GSEB Class 12 Maths Solutions Chapter 3 Matrices Exercise 3.2

Question 1. Let \( A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} \), \( C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} \). Find each of the following:
(i) A + B
(ii) A-B
(iii) 3A-C
(iv) AB
(v) BA
Answer:
(i) \( A + B = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} 2+1 & 4+3 \\ 3+(-2) & 2+5 \end{bmatrix} = \begin{bmatrix} 3 & 7 \\ 1 & 7 \end{bmatrix} \)
(ii) \( A - B = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} 2-1 & 4-3 \\ 3-(-2) & 2-5 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 5 & -3 \end{bmatrix} \)
(iii) \( 3A - C = 3 \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} - \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 3 \times 2 & 3 \times 4 \\ 3 \times 3 & 3 \times 2 \end{bmatrix} - \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix} - \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 6-(-2) & 12-5 \\ 9-3 & 6-4 \end{bmatrix} = \begin{bmatrix} 8 & 7 \\ 6 & 2 \end{bmatrix} \)
(iv) \( AB = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} \)
Let \( AB = [C_{ij}]_{2 \times 2} \)
\( C_{11} \) = (1st row of A) (1st column of B) \( = \begin{bmatrix} 2 & 4 \end{bmatrix} \begin{bmatrix} 1 \\ -2 \end{bmatrix} = 2 \times 1 + 4 \times (-2) = 2 - 8 = -6 \)
\( C_{12} \) = (1st row of A) (2nd column of B) \( = \begin{bmatrix} 2 & 4 \end{bmatrix} \begin{bmatrix} 3 \\ 5 \end{bmatrix} = 2 \times 3 + 4 \times 5 = 6 + 20 = 26 \)
\( C_{21} \) = (2nd row of A) (1st column of B) \( = \begin{bmatrix} 3 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ -2 \end{bmatrix} = 3 \times 1 + 2 \times (-2) = 3 - 4 = -1 \)
\( C_{22} \) = (2nd row of A) (2nd column of B) \( = \begin{bmatrix} 3 & 2 \end{bmatrix} \begin{bmatrix} 3 \\ 5 \end{bmatrix} = 3 \times 3 + 2 \times 5 = 9 + 10 = 19 \)
So, \( AB = [C_{ij}]_{2 \times 2} = \begin{bmatrix} -6 & 26 \\ -1 & 19 \end{bmatrix} \)
(v) \( BA = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} = C' \)
\( C'_{11} = 1 \times 2 + 3 \times 3 = 2+9=11 \)
\( C'_{12} = 1 \times 4 + 3 \times 2 = 4+6=10 \)
\( C'_{21} = (-2) \times 2 + 5 \times 3 = -4+15=11 \)
\( C'_{22} = (-2) \times 4 + 5 \times 2 = -8+10=2 \)
So, \( BA = \begin{bmatrix} 11 & 10 \\ 11 & 2 \end{bmatrix} \)
In simple words: We performed various operations like addition, subtraction, scalar multiplication, and matrix multiplication on the given matrices. Each operation involved combining the elements in specific ways according to matrix rules to get the resulting matrices.

Exam Tip: Remember to perform matrix operations element-wise for addition and subtraction, but row-by-column for multiplication. Pay close attention to signs and order of operations.

