matrix X, we can subtract the known matrix from the right-hand side:
\( X = \left[\begin{array}{III} 2 & 1 & 2 \\ 3 & 2 & 3 \\ 4 & 3 & 4 \end{array}\right] - \left[\begin{array}{III} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}\right] \)

Now, we perform the subtraction element-wise:
\( X = \left[\begin{array}{ccc} 2-0 & 1-0 & 2-1 \\ 3-0 & 2-1 & 3-0 \\ 4-1 & 3-0 & 4-0 \end{array}\right] \)

Simplifying each element gives us:
\( X = \left[\begin{array}{III} 2 & 1 & 1 \\ 3 & 1 & 3 \\ 3 & 3 & 4 \end{array}\right] \)
In simple words: To find the unknown matrix X, we move the known matrix on the left side to the right side of the equation. This means we subtract it from the matrix already on the right side. Then, we subtract the numbers that are in the same spots in both matrices to get our final answer.

🎯 Exam Tip: When solving matrix equations, treat the matrices like single variables in algebra. Always remember to subtract or add corresponding elements carefully.

Question 4. If \( \left[\begin{array}{rr} 2 & -3 \\ 4 & 0 \end{array}\right]-\left[\begin{array}{ll} x_1 & x_2 \\ y_1 & y_2 \end{array}\right]=\left[\begin{array}{rr} -3 & 4 \\ 5 & -1 \end{array}\right] \) determine \( x_1, x_2, y_1 \) and \( y_2 \).
Answer:
Given the matrix equation:
\( \left[\begin{array}{rr} 2 & -3 \\ 4 & 0 \end{array}\right]-\left[\begin{array}{ll} x_1 & x_2 \\ y_1 & y_2 \end{array}\right]=\left[\begin{array}{rr} -3 & 4 \\ 5 & -1 \end{array}\right] \)

First, perform the subtraction on the left side:
\( \left[\begin{array}{cc} 2-x_1 & -3-x_2 \\ 4-y_1 & 0-y_2 \end{array}\right]=\left[\begin{array}{rr} -3 & 4 \\ 5 & -1 \end{array}\right] \)

Since the two matrices are equal, their corresponding elements must be equal. We set up equations for each position:
For the top-left element:
\( 2 - x_1 = -3 \)
\( \implies -x_1 = -3 - 2 \)
\( \implies -x_1 = -5 \)
\( \implies x_1 = 5 \)

For the top-right element:
\( -3 - x_2 = 4 \)
\( \implies -x_2 = 4 + 3 \)
\( \implies -x_2 = 7 \)
\( \implies x_2 = -7 \)

For the bottom-left element:
\( 4 - y_1 = 5 \)
\( \implies -y_1 = 5 - 4 \)
\( \implies -y_1 = 1 \)
\( \implies y_1 = -1 \)

For the bottom-right element:
\( 0 - y_2 = -1 \)
\( \implies -y_2 = -1 \)
\( \implies y_2 = 1 \)

So, the values are \( x_1 = 5, x_2 = -7, y_1 = -1, y_2 = 1 \).
In simple words: When two matrices are said to be equal, it means that the numbers in the same exact spot in both matrices must be the same. So, we do the subtraction first, then we match up each number from the left matrix with its partner on the right matrix. This gives us small math problems to solve for each unknown letter.

🎯 Exam Tip: Always equate corresponding elements when two matrices are equal. Be careful with negative signs during subtraction and when solving for the variables.

Question 5. If \( A = \left[\begin{array}{rr} 1 & 2 \\ -1 & 8 \\ 4 & 9 \end{array}\right] \), construct a matrix X such that \( X + A = 0 \).
Answer:
Given matrix \( A = \left[\begin{array}{rr} 1 & 2 \\ -1 & 8 \\ 4 & 9 \end{array}\right] \). We need to find matrix X such that \( X + A = 0 \).

Here, 0 represents the zero matrix of the same dimensions as A, which is a \( 3 \times 2 \) matrix with all elements being zero:
\( 0 = \left[\begin{array}{rr} 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{array}\right] \)

From the equation \( X + A = 0 \), we can rearrange to solve for X:
\( X = 0 - A \)
\( X = -A \)

To find -A, we multiply each element of A by -1:
\( X = - \left[\begin{array}{rr} 1 & 2 \\ -1 & 8 \\ 4 & 9 \end{array}\right] = \left[\begin{array}{rr} -1 & -2 \\ -(-1) & -8 \\ -4 & -9 \end{array}\right] = \left[\begin{array}{rr} -1 & -2 \\ 1 & -8 \\ -4 & -9 \end{array}\right] \)
In simple words: When you need to find a matrix X that, when added to matrix A, results in a zero matrix, it means X must be the negative of A. You get the negative of a matrix by changing the sign of every single number inside it.

🎯 Exam Tip: The zero matrix (0) acts as an additive identity in matrix algebra. Its elements are all zeros, and its dimensions must match those of the other matrices in the equation.

Question 6. For \( A = \left[\begin{array}{rrr} 2 & 2 & 2 \\ 2 & 1 & -3 \\ 1 & 0 & 4 \end{array}\right] \), \( B = \left[\begin{array}{rrr} 3 & 3 & 3 \\ 3 & 0 & 5 \\ 6 & 9 & -1 \end{array}\right] \), \( C = \left[\begin{array}{rrr} 4 & 4 & 4 \\ 5 & -1 & 4 \\ 7 & 8 & -1 \end{array}\right] \) compute
(a) 3A – 6B + 9C
(b) 7A – 2B – C
Answer:
(a) To compute 3A – 6B + 9C, we first perform the scalar multiplication for each matrix:
\( 3A = 3 \left[\begin{array}{rrr} 2 & 2 & 2 \\ 2 & 1 & -3 \\ 1 & 0 & 4 \end{array}\right] = \left[\begin{array}{rrr} 3 \times 2 & 3 \times 2 & 3 \times 2 \\ 3 \times 2 & 3 \times 1 & 3 \times (-3) \\ 3 \times 1 & 3 \times 0 & 3 \times 4 \end{array}\right] = \left[\begin{array}{rrr} 6 & 6 & 6 \\ 6 & 3 & -9 \\ 3 & 0 & 12 \end{array}\right] \)