Question 2. Compute the following:
(i) \( \begin{bmatrix} a & b \\ -b & a \end{bmatrix} + \begin{bmatrix} a & b \\ b & a \end{bmatrix} \)
(ii) \( \begin{bmatrix} a^{2}+b^{2} & b^{2}+c^{2} \\ a^{2}+c^{2} & a^{2}+b^{2} \end{bmatrix} + \begin{bmatrix} 2 a b & 2 b c \\ -2 a c & -2 a b \end{bmatrix} \)
(iii) \( \begin{bmatrix} -1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \end{bmatrix} + \begin{bmatrix} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \end{bmatrix} \)
(iv) \( \begin{bmatrix} \cos ^{2} x & \sin ^{2} x \\ \sin ^{2} x & \cos ^{2} x \end{bmatrix} + \begin{bmatrix} \sin ^{2} x & \cos ^{2} x \\ \cos ^{2} x & \sin ^{2} x \end{bmatrix} \)
Answer:
(i) \( \begin{bmatrix} a & b \\ -b & a \end{bmatrix} + \begin{bmatrix} a & b \\ b & a \end{bmatrix} = \begin{bmatrix} a+a & b+b \\ -b+b & a+a \end{bmatrix} = \begin{bmatrix} 2a & 2b \\ 0 & 2a \end{bmatrix} \)
(ii) \( \begin{bmatrix} a^{2}+b^{2} & b^{2}+c^{2} \\ a^{2}+c^{2} & a^{2}+b^{2} \end{bmatrix} + \begin{bmatrix} 2 a b & 2 b c \\ -2 a c & -2 a b \end{bmatrix} = \begin{bmatrix} a^{2}+b^{2}+2ab & b^{2}+c^{2}+2bc \\ a^{2}+c^{2}-2ac & a^{2}+b^{2}-2ab \end{bmatrix} = \begin{bmatrix} (a+b)^{2} & (b+c)^{2} \\ (a-c)^{2} & (a-b)^{2} \end{bmatrix} \)
(iii) \( \begin{bmatrix} -1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \end{bmatrix} + \begin{bmatrix} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \end{bmatrix} = \begin{bmatrix} -1+12 & 4+7 & -6+6 \\ 8+8 & 5+0 & 16+5 \\ 2+3 & 8+2 & 5+4 \end{bmatrix} = \begin{bmatrix} 11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9 \end{bmatrix} \)
(iv) \( \begin{bmatrix} \cos ^{2} x & \sin ^{2} x \\ \sin ^{2} x & \cos ^{2} x \end{bmatrix} + \begin{bmatrix} \sin ^{2} x & \cos ^{2} x \\ \cos ^{2} x & \sin ^{2} x \end{bmatrix} = \begin{bmatrix} \cos ^{2} x + \sin ^{2} x & \sin ^{2} x + \cos ^{2} x \\ \sin ^{2} x + \cos ^{2} x & \cos ^{2} x + \sin ^{2} x \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \)
In simple words: For each part, we added the corresponding elements of the matrices. This means we combined the numbers in the same positions from both matrices. For part (ii), we used algebraic identities to simplify the sums. For part (iv), we applied the trigonometric identity \( \sin^2 x + \cos^2 x = 1 \).

Exam Tip: Recall basic algebraic identities like \( (a+b)^2 \) and \( (a-b)^2 \) as well as trigonometric identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) as they are often used to simplify matrix elements.

Question 3. Compute the indicated products:
(i) \( \begin{bmatrix} a & b \\ -b & a \end{bmatrix} \begin{bmatrix} a & -b \\ b & a \end{bmatrix} \)
(ii) \( \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \begin{bmatrix} 2 & 3 & 4 \end{bmatrix} \)
(iii) \( \begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix} \)
(iv) \( \begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{bmatrix} \)
(v) \( \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\ -1 & 2 & 1 \end{bmatrix} \)
(vi) \( \begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{bmatrix} \)
Answer:
(i) \( \begin{bmatrix} a & b \\ -b & a \end{bmatrix} \begin{bmatrix} a & -b \\ b & a \end{bmatrix} = \begin{bmatrix} a \times a + b \times b & a \times (-b) + b \times a \\ (-b) \times a + a \times b & (-b) \times (-b) + a \times a \end{bmatrix} = \begin{bmatrix} a^2+b^2 & -ab+ab \\ -ab+ab & b^2+a^2 \end{bmatrix} = \begin{bmatrix} a^2+b^2 & 0 \\ 0 & a^2+b^2 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \begin{bmatrix} 2 & 3 & 4 \end{bmatrix} = \begin{bmatrix} 1 \times 2 & 1 \times 3 & 1 \times 4 \\ 2 \times 2 & 2 \times 3 & 2 \times 4 \\ 3 \times 2 & 3 \times 3 & 3 \times 4 \end{bmatrix} = \begin{bmatrix} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{bmatrix} \)
(iii) \( \begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix} = \begin{bmatrix} 1 \times 1 + (-2) \times 2 & 1 \times 2 + (-2) \times 3 & 1 \times 3 + (-2) \times 1 \\ 2 \times 1 + 3 \times 2 & 2 \times 2 + 3 \times 3 & 2 \times 3 + 3 \times 1 \end{bmatrix} = \begin{bmatrix} 1-4 & 2-6 & 3-2 \\ 2+6 & 4+9 & 6+3 \end{bmatrix} = \begin{bmatrix} -3 & -4 & 1 \\ 8 & 13 & 9 \end{bmatrix} \)
(iv) \( \begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{bmatrix} \)
\( = \begin{bmatrix} 2 \times 1+3 \times 0+4 \times 3 & 2 \times (-3)+3 \times 2+4 \times 0 & 2 \times 5+3 \times 4+4 \times 5 \\ 3 \times 1+4 \times 0+5 \times 3 & 3 \times (-3)+4 \times 2+5 \times 0 & 3 \times 5+4 \times 4+5 \times 5 \\ 4 \times 1+5 \times 0+6 \times 3 & 4 \times (-3)+5 \times 2+6 \times 0 & 4 \times 5+5 \times 4+6 \times 5 \end{bmatrix} \)
\( = \begin{bmatrix} 2+0+12 & -6+6+0 & 10+12+20 \\ 3+0+15 & -9+8+0 & 15+16+25 \\ 4+0+18 & -12+10+0 & 20+20+30 \end{bmatrix} = \begin{bmatrix} 14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70 \end{bmatrix} \)
(v) \( \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\ -1 & 2 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 2 \times 1+1 \times (-1) & 2 \times 0+1 \times 2 & 2 \times 1+1 \times 1 \\ 3 \times 1+2 \times (-1) & 3 \times 0+2 \times 2 & 3 \times 1+2 \times 1 \\ (-1) \times 1+1 \times (-1) & (-1) \times 0+1 \times 2 & (-1) \times 1+1 \times 1 \end{bmatrix} \)
\( = \begin{bmatrix} 2-1 & 0+2 & 2+1 \\ 3-2 & 0+4 & 3+2 \\ -1-1 & 0+2 & -1+1 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 4 & 5 \\ -2 & 2 & 0 \end{bmatrix} \)
(vi) \( \begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 3 \times 2+(-1) \times 1+3 \times 3 & 3 \times (-3)+(-1) \times 0+3 \times 1 \\ (-1) \times 2+0 \times 1+2 \times 3 & (-1) \times (-3)+0 \times 0+2 \times 1 \end{bmatrix} \)
\( = \begin{bmatrix} 6-1+9 & -9+0+3 \\ -2+0+6 & 3+0+2 \end{bmatrix} = \begin{bmatrix} 14 & -6 \\ 4 & 5 \end{bmatrix} \)
In simple words: We computed the product of the given matrices by multiplying rows of the first matrix with columns of the second matrix. This involves summing the products of corresponding elements from each row-column pair.

Exam Tip: Always check the dimensions of the matrices before multiplying. For multiplication \( A \times B \), the number of columns in A must equal the number of rows in B. The resulting matrix will have dimensions (rows of A) x (columns of B).

Question 4. If \( A = \begin{bmatrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{bmatrix} \), \( B = \begin{bmatrix} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{bmatrix} \) and \( C = \begin{bmatrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{bmatrix} \), then compute (A + B) and (B – C). Also, verify that A + (B – C) = (A + B) – C.
Answer:
First, we calculate \( A + B \):
\( A + B = \begin{bmatrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{bmatrix} + \begin{bmatrix} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 1+3 & 2-1 & -3+2 \\ 5+4 & 0+2 & 2+5 \\ 1+2 & -1+0 & 1+3 \end{bmatrix} = \begin{bmatrix} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{bmatrix} \quad ...(1) \)
Next, we calculate \( B - C \):
\( B - C = \begin{bmatrix} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{bmatrix} - \begin{bmatrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{bmatrix} = \begin{bmatrix} 3-4 & -1-1 & 2-2 \\ 4-0 & 2-3 & 5-2 \\ 2-1 & 0-(-2) & 3-3 \end{bmatrix} = \begin{bmatrix} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{bmatrix} \quad ...(2) \)
Now, we will verify \( A + (B - C) = (A + B) - C \).
First, calculate \( A + (B - C) \) using the value of \( B - C \) from (2):
\( A + (B - C) = \begin{bmatrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{bmatrix} + \begin{bmatrix} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{bmatrix} = \begin{bmatrix} 1-1 & 2-2 & -3+0 \\ 5+4 & 0-1 & 2+3 \\ 1+1 & -1+2 & 1+0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix} \quad ...(3) \)
Next, calculate \( (A + B) - C \) using the value of \( A + B \) from (1):
\( (A + B) - C = \begin{bmatrix} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{bmatrix} - \begin{bmatrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{bmatrix} = \begin{bmatrix} 4-4 & 1-1 & -1-2 \\ 9-0 & 2-3 & 7-2 \\ 3-1 & -1+2 & 4-3 \end{bmatrix} = \begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix} \quad ...(4) \)
From (3) and (4), we observe that \( A + (B - C) = (A + B) - C \). Thus, the identity is proven.
In simple words: We added A and B, and subtracted C from B separately. Then we checked if adding A to (B minus C) gives the same result as taking (A plus B) and subtracting C. Both calculations produced the same final matrix, verifying the given relationship.