\( 6B = 6 \left[\begin{array}{rrr} 3 & 3 & 3 \\ 3 & 0 & 5 \\ 6 & 9 & -1 \end{array}\right] = \left[\begin{array}{rrr} 6 \times 3 & 6 \times 3 & 6 \times 3 \\ 6 \times 3 & 6 \times 0 & 6 \times 5 \\ 6 \times 6 & 6 \times 9 & 6 \times (-1) \end{array}\right] = \left[\begin{array}{rrr} 18 & 18 & 18 \\ 18 & 0 & 30 \\ 36 & 54 & -6 \end{array}\right] \)

\( 9C = 9 \left[\begin{array}{rrr} 4 & 4 & 4 \\ 5 & -1 & 4 \\ 7 & 8 & -1 \end{array}\right] = \left[\begin{array}{rrr} 9 \times 4 & 9 \times 4 & 9 \times 4 \\ 9 \times 5 & 9 \times (-1) & 9 \times 4 \\ 9 \times 7 & 9 \times 8 & 9 \times (-1) \end{array}\right] = \left[\begin{array}{rrr} 36 & 36 & 36 \\ 45 & -9 & 36 \\ 63 & 72 & -9 \end{array}\right] \)

Now, we combine these matrices: \( 3A - 6B + 9C \)
\( = \left[\begin{array}{rrr} 6 & 6 & 6 \\ 6 & 3 & -9 \\ 3 & 0 & 12 \end{array}\right] - \left[\begin{array}{rrr} 18 & 18 & 18 \\ 18 & 0 & 30 \\ 36 & 54 & -6 \end{array}\right] + \left[\begin{array}{rrr} 36 & 36 & 36 \\ 45 & -9 & 36 \\ 63 & 72 & -9 \end{array}\right] \)
\( = \left[\begin{array}{rrr} 6-18+36 & 6-18+36 & 6-18+36 \\ 6-18+45 & 3-0-9 & -9-30+36 \\ 3-36+63 & 0-54+72 & 12-(-6)+(-9) \end{array}\right] \)
\( = \left[\begin{array}{rrr} 24 & 24 & 24 \\ 33 & -6 & -3 \\ 30 & 18 & 9 \end{array}\right] \)

(b) To compute 7A – 2B – C, we first perform the scalar multiplication for A and B:
\( 7A = 7 \left[\begin{array}{rrr} 2 & 2 & 2 \\ 2 & 1 & -3 \\ 1 & 0 & 4 \end{array}\right] = \left[\begin{array}{rrr} 14 & 14 & 14 \\ 14 & 7 & -21 \\ 7 & 0 & 28 \end{array}\right] \)

\( 2B = 2 \left[\begin{array}{rrr} 3 & 3 & 3 \\ 3 & 0 & 5 \\ 6 & 9 & -1 \end{array}\right] = \left[\begin{array}{rrr} 6 & 6 & 6 \\ 6 & 0 & 10 \\ 12 & 18 & -2 \end{array}\right] \)

Now, we combine these matrices: \( 7A - 2B - C \)
\( = \left[\begin{array}{rrr} 14 & 14 & 14 \\ 14 & 7 & -21 \\ 7 & 0 & 28 \end{array}\right] - \left[\begin{array}{rrr} 6 & 6 & 6 \\ 6 & 0 & 10 \\ 12 & 18 & -2 \end{array}\right] - \left[\begin{array}{rrr} 4 & 4 & 4 \\ 5 & -1 & 4 \\ 7 & 8 & -1 \end{array}\right] \)
\( = \left[\begin{array}{rrr} 14-6-4 & 14-6-4 & 14-6-4 \\ 14-6-5 & 7-0-(-1) & -21-10-4 \\ 7-12-7 & 0-18-8 & 28-(-2)-(-1) \end{array}\right] \)
\( = \left[\begin{array}{rrr} 4 & 4 & 4 \\ 3 & 8 & -35 \\ -12 & -26 & 31 \end{array}\right] \)
In simple words: For each problem, we first multiplied each matrix by the number given outside it (called scalar multiplication). This means we multiplied every number inside the matrix by that scalar. Then, we added or subtracted these new matrices by combining the numbers in the same positions.

🎯 Exam Tip: When performing multiple matrix operations (scalar multiplication, addition, subtraction), it's often best to perform scalar multiplication first, then handle addition/subtraction carefully, element by element, paying close attention to signs.

Question 7. For \( A = \left[\begin{array}{rr} 1 & -2 \\ 1 & 0 \end{array}\right] \) and \( B = \left[\begin{array}{rr} 2 & 1 \\ -1 & 1 \end{array}\right] \), solve each over \( S_{2\times2} \):
(i) X + 2A = B
(ii) X – A = 3B
(iii) 2X - 3A = 2B – X
Answer:
Given matrices \( A = \left[\begin{array}{rr} 1 & -2 \\ 1 & 0 \end{array}\right] \) and \( B = \left[\begin{array}{rr} 2 & 1 \\ -1 & 1 \end{array}\right] \).

(i) Solve X + 2A = B
First, rearrange the equation to solve for X:
\( X = B - 2A \)
Calculate \( 2A \):
\( 2A = 2 \left[\begin{array}{rr} 1 & -2 \\ 1 & 0 \end{array}\right] = \left[\begin{array}{rr} 2 & -4 \\ 2 & 0 \end{array}\right] \)
Now, substitute and subtract:
\( X = \left[\begin{array}{rr} 2 & 1 \\ -1 & 1 \end{array}\right] - \left[\begin{array}{rr} 2 & -4 \\ 2 & 0 \end{array}\right] \)
\( X = \left[\begin{array}{rr} 2-2 & 1-(-4) \\ -1-2 & 1-0 \end{array}\right] = \left[\begin{array}{rr} 0 & 5 \\ -3 & 1 \end{array}\right] \)

(ii) Solve X – A = 3B
First, rearrange the equation to solve for X:
\( X = 3B + A \)
Calculate \( 3B \):
\( 3B = 3 \left[\begin{array}{rr} 2 & 1 \\ -1 & 1 \end{array}\right] = \left[\begin{array}{rr} 6 & 3 \\ -3 & 3 \end{array}\right] \)
Now, substitute and add:
\( X = \left[\begin{array}{rr} 6 & 3 \\ -3 & 3 \end{array}\right] + \left[\begin{array}{rr} 1 & -2 \\ 1 & 0 \end{array}\right] \)
\( X = \left[\begin{array}{rr} 6+1 & 3+(-2) \\ -3+1 & 3+0 \end{array}\right] = \left[\begin{array}{rr} 7 & 1 \\ -2 & 3 \end{array}\right] \)

(iii) Solve 2X - 3A = 2B – X
First, rearrange the equation to gather all X terms on one side and other terms on the other side:
\( 2X + X = 2B + 3A \)
\( 3X = 2B + 3A \)
So, \( X = \frac{1}{3}(2B + 3A) \)
Calculate \( 2B \):
\( 2B = 2 \left[\begin{array}{rr} 2 & 1 \\ -1 & 1 \end{array}\right] = \left[\begin{array}{rr} 4 & 2 \\ -2 & 2 \end{array}\right] \)
Calculate \( 3A \):
\( 3A = 3 \left[\begin{array}{rr} 1 & -2 \\ 1 & 0 \end{array}\right] = \left[\begin{array}{rr} 3 & -6 \\ 3 & 0 \end{array}\right] \)
Now, add \( 2B + 3A \):
\( 2B + 3A = \left[\begin{array}{rr} 4 & 2 \\ -2 & 2 \end{array}\right] + \left[\begin{array}{rr} 3 & -6 \\ 3 & 0 \end{array}\right] = \left[\begin{array}{rr} 4+3 & 2-6 \\ -2+3 & 2+0 \end{array}\right] = \left[\begin{array}{rr} 7 & -4 \\ 1 & 2 \end{array}\right] \)
Finally, multiply by \( \frac{1}{3} \) to find X:
\( X = \frac{1}{3} \left[\begin{array}{rr} 7 & -4 \\ 1 & 2 \end{array}\right] = \left[\begin{array}{rr} \frac{7}{3} & -\frac{4}{3} \\ \frac{1}{3} & \frac{2}{3} \end{array}\right] \)
In simple words: We solved these matrix puzzles by using the same rules we use for regular numbers. First, we got X by itself on one side of the equal sign. Then, we did the multiplication (scalar multiplication) and addition or subtraction for the numbers in the matrices, always matching up numbers in the same spots. If we needed to divide by a number, we divided every number in the matrix by that number.

🎯 Exam Tip: Remember to treat matrix equations algebraically by moving terms. Scalar multiplication must be applied to every element, and matrix addition/subtraction also involves corresponding elements. For fractional scalars, apply the fraction to each element.

Question 8. If \( A = \left[\begin{array}{rrr} 1 & 2 & 3 \\ -2 & 5 & 7 \end{array}\right] \) and \( 2A – 3B = \left[\begin{array}{rrr} 4 & 5 & -9 \\ 1 & 2 & 3 \end{array}\right] \), find B.
Answer:
Given \( A = \left[\begin{array}{rrr} 1 & 2 & 3 \\ -2 & 5 & 7 \end{array}\right] \) and the equation \( 2A – 3B = \left[\begin{array}{rrr} 4 & 5 & -9 \\ 1 & 2 & 3 \end{array}\right] \).

First, rearrange the equation to solve for B:
\( 2A - \left[\begin{array}{rrr} 4 & 5 & -9 \\ 1 & 2 & 3 \end{array}\right] = 3B \)
So, \( 3B = 2A - \left[\begin{array}{rrr} 4 & 5 & -9 \\ 1 & 2 & 3 \end{array}\right] \)

Now, calculate \( 2A \):
\( 2A = 2 \left[\begin{array}{rrr} 1 & 2 & 3 \\ -2 & 5 & 7 \end{array}\right] = \left[\begin{array}{rrr} 2 \times 1 & 2 \times 2 & 2 \times 3 \\ 2 \times (-2) & 2 \times 5 & 2 \times 7 \end{array}\right] = \left[\begin{array}{rrr} 2 & 4 & 6 \\ -4 & 10 & 14 \end{array}\right] \)

Next, substitute \( 2A \) into the equation for \( 3B \):
\( 3B = \left[\begin{array}{rrr} 2 & 4 & 6 \\ -4 & 10 & 14 \end{array}\right] - \left[\begin{array}{rrr} 4 & 5 & -9 \\ 1 & 2 & 3 \end{array}\right] \)

Perform the subtraction element-wise:
\( 3B = \left[\begin{array}{rrr} 2-4 & 4-5 & 6-(-9) \\ -4-1 & 10-2 & 14-3 \end{array}\right] = \left[\begin{array}{rrr} -2 & -1 & 15 \\ -5 & 8 & 11 \end{array}\right] \)

Finally, to find B, multiply the resulting matrix by \( \frac{1}{3} \):
\( B = \frac{1}{3} \left[\begin{array}{rrr} -2 & -1 & 15 \\ -5 & 8 & 11 \end{array}\right] = \left[\begin{array}{rrr} -\frac{2}{3} & -\frac{1}{3} & \frac{15}{3} \\ -\frac{5}{3} & \frac{8}{3} & \frac{11}{3} \end{array}\right] = \left[\begin{array}{rrr} -\frac{2}{3} & -\frac{1}{3} & 5 \\ -\frac{5}{3} & \frac{8}{3} & \frac{11}{3} \end{array}\right] \)
In simple words: We used the given equation to find matrix B. First, we moved the parts of the equation around so that 3B was by itself. Then, we multiplied matrix A by 2, and subtracted the given matrix from it. After we had the matrix for 3B, we divided every number in that matrix by 3 to get our final matrix B.

🎯 Exam Tip: When solving for an unknown matrix in an equation, use algebraic principles to isolate it. Remember that scalar division involves dividing every element of the matrix by the scalar.

Question 9. If \( A = \left[\begin{array}{rrr} 1 & -3 & 2 \\ 2 & 0 & 2 \end{array}\right] \), \( B = \left[\begin{array}{rrr} 2 & -1 & -1 \\ 1 & 0 & -1 \end{array}\right] \), find the matrix C such that A + B + C is a zero matrix.
Answer:
Given \( A = \left[\begin{array}{rrr} 1 & -3 & 2 \\ 2 & 0 & 2 \end{array}\right] \) and \( B = \left[\begin{array}{rrr} 2 & -1 & -1 \\ 1 & 0 & -1 \end{array}\right] \).
We need to find matrix C such that A + B + C = O, where O is the zero matrix.