Exam Tip: Matrix addition and subtraction are associative, meaning \( A + (B + C) = (A + B) + C \). This property is demonstrated in this problem. Ensure you perform each step accurately to avoid errors in verification.

Question 5. If \( A = \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix} \) and \( B = \begin{bmatrix} \frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix} \), then compute 3A – 5B.
Answer:
To compute \( 3A - 5B \), first we find \( 3A \) and \( 5B \):
\( 3A = 3 \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix} = \begin{bmatrix} 3 \times \frac{2}{3} & 3 \times 1 & 3 \times \frac{5}{3} \\ 3 \times \frac{1}{3} & 3 \times \frac{2}{3} & 3 \times \frac{4}{3} \\ 3 \times \frac{7}{3} & 3 \times 2 & 3 \times \frac{2}{3} \end{bmatrix} = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix} \)
\( 5B = 5 \begin{bmatrix} \frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix} = \begin{bmatrix} 5 \times \frac{2}{5} & 5 \times \frac{3}{5} & 5 \times 1 \\ 5 \times \frac{1}{5} & 5 \times \frac{2}{5} & 5 \times \frac{4}{5} \\ 5 \times \frac{7}{5} & 5 \times \frac{6}{5} & 5 \times \frac{2}{5} \end{bmatrix} = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix} \)
Now, we perform the subtraction:
\( 3A - 5B = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix} - \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix} = \begin{bmatrix} 2-2 & 3-3 & 5-5 \\ 1-1 & 2-2 & 4-4 \\ 7-7 & 6-6 & 2-2 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \)
In simple words: First, we multiplied matrix A by 3 and matrix B by 5, which removed the fractions. Then, we subtracted the resulting matrix \(5B\) from \(3A\). Since both matrices were identical after scalar multiplication, their difference was a zero matrix.

Exam Tip: When performing scalar multiplication on a matrix with fractional elements, the scalar often simplifies the fractions. Always check for common factors to ease calculations.

Question 6. Simplify : \( \cos \theta \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} + \sin \theta \begin{bmatrix} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{bmatrix} \)
Answer:
First, distribute the scalar factors into each matrix:
\( \cos \theta \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} = \begin{bmatrix} \cos^2 \theta & \cos \theta \sin \theta \\ -\cos \theta \sin \theta & \cos^2 \theta \end{bmatrix} \)
\( \sin \theta \begin{bmatrix} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{bmatrix} = \begin{bmatrix} \sin^2 \theta & -\sin \theta \cos \theta \\ \sin \theta \cos \theta & \sin^2 \theta \end{bmatrix} \)
Next, add the two resulting matrices:
\( \begin{bmatrix} \cos^2 \theta & \cos \theta \sin \theta \\ -\cos \theta \sin \theta & \cos^2 \theta \end{bmatrix} + \begin{bmatrix} \sin^2 \theta & -\sin \theta \cos \theta \\ \sin \theta \cos \theta & \sin^2 \theta \end{bmatrix} \)
\( = \begin{bmatrix} \cos^2 \theta + \sin^2 \theta & \cos \theta \sin \theta - \sin \theta \cos \theta \\ -\cos \theta \sin \theta + \sin \theta \cos \theta & \cos^2 \theta + \sin^2 \theta \end{bmatrix} \)
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\( = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
In simple words: We multiplied each matrix by its scalar (cos \(\theta\) or sin \(\theta\)) and then added them together. We used the basic trigonometry rule \(\sin^2 \theta + \cos^2 \theta = 1\) to simplify the elements and arrive at the identity matrix.

Exam Tip: This problem frequently appears to test your understanding of both matrix scalar multiplication/addition and fundamental trigonometric identities. Remember that \( \cos \theta \sin \theta = \sin \theta \cos \theta \).