Rearrange the equation to solve for C:
\( C = O - A - B \)
\( C = -(A+B) \)

First, calculate the sum of A and B:
\( A+B = \left[\begin{array}{rrr} 1 & -3 & 2 \\ 2 & 0 & 2 \end{array}\right] + \left[\begin{array}{rrr} 2 & -1 & -1 \\ 1 & 0 & -1 \end{array}\right] \)
\( = \left[\begin{array}{rrr} 1+2 & -3+(-1) & 2+(-1) \\ 2+1 & 0+0 & 2+(-1) \end{array}\right] \)
\( = \left[\begin{array}{rrr} 3 & -4 & 1 \\ 3 & 0 & 1 \end{array}\right] \)

Now, find the negative of this sum to get C:
\( C = - \left[\begin{array}{rrr} 3 & -4 & 1 \\ 3 & 0 & 1 \end{array}\right] = \left[\begin{array}{rrr} -3 & -(-4) & -1 \\ -3 & -0 & -1 \end{array}\right] = \left[\begin{array}{rrr} -3 & 4 & -1 \\ -3 & 0 & -1 \end{array}\right] \)
In simple words: We want to find a matrix C that makes the total sum of A, B, and C equal to a matrix of all zeros. To do this, we first add A and B together. Then, we find the "opposite" of this combined matrix, which means changing the sign of every number inside it. That "opposite" matrix is C.

🎯 Exam Tip: When a matrix sum is equal to the zero matrix, the unknown matrix is the negative of the sum of the other matrices. Ensure all matrices have compatible dimensions for addition and subtraction.

Question 10. Find the value of k, a non-zero scalar, if \( 2\left[\begin{array}{rrr} 1 & 2 & 3 \\ -1 & -3 & 2 \end{array}\right]+k\left[\begin{array}{III} 1 & 0 & 2 \\ 3 & 4 & 5 \end{array}\right]=\left[\begin{array}{lll} 4 & 4 & 10 \\ 4 & 2 & 14 \end{array}\right] \)
Answer:
Given the matrix equation:
\( 2\left[\begin{array}{rrr} 1 & 2 & 3 \\ -1 & -3 & 2 \end{array}\right]+k\left[\begin{array}{III} 1 & 0 & 2 \\ 3 & 4 & 5 \end{array}\right]=\left[\begin{array}{lll} 4 & 4 & 10 \\ 4 & 2 & 14 \end{array}\right] \)

First, perform the scalar multiplication on the left side:
\( \left[\begin{array}{rrr} 2 \times 1 & 2 \times 2 & 2 \times 3 \\ 2 \times (-1) & 2 \times (-3) & 2 \times 2 \end{array}\right] + \left[\begin{array}{rrr} k \times 1 & k \times 0 & k \times 2 \\ k \times 3 & k \times 4 & k \times 5 \end{array}\right] = \left[\begin{array}{lll} 4 & 4 & 10 \\ 4 & 2 & 14 \end{array}\right] \)

This simplifies to:
\( \left[\begin{array}{rrr} 2 & 4 & 6 \\ -2 & -6 & 4 \end{array}\right] + \left[\begin{array}{rrr} k & 0 & 2k \\ 3k & 4k & 5k \end{array}\right] = \left[\begin{array}{lll} 4 & 4 & 10 \\ 4 & 2 & 14 \end{array}\right] \)

Now, add the two matrices on the left side:
\( \left[\begin{array}{ccc} 2+k & 4+0 & 6+2k \\ -2+3k & -6+4k & 4+5k \end{array}\right] = \left[\begin{array}{lll} 4 & 4 & 10 \\ 4 & 2 & 14 \end{array}\right] \)

Since the two matrices are equal, their corresponding elements must be equal. We can pick any element to solve for k:
Using the top-left element:
\( 2+k = 4 \)
\( \implies k = 4-2 \)
\( \implies k = 2 \)

Using the top-middle element:
\( 4+0 = 4 \)
\( \implies 4 = 4 \) (This statement is consistent, but doesn't help find k.)

Using the top-right element:
\( 6+2k = 10 \)
\( \implies 2k = 10-6 \)
\( \implies 2k = 4 \)
\( \implies k = 2 \)

Using the bottom-left element:
\( -2+3k = 4 \)
\( \implies 3k = 4+2 \)
\( \implies 3k = 6 \)
\( \implies k = 2 \)

Using the bottom-middle element:
\( -6+4k = 2 \)
\( \implies 4k = 2+6 \)
\( \implies 4k = 8 \)
\( \implies k = 2 \)

Using the bottom-right element:
\( 4+5k = 14 \)
\( \implies 5k = 14-4 \)
\( \implies 5k = 10 \)
\( \implies k = 2 \)

All equations consistently give \( k = 2 \).
In simple words: We have an equation with matrices and a missing number 'k'. We first multiplied the numbers outside the matrices with all the numbers inside them. Then, we added the matrices on the left side. Since the two sides of the equation must be exactly equal, we picked any matching number from the left and right matrices and set them equal to each other. Solving that simple math problem helped us find the value of 'k'.

🎯 Exam Tip: When finding an unknown scalar in a matrix equation, perform all matrix operations on the left side first to get a single matrix. Then, equate any corresponding element of this matrix to the element in the same position on the right side to form an algebraic equation and solve for the scalar. It's a good practice to verify with another element.

Question 11. If A = diag (1 -4 8), B = diag (- 2 3 5), C = diag (-3 7 10), find
(i) 2A + 3B
(ii) B + 2C – A
(iii) 3A – B + 4C.
Answer:
First, let's write out the diagonal matrices in their full form:
\( A = \text{diag}(1, -4, 8) = \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 8 \end{array}\right] \)
\( B = \text{diag}(-2, 3, 5) = \left[\begin{array}{rrr} -2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{array}\right] \)
\( C = \text{diag}(-3, 7, 10) = \left[\begin{array}{rrr} -3 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 10 \end{array}\right] \)

(i) Compute 2A + 3B:
\( 2A = 2 \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 8 \end{array}\right] = \left[\begin{array}{rrr} 2 & 0 & 0 \\ 0 & -8 & 0 \\ 0 & 0 & 16 \end{array}\right] \)
\( 3B = 3 \left[\begin{array}{rrr} -2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{array}\right] = \left[\begin{array}{rrr} -6 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 15 \end{array}\right] \)
\( 2A + 3B = \left[\begin{array}{rrr} 2 & 0 & 0 \\ 0 & -8 & 0 \\ 0 & 0 & 16 \end{array}\right] + \left[\begin{array}{rrr} -6 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 15 \end{array}\right] = \left[\begin{array}{rrr} 2-6 & 0 & 0 \\ 0 & -8+9 & 0 \\ 0 & 0 & 16+15 \end{array}\right] = \left[\begin{array}{rrr} -4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 31 \end{array}\right] \)
This can also be written as \( \text{diag}(-4, 1, 31) \).

(ii) Compute B + 2C – A:
\( 2C = 2 \left[\begin{array}{rrr} -3 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 10 \end{array}\right] = \left[\begin{array}{rrr} -6 & 0 & 0 \\ 0 & 14 & 0 \\ 0 & 0 & 20 \end{array}\right] \)
\( B + 2C - A = \left[\begin{array}{rrr} -2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{array}\right] + \left[\begin{array}{rrr} -6 & 0 & 0 \\ 0 & 14 & 0 \\ 0 & 0 & 20 \end{array}\right] - \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 8 \end{array}\right] \)
\( = \left[\begin{array}{rrr} -2-6-1 & 0 & 0 \\ 0 & 3+14-(-4) & 0 \\ 0 & 0 & 5+20-8 \end{array}\right] = \left[\begin{array}{rrr} -9 & 0 & 0 \\ 0 & 21 & 0 \\ 0 & 0 & 17 \end{array}\right] \)
This can also be written as \( \text{diag}(-9, 21, 17) \).

(iii) Compute 3A – B + 4C:
\( 3A = 3 \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 8 \end{array}\right] = \left[\begin{array}{rrr} 3 & 0 & 0 \\ 0 & -12 & 0 \\ 0 & 0 & 24 \end{array}\right] \)
\( 4C = 4 \left[\begin{array}{rrr} -3 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 10 \end{array}\right] = \left[\begin{array}{rrr} -12 & 0 & 0 \\ 0 & 28 & 0 \\ 0 & 0 & 40 \end{array}\right] \)
\( 3A - B + 4C = \left[\begin{array}{rrr} 3 & 0 & 0 \\ 0 & -12 & 0 \\ 0 & 0 & 24 \end{array}\right] - \left[\begin{array}{rrr} -2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{array}\right] + \left[\begin{array}{rrr} -12 & 0 & 0 \\ 0 & 28 & 0 \\ 0 & 0 & 40 \end{array}\right] \)
\( = \left[\begin{array}{rrr} 3-(-2)+(-12) & 0 & 0 \\ 0 & -12-3+28 & 0 \\ 0 & 0 & 24-5+40 \end{array}\right] = \left[\begin{array}{rrr} -7 & 0 & 0 \\ 0 & 13 & 0 \\ 0 & 0 & 59 \end{array}\right] \)
This can also be written as \( \text{diag}(-7, 13, 59) \).
In simple words: First, we wrote out the diagonal matrices fully, which means putting the given numbers on the main diagonal and zeros everywhere else. Then, for each problem, we multiplied the matrices by any numbers outside them (scalar multiplication) and combined them by adding or subtracting the numbers in the same positions. Because they are diagonal matrices, only the numbers on the main diagonal are affected by these operations.

🎯 Exam Tip: For diagonal matrices, scalar multiplication and addition/subtraction simply involve applying the operations directly to the diagonal elements. Non-diagonal elements remain zero. You can perform these operations either by writing out the full matrices or by directly manipulating the diagonal entries, then converting back to diag notation if required.

Question 12. Solve the matrix equation : \( 2\left[\begin{array}{rr} x & y \\ z & t \end{array}\right]+3\left[\begin{array}{rr} 1 & -1 \\ 0 & 2 \end{array}\right]=3\left[\begin{array}{ll} 3 & 5 \\ 4 & 6 \end{array}\right] \)
Answer:
Given the matrix equation:
\( 2\left[\begin{array}{rr} x & y \\ z & t \end{array}\right]+3\left[\begin{array}{rr} 1 & -1 \\ 0 & 2 \end{array}\right]=3\left[\begin{array}{ll} 3 & 5 \\ 4 & 6 \end{array}\right] \)

First, perform the scalar multiplications on both sides of the equation:
\( \left[\begin{array}{rr} 2x & 2y \\ 2z & 2t \end{array}\right] + \left[\begin{array}{rr} 3 \times 1 & 3 \times (-1) \\ 3 \times 0 & 3 \times 2 \end{array}\right] = \left[\begin{array}{rr} 3 \times 3 & 3 \times 5 \\ 3 \times 4 & 3 \times 6 \end{array}\right] \)

This simplifies to:
\( \left[\begin{array}{rr} 2x & 2y \\ 2z & 2t \end{array}\right] + \left[\begin{array}{rr} 3 & -3 \\ 0 & 6 \end{array}\right] = \left[\begin{array}{rr} 9 & 15 \\ 12 & 18 \end{array}\right] \)

Now, add the two matrices on the left side:
\( \left[\begin{array}{rr} 2x+3 & 2y-3 \\ 2z+0 & 2t+6 \end{array}\right] = \left[\begin{array}{rr} 9 & 15 \\ 12 & 18 \end{array}\right] \)

Since the two matrices are equal, their corresponding elements must be equal. We set up equations for each position:
For the top-left element:
\( 2x+3 = 9 \)
\( 2x = 9-3 \)
\( 2x = 6 \)
\( x = 3 \)

For the top-right element:
\( 2y-3 = 15 \)
\( 2y = 15+3 \)
\( 2y = 18 \)
\( y = 9 \)

For the bottom-left element:
\( 2z+0 = 12 \)
\( 2z = 12 \)
\( z = 6 \)

For the bottom-right element:
\( 2t+6 = 18 \)
\( 2t = 18-6 \)
\( 2t = 12 \)
\( t = 6 \)

So, the solution is \( x=3, y=9, z=6, t=6 \).
In simple words: To solve this puzzle, we first multiplied the numbers outside each matrix by every number inside it. Then, we added the matrices on the left side of the equal sign. Since the two matrices on either side must be exactly the same, we matched up each number in the left matrix with the number in the same spot in the right matrix. This gave us four small math problems, which we solved one by one to find x, y, z, and t.

🎯 Exam Tip: Break down complex matrix equations into simpler steps: first scalar multiplication, then matrix addition/subtraction. Finally, equate corresponding elements to form a system of linear equations, which can then be solved for the unknown variables.