Question 7. Find X and Y, if
(i) \( X + Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} \) and \( X - Y = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \)
(ii) \( 2X + 3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} \) and \( 3X + 2Y = \begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix} \)
Answer:
(i) Given:
\( X + Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} \quad ...(1) \)
\( X - Y = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \quad ...(2) \)
Adding (1) and (2):
\( (X + Y) + (X - Y) = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \)
\( 2X = \begin{bmatrix} 7+3 & 0+0 \\ 2+0 & 5+3 \end{bmatrix} = \begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix} \)
To find X, multiply by \( \frac{1}{2} \):
\( X = \frac{1}{2} \begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix} = \begin{bmatrix} \frac{10}{2} & \frac{0}{2} \\ \frac{2}{2} & \frac{8}{2} \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix} \)
Subtracting (2) from (1):
\( (X + Y) - (X - Y) = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \)
\( 2Y = \begin{bmatrix} 7-3 & 0-0 \\ 2-0 & 5-3 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 2 & 2 \end{bmatrix} \)
To find Y, multiply by \( \frac{1}{2} \):
\( Y = \frac{1}{2} \begin{bmatrix} 4 & 0 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} \frac{4}{2} & \frac{0}{2} \\ \frac{2}{2} & \frac{2}{2} \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} \)
(ii) Given:
\( 2X + 3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} \quad ...(1) \)
\( 3X + 2Y = \begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix} \quad ...(2) \)
To solve this system, we can multiply (1) by 2 and (2) by 3:
Multiplying (1) by 2: \( 2(2X + 3Y) = 2 \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} \)
\( 4X + 6Y = \begin{bmatrix} 4 & 6 \\ 8 & 0 \end{bmatrix} \quad ...(3) \)
Multiplying (2) by 3: \( 3(3X + 2Y) = 3 \begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix} \)
\( 9X + 6Y = \begin{bmatrix} 6 & -6 \\ -3 & 15 \end{bmatrix} \quad ...(4) \)
Subtracting (3) from (4) to eliminate Y:
\( (9X + 6Y) - (4X + 6Y) = \begin{bmatrix} 6 & -6 \\ -3 & 15 \end{bmatrix} - \begin{bmatrix} 4 & 6 \\ 8 & 0 \end{bmatrix} \)
\( 5X = \begin{bmatrix} 6-4 & -6-6 \\ -3-8 & 15-0 \end{bmatrix} = \begin{bmatrix} 2 & -12 \\ -11 & 15 \end{bmatrix} \)
To find X, multiply by \( \frac{1}{5} \):
\( X = \frac{1}{5} \begin{bmatrix} 2 & -12 \\ -11 & 15 \end{bmatrix} = \begin{bmatrix} \frac{2}{5} & -\frac{12}{5} \\ -\frac{11}{5} & 3 \end{bmatrix} \)
Substitute the value of X into (1) to find Y:
\( 2 \begin{bmatrix} \frac{2}{5} & -\frac{12}{5} \\ -\frac{11}{5} & 3 \end{bmatrix} + 3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} \)
\( \begin{bmatrix} \frac{4}{5} & -\frac{24}{5} \\ -\frac{22}{5} & 6 \end{bmatrix} + 3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} \)
\( 3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} - \begin{bmatrix} \frac{4}{5} & -\frac{24}{5} \\ -\frac{22}{5} & 6 \end{bmatrix} \)
\( 3Y = \begin{bmatrix} 2-\frac{4}{5} & 3-(-\frac{24}{5}) \\ 4-(-\frac{22}{5}) & 0-6 \end{bmatrix} = \begin{bmatrix} \frac{10-4}{5} & \frac{15+24}{5} \\ \frac{20+22}{5} & -6 \end{bmatrix} = \begin{bmatrix} \frac{6}{5} & \frac{39}{5} \\ \frac{42}{5} & -6 \end{bmatrix} \)
To find Y, multiply by \( \frac{1}{3} \):
\( Y = \frac{1}{3} \begin{bmatrix} \frac{6}{5} & \frac{39}{5} \\ \frac{42}{5} & -6 \end{bmatrix} = \begin{bmatrix} \frac{6}{15} & \frac{39}{15} \\ \frac{42}{15} & \frac{-6}{3} \end{bmatrix} = \begin{bmatrix} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2 \end{bmatrix} \)
In simple words: For part (i), we treated the matrix equations like simple algebraic equations. We added and subtracted the equations to find matrices X and Y separately. For part (ii), we used a similar method to solve simultaneous equations. We scaled the equations and then subtracted them to isolate one variable, then substituted it back to find the other.

Exam Tip: When solving simultaneous matrix equations, you can use methods similar to solving linear equations, such as elimination or substitution. Remember to apply scalar multiplication to every element of the matrix.

Question 8. Find X, if \( Y = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} \) and \( 2X + Y = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} \).
Answer:
Given the equations:
\( Y = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} \)
\( 2X + Y = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} \)
Substitute the matrix Y into the second equation:
\( 2X + \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} \)
To find 2X, subtract Y from both sides:
\( 2X = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} \)
\( 2X = \begin{bmatrix} 1-3 & 0-2 \\ -3-1 & 2-4 \end{bmatrix} \)
\( 2X = \begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix} \)
To find X, multiply the matrix by \( \frac{1}{2} \):
\( X = \frac{1}{2} \begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix} \)
\( X = \begin{bmatrix} \frac{-2}{2} & \frac{-2}{2} \\ \frac{-4}{2} & \frac{-2}{2} \end{bmatrix} \)
\( X = \begin{bmatrix} -1 & -1 \\ -2 & -1 \end{bmatrix} \)
In simple words: We substituted the given matrix Y into the equation \(2X + Y\). Then we rearranged the equation to find \(2X\) by subtracting Y. Finally, we divided the elements of \(2X\) by 2 to get the matrix X.

Exam Tip: Treat matrix equations similar to scalar algebraic equations when isolating a variable. Remember to perform addition, subtraction, or scalar multiplication uniformly across all elements of the matrix.

Question 9. Find x and y, if \( 2\left[\begin{array}{ll} 1 & 3 \\ 0 & x \end{array}\right] + \left[\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}\right] = \left[\begin{array}{ll} 5 & 6 \\ 1 & 8 \end{array}\right]. \)
Answer: First, let's look at the Left Hand Side (L.H.S.) of the equation:
\[ L.H.S. = 2\left[\begin{array}{ll} 1 & 3 \\ 0 & x \end{array}\right] + \left[\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}\right] \]
\[ = \left[\begin{array}{ll} 2 \times 1 & 2 \times 3 \\ 2 \times 0 & 2 \times x \end{array}\right] + \left[\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}\right] \]
\[ = \left[\begin{array}{ll} 2 & 6 \\ 0 & 2x \end{array}\right] + \left[\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}\right] \]
\[ = \left[\begin{array}{ll} 2+y & 6+0 \\ 0+1 & 2x+2 \end{array}\right] \]
\[ = \left[\begin{array}{ll} 2+y & 6 \\ 1 & 2x+2 \end{array}\right] \]
Now, the Right Hand Side (R.H.S.) is given as:
\[ R.H.S. = \left[\begin{array}{ll} 5 & 6 \\ 1 & 8 \end{array}\right] \]
By setting the corresponding elements of the matrices equal, we can find x and y.
From the first element in the first row: \( 2 + y = 5 \)
\( \implies y = 5 - 2 \)
\( \implies y = 3 \)
From the second element in the second row: \( 2x + 2 = 8 \)
\( \implies 2x = 8 - 2 \)
\( \implies 2x = 6 \)
\( \implies x = 3 \)
Thus, the values for x and y are 3 and 3 respectively.
In simple words: We first multiply the first matrix by 2, then add the two matrices together. After that, we match the numbers in the new matrix with the numbers in the given result matrix. This helps us solve for x and y.

Exam Tip: Remember that when two matrices are equal, their corresponding elements must be equal. This allows you to set up a system of equations to solve for unknowns.