Question 13. Find x, y, z and w if \( 3\left[\begin{array}{cc} x & y \\ z & w \end{array}\right]=\left[\begin{array}{rr} x & 6 \\ -1 & 2 w \end{array}\right]+\left[\begin{array}{cc} 4 & x+y \\ z+w & 3 \end{array}\right] \)
Answer:
Given the matrix equation:
\( 3\left[\begin{array}{cc} x & y \\ z & w \end{array}\right]=\left[\begin{array}{rr} x & 6 \\ -1 & 2 w \end{array}\right]+\left[\begin{array}{cc} 4 & x+y \\ z+w & 3 \end{array}\right] \)

First, perform the scalar multiplication on the left side and matrix addition on the right side:
\( \left[\begin{array}{cc} 3x & 3y \\ 3z & 3w \end{array}\right] = \left[\begin{array}{cc} x+4 & 6+(x+y) \\ -1+(z+w) & 2w+3 \end{array}\right] \)

This simplifies to:
\( \left[\begin{array}{cc} 3x & 3y \\ 3z & 3w \end{array}\right] = \left[\begin{array}{cc} x+4 & x+y+6 \\ z+w-1 & 2w+3 \end{array}\right] \)

Since the two matrices are equal, their corresponding elements must be equal. We set up equations for each position:
1. For the top-left element:
\( 3x = x+4 \)
\( 3x-x = 4 \)
\( 2x = 4 \)
\( x = 2 \)

2. For the bottom-right element (since it only has 'w'):
\( 3w = 2w+3 \)
\( 3w-2w = 3 \)
\( w = 3 \)

3. For the top-right element (now we know x):
\( 3y = x+y+6 \)
Substitute \( x=2 \):
\( 3y = 2+y+6 \)
\( 3y-y = 8 \)
\( 2y = 8 \)
\( y = 4 \)

4. For the bottom-left element (now we know w):
\( 3z = z+w-1 \)
Substitute \( w=3 \):
\( 3z = z+3-1 \)
\( 3z = z+2 \)
\( 3z-z = 2 \)
\( 2z = 2 \)
\( z = 1 \)

So, the values are \( x=2, y=4, z=1, w=3 \).
In simple words: We had an equation where one matrix was multiplied by 3 on the left, and two matrices were added on the right. We did these operations first. Then, since the two resulting matrices must be identical, we matched each number in the left matrix to its partner in the right matrix. This gave us four separate equations, which we solved one by one to find the values of x, y, z, and w. It's often easiest to solve for the variables that appear in fewer equations first.

🎯 Exam Tip: When solving matrix equations with multiple variables, always simplify both sides first. Then, equate corresponding elements to form a system of equations. Strategically choose which equation to solve first (e.g., those with only one variable) to simplify the process.

Question 14. If \( A = \left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{array}\right] \) and \( B = \left[\begin{array}{rr} -3 & -2 \\ 1 & -5 \\ 4 & 3 \end{array}\right] \), then find a matrix C = \( \left[\begin{array}{ll} p & q \\ r & s \\ t & u \end{array}\right] \) such that A + B – C = 0.
Answer:
Given \( A = \left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{array}\right] \) and \( B = \left[\begin{array}{rr} -3 & -2 \\ 1 & -5 \\ 4 & 3 \end{array}\right] \).
We need to find matrix C such that \( A + B – C = 0 \), where 0 is the zero matrix.

Rearrange the equation to solve for C:
\( A + B = C \)

Now, we simply need to add matrices A and B to find C:
\( C = A + B = \left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{array}\right] + \left[\begin{array}{rr} -3 & -2 \\ 1 & -5 \\ 4 & 3 \end{array}\right] \)

Perform the addition element-wise:
\( C = \left[\begin{array}{rr} 1+(-3) & 2+(-2) \\ 3+1 & 4+(-5) \\ 5+4 & 6+3 \end{array}\right] \)

Simplifying each element gives us:
\( C = \left[\begin{array}{rr} -2 & 0 \\ 4 & -1 \\ 9 & 9 \end{array}\right] \)

Comparing this to \( C = \left[\begin{array}{ll} p & q \\ r & s \\ t & u \end{array}\right] \), we get:
\( p = -2 \)
\( q = 0 \)
\( r = 4 \)
\( s = -1 \)
\( t = 9 \)
\( u = 9 \)
In simple words: To make the sum of A and B minus C equal to zero, C must be exactly equal to the sum of A and B. So, we just added matrix A and matrix B by combining the numbers in the same positions. The resulting matrix is C.

🎯 Exam Tip: When given an equation like A + B - C = 0, the simplest approach is to isolate C by performing standard algebraic rearrangement. Then, carry out the matrix operations (in this case, addition) element by element.

Question 15. Let \( A = \left[\begin{array}{III} 2 & 3 & 5 \\ 1 & 0 & 2 \\ 3 & 4 & 5 \end{array}\right] \), find a matrix B such that A + B – 4I = 0.
Answer:
Given \( A = \left[\begin{array}{III} 2 & 3 & 5 \\ 1 & 0 & 2 \\ 3 & 4 & 5 \end{array}\right] \). We need to find matrix B such that \( A + B – 4I = 0 \).
Here, I is the identity matrix of the same dimension as A, which is a \( 3 \times 3 \) matrix. The identity matrix has 1s on its main diagonal and 0s elsewhere:
\( I = \left[\begin{array}{III} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \)

Rearrange the given equation to solve for B:
\( B = 4I - A \)

First, calculate \( 4I \):
\( 4I = 4 \left[\begin{array}{III} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] = \left[\begin{array}{rrr} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array}\right] \)

Now, substitute \( 4I \) and A into the equation for B and perform the subtraction:
\( B = \left[\begin{array}{rrr} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array}\right] - \left[\begin{array}{III} 2 & 3 & 5 \\ 1 & 0 & 2 \\ 3 & 4 & 5 \end{array}\right] \)

Perform the subtraction element-wise:
\( B = \left[\begin{array}{rrr} 4-2 & 0-3 & 0-5 \\ 0-1 & 4-0 & 0-2 \\ 0-3 & 0-4 & 4-5 \end{array}\right] \)

Simplifying each element gives us:
\( B = \left[\begin{array}{rrr} 2 & -3 & -5 \\ -1 & 4 & -2 \\ -3 & -4 & -1 \end{array}\right] \)
In simple words: We want to find matrix B so that when it's added to A and then we take away 4 times the identity matrix (I), the result is a zero matrix. We first rearranged the equation to get B by itself. This meant calculating 4 times the identity matrix and then subtracting matrix A from it. We did these operations by combining the numbers in the same positions.

🎯 Exam Tip: The identity matrix (I) is crucial in matrix algebra; it acts like the number '1'. For matrix addition/subtraction, ensure 'I' has the same dimensions as the other matrices involved. Scalar multiplication of 'I' means multiplying only its diagonal elements by the scalar.