Question 10. Solve the equation for x, y, z and t, if, \( 2\left[\begin{array}{ll} x & z \\ y & t \end{array}\right] +3\left[\begin{array}{ll} 1 & -1 \\ 0 & 2 \end{array}\right] = 3\left[\begin{array}{ll} 3 & 5 \\ 4 & 6 \end{array}\right]. \)
Answer: Let's begin by writing out the given matrix equation:
\[ 2\left[\begin{array}{ll} x & z \\ y & t \end{array}\right] +3\left[\begin{array}{ll} 1 & -1 \\ 0 & 2 \end{array}\right] = 3\left[\begin{array}{ll} 3 & 5 \\ 4 & 6 \end{array}\right] \]
Now, we perform the scalar multiplications:
\[ \left[\begin{array}{ll} 2x & 2z \\ 2y & 2t \end{array}\right]+\left[\begin{array}{ll} 3 \times 1 & 3 \times (-1) \\ 3 \times 0 & 3 \times 2 \end{array}\right] = \left[\begin{array}{ll} 3 \times 3 & 3 \times 5 \\ 3 \times 4 & 3 \times 6 \end{array}\right] \]
\[ \left[\begin{array}{ll} 2x & 2z \\ 2y & 2t \end{array}\right]+\left[\begin{array}{ll} 3 & -3 \\ 0 & 6 \end{array}\right] = \left[\begin{array}{ll} 9 & 15 \\ 12 & 18 \end{array}\right] \]
Next, we add the matrices on the left side:
\[ \left[\begin{array}{cc} 2x+3 & 2z-3 \\ 2y+0 & 2t+6 \end{array}\right] = \left[\begin{array}{cc} 9 & 15 \\ 12 & 18 \end{array}\right] \]
\[ \left[\begin{array}{cc} 2x+3 & 2z-3 \\ 2y & 2t+6 \end{array}\right] = \left[\begin{array}{cc} 9 & 15 \\ 12 & 18 \end{array}\right] \]
By equating the corresponding elements of the matrices, we get a system of equations:
1. \( 2x + 3 = 9 \)
2. \( 2z - 3 = 15 \)
3. \( 2y = 12 \)
4. \( 2t + 6 = 18 \)
Let's solve each equation:
From (1): \( 2x = 9 - 3 \)
\( \implies 2x = 6 \)
\( \implies x = 3 \)
From (2): \( 2z = 15 + 3 \)
\( \implies 2z = 18 \)
\( \implies z = 9 \)
From (3): \( 2y = 12 \)
\( \implies y = 6 \)
From (4): \( 2t = 18 - 6 \)
\( \implies 2t = 12 \)
\( \implies t = 6 \)
Therefore, the values are x = 3, y = 6, z = 9, and t = 6.
In simple words: First, multiply the numbers outside the matrices with each element inside. Then, add the two matrices on the left side. Finally, match each number in the resulting matrix to the corresponding number in the matrix on the right to find the values of x, y, z, and t.

Exam Tip: Carefully perform scalar multiplication and matrix addition/subtraction. Errors often occur in arithmetic during these steps. Always ensure you equate corresponding elements correctly.

Question 11. If \( x\left[\begin{array}{l} 2 \\ 3 \end{array}\right] + y\left[\begin{array}{l} -1 \\ 1 \end{array}\right] = \left[\begin{array}{l} 10 \\ 5 \end{array}\right], \) then find the values of x and y.
Answer: We are given the matrix equation:
\[ x\left[\begin{array}{l} 2 \\ 3 \end{array}\right] + y\left[\begin{array}{l} -1 \\ 1 \end{array}\right] = \left[\begin{array}{l} 10 \\ 5 \end{array}\right] \]
First, perform the scalar multiplications:
\[ \left[\begin{array}{l} 2x \\ 3x \end{array}\right] + \left[\begin{array}{l} -y \\ y \end{array}\right] = \left[\begin{array}{l} 10 \\ 5 \end{array}\right] \]
Now, add the matrices on the left side:
\[ \left[\begin{array}{l} 2x-y \\ 3x+y \end{array}\right] = \left[\begin{array}{l} 10 \\ 5 \end{array}\right] \]
By equating the corresponding elements, we obtain a system of linear equations:
1. \( 2x - y = 10 \)
2. \( 3x + y = 5 \)
To solve for x and y, we can add equation (1) and equation (2):
\( (2x - y) + (3x + y) = 10 + 5 \)
\( \implies 5x = 15 \)
\( \implies x = \frac{15}{5} \)
\( \implies x = 3 \)
Now, substitute the value of \( x = 3 \) into equation (1):
\( 2(3) - y = 10 \)
\( \implies 6 - y = 10 \)
\( \implies -y = 10 - 6 \)
\( \implies -y = 4 \)
\( \implies y = -4 \)
Therefore, the values are \( x = 3 \) and \( y = -4 \).
In simple words: We multiply the variable outside each matrix with its elements. Then