Question 11. If A = diag (1, -4, 8), B = diag (-2, 3, 5), C = diag (-3, 7, 10), find
(i) 2A + 3B
(ii) B + 2C – A
(iii) 3A – B + 4C

Answer:
Given the diagonal matrices:
\( A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 8 \end{bmatrix} \), \( B = \begin{bmatrix} -2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix} \), \( C = \begin{bmatrix} -3 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 10 \end{bmatrix} \)

(i) \( 2A + 3B \)
First, we multiply matrix A by 2 and matrix B by 3. This means we multiply each number inside matrix A by 2, and each number inside matrix B by 3. Then, we add the two new matrices together by adding the numbers that are in the same spot in both matrices. The final result is a diagonal matrix where the elements are -4, 1, and 31. This is a common operation for diagonal matrices, simplifying to operations on their diagonal elements.
\( = 2\begin{bmatrix} 1 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 8 \end{bmatrix} + 3\begin{bmatrix} -2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix} \)
\( = \begin{bmatrix} 2 & 0 & 0 \\ 0 & -8 & 0 \\ 0 & 0 & 16 \end{bmatrix} + \begin{bmatrix} -6 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 15 \end{bmatrix} \)
\( = \begin{bmatrix} 2-6 & 0 & 0 \\ 0 & -8+9 & 0 \\ 0 & 0 & 16+15 \end{bmatrix} \)
\( = \begin{bmatrix} -4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 31 \end{bmatrix} \)
\( = \text{diag } (-4, 1, 31) \)

(ii) \( B + 2C - A \)
To find this, we first multiply matrix C by 2. Then, we add matrix B to this new matrix (2C). Finally, we subtract matrix A from the result. To add or subtract matrices, we perform the operation on the numbers located in the corresponding positions. This sequential application of matrix operations ensures correctness. This process gives us a new diagonal matrix with elements -9, 21, and 17.
\( = \begin{bmatrix} -2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix} + 2\begin{bmatrix} -3 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 10 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 8 \end{bmatrix} \)
\( = \begin{bmatrix} -2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix} + \begin{bmatrix} -6 & 0 & 0 \\ 0 & 14 & 0 \\ 0 & 0 & 20 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 8 \end{bmatrix} \)
\( = \begin{bmatrix} -2-6 & 0 & 0 \\ 0 & 3+14 & 0 \\ 0 & 0 & 5+20 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 8 \end{bmatrix} \)
\( = \begin{bmatrix} -8 & 0 & 0 \\ 0 & 17 & 0 \\ 0 & 0 & 25 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 8 \end{bmatrix} \)
\( = \begin{bmatrix} -8-1 & 0 & 0 \\ 0 & 17-(-4) & 0 \\ 0 & 0 & 25-8 \end{bmatrix} \)
\( = \begin{bmatrix} -9 & 0 & 0 \\ 0 & 21 & 0 \\ 0 & 0 & 17 \end{bmatrix} \)
\( = \text{diag } (-9, 21, 17) \)

(iii) \( 3A - B + 4C \)
We begin by multiplying matrix A by 3 and matrix C by 4. Next, we subtract matrix B from the result of 3A. Then, we add the result of 4C to that value. Each multiplication and addition/subtraction is done element by element, meaning numbers in the same positions are operated on. This method ensures that the diagonal structure is maintained throughout the calculation. The final matrix is a diagonal matrix containing -7, 13, and 59.
\( = 3\begin{bmatrix} 1 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 8 \end{bmatrix} - \begin{bmatrix} -2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix} + 4\begin{bmatrix} -3 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 10 \end{bmatrix} \)
\( = \begin{bmatrix} 3 & 0 & 0 \\ 0 & -12 & 0 \\ 0 & 0 & 24 \end{bmatrix} - \begin{bmatrix} -2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix} + \begin{bmatrix} -12 & 0 & 0 \\ 0 & 28 & 0 \\ 0 & 0 & 40 \end{bmatrix} \)
\( = \begin{bmatrix} 3-(-2) & 0 & 0 \\ 0 & -12-3 & 0 \\ 0 & 0 & 24-5 \end{bmatrix} + \begin{bmatrix} -12 & 0 & 0 \\ 0 & 28 & 0 \\ 0 & 0 & 40 \end{bmatrix} \)
\( = \begin{bmatrix} 5 & 0 & 0 \\ 0 & -15 & 0 \\ 0 & 0 & 19 \end{bmatrix} + \begin{bmatrix} -12 & 0 & 0 \\ 0 & 28 & 0 \\ 0 & 0 & 40 \end{bmatrix} \)
\( = \begin{bmatrix} 5-12 & 0 & 0 \\ 0 & -15+28 & 0 \\ 0 & 0 & 19+40 \end{bmatrix} \)
\( = \begin{bmatrix} -7 & 0 & 0 \\ 0 & 13 & 0 \\ 0 & 0 & 59 \end{bmatrix} \)
\( = \text{diag } (-7, 13, 59) \)
In simple words: For diagonal matrices, these operations mostly affect the diagonal numbers. We multiply each matrix by its scalar, then add or subtract the numbers that are in the same exact spot in each matrix. This gives us new numbers for the diagonal of the final matrix.

🎯 Exam Tip: Remember that scalar multiplication involves multiplying every element in the matrix by the scalar, and matrix addition/subtraction requires adding/subtracting corresponding elements. This is especially simple for diagonal matrices as only diagonal elements are non-zero.

Question 12. Solve the matrix equation : \( 2\begin{bmatrix} x & y \\ z & t \end{bmatrix}+3\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}=3\begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix} \)
Answer: To solve this matrix equation, we first perform the scalar multiplications. We multiply each element inside the first matrix by 2, and each element inside the second matrix by 3. The right side of the equation also gets multiplied by 3. After these multiplications, we combine the matrices on the left side by adding the numbers in corresponding positions. Once both sides are single matrices, we set the elements in the same positions equal to each other. This creates four separate equations that help us find the values of x, y, z, and t, ensuring a direct solution.
Given: \( 2\begin{bmatrix} x & y \\ z & t \end{bmatrix}+3\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}=3\begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix} \)
\( \implies \begin{bmatrix} 2x & 2y \\ 2z & 2t \end{bmatrix}+\begin{bmatrix} 3 & -3 \\ 0 & 6 \end{bmatrix}=\begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix} \)
\( \implies \begin{bmatrix} 2x+3 & 2y-3 \\ 2z & 2t+6 \end{bmatrix}=\begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix} \)
Thus, their corresponding elements are equal.
\( 2x + 3 = 9 \)
\( \implies 2x = 6 \)
\( \implies x = 3 \)
\( 2y - 3 = 15 \)
\( \implies 2y = 18 \)
\( \implies y = 9 \)
\( 2z = 12 \)
\( \implies z = 6 \)
\( 2t + 6 = 18 \)
\( \implies 2t = 12 \)
\( \implies t = 6 \)
So, \( x = 3, y = 9, z = 6, t = 6 \).
In simple words: First, multiply numbers outside the matrices with all numbers inside them. Then, add the matrices on the left side. Finally, match up the numbers in the same spots on both sides to find x, y, z, and t.

🎯 Exam Tip: When equating matrices, always make sure the dimensions (number of rows and columns) are identical. Then, match each element in its corresponding position to form algebraic equations and solve them carefully.

Question 13. Find x, y, z and w if \( 3\begin{bmatrix} x & y \\ z & w \end{bmatrix}=\begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix}+\begin{bmatrix} 4 & x+y \\ z+w & 3 \end{bmatrix} \)
Answer: To solve for x, y, z, and w, we first perform the scalar multiplication on the left side, multiplying each element in the matrix by 3. On the right side, we add the two matrices together by adding their corresponding elements. Once both sides of the equation are single matrices, we can set the elements in the same positions equal to each other. This gives us a system of four equations which we solve one by one to find the values of x, y, z, and w, making sure to substitute known values as we proceed.
Given: \( 3\begin{bmatrix} x & y \\ z & w \end{bmatrix}=\begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix}+\begin{bmatrix} 4 & x+y \\ z+w & 3 \end{bmatrix} \)
\( \implies \begin{bmatrix} 3x & 3y \\ 3z & 3w \end{bmatrix}=\begin{bmatrix} x+4 & 6+x+y \\ -1+z+w & 2w+3 \end{bmatrix} \)
Thus, their corresponding entries are equal.

1. \( 3x = x+4 \)
\( \implies 2x = 4 \)
\( \implies x = 2 \)

2. \( 3y = 6+x+y \)
Substitute \( x=2 \): \( 3y = 6+2+y \)
\( \implies 3y = 8+y \)
\( \implies 2y = 8 \)
\( \implies y = 4 \)

3. \( 3z = -1+z+w \)
\( \implies 2z = w-1 \) ... (1)

4. \( 3w = 2w+3 \)
\( \implies w = 3 \)

From (1), substitute \( w=3 \):
\( 2z = 3-1 \)
\( \implies 2z = 2 \)
\( \implies z = 1 \)
Thus, \( x = 2, y = 4, z = 1, w = 3 \).
In simple words: Multiply the left matrix by 3. Add the two matrices on the right. Then, match up numbers in the same locations to make small math problems and solve for x, y, z, and w, using the values you find along the way.

🎯 Exam Tip: Solve for the variables that appear in only one equation first (like 'x' and 'w' here) before tackling equations with multiple unknowns (like 'z'), as this simplifies the process. Always check your answers by plugging them back into the original equation.

Question 14. If \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix} \) and \( B = \begin{bmatrix} -3 & -2 \\ 1 & -5 \\ 4 & 3 \end{bmatrix} \), then find a matrix C = \( \begin{bmatrix} p & q \\ r & s \\ t & u \end{bmatrix} \) such that A + B - C = 0.
Answer: We are given an equation A + B - C = 0, and we need to find matrix C. We can rearrange the equation to \( C = A + B \). This means we just need to add matrix A and matrix B together. To add matrices, we add the numbers that are in the exact same position in both matrices. By doing this for each position, we combine corresponding elements. This gives us the elements of matrix C: p, q, r, s, t, and u.
Given A + B - C = 0
\( \implies C = A + B \)
\( C = \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix} + \begin{bmatrix} -3 & -2 \\ 1 & -5 \\ 4 & 3 \end{bmatrix} \)
\( C = \begin{bmatrix} 1+(-3) & 2+(-2) \\ 3+1 & 4+(-5) \\ 5+4 & 6+3 \end{bmatrix} \)
\( C = \begin{bmatrix} 1-3 & 2-2 \\ 3+1 & 4-5 \\ 5+4 & 6+3 \end{bmatrix} \)
\( C = \begin{bmatrix} -2 & 0 \\ 4 & -1 \\ 9 & 9 \end{bmatrix} \)
Therefore, \( p = -2, q = 0, r = 4, s = -1, t = 9, u = 9 \).
In simple words: Since A plus B minus C equals zero, we can say C equals A plus B. So, we add the numbers in the same spots from matrix A and matrix B to find C.

🎯 Exam Tip: When solving matrix equations, always isolate the unknown matrix first (e.g., C = A + B). This simplifies the problem into a direct computation rather than solving element-wise equations for each variable.

Question 15. Let \( A = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 0 & 2 \\ 3 & 4 & 5 \end{bmatrix} \), find a matrix B such that A + B - 4I = 0.
Answer: We are given the equation A + B - 4I = 0, where A is a given matrix and I is the identity matrix of the same size. The identity matrix I has ones on its main diagonal and zeros everywhere else; it acts like the number '1' in scalar algebra for matrix operations. We need to find matrix B. Rearranging the equation, we get \( B = 4I - A \). First, we multiply the identity matrix I by 4. This means all the '1's on the diagonal become '4's. Then, we subtract matrix A from this new matrix (4I) by subtracting corresponding elements. This calculation gives us the final matrix B.
Given A + B - 4I = 0
Where I is the identity matrix of the same order as A (which is 3x3).
So, \( I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
From the equation A + B - 4I = 0, we get:
\( B = 4I - A \)
\( B = 4\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 3 & 5 \\ 1 & 0 & 2 \\ 3 & 4 & 5 \end{bmatrix} \)
\( B = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 3 & 5 \\ 1 & 0 & 2 \\ 3 & 4 & 5 \end{bmatrix} \)
\( B = \begin{bmatrix} 4-2 & 0-3 & 0-5 \\ 0-1 & 4-0 & 0-2 \\ 0-3 & 0-4 & 4-5 \end{bmatrix} \)
\( B = \begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & -2 \\ -3 & -4 & -1 \end{bmatrix} \)
In simple words: We need to find B from the equation A + B - 4I = 0. We change it to B = 4I - A. First, multiply the identity matrix by 4. Then, subtract matrix A from this result, number by number, in the same spots.

🎯 Exam Tip: Always remember that 'I' stands for the identity matrix. Its dimensions must match the matrices in the equation. When a scalar multiplies 'I', only the diagonal elements change value from 1 to the scalar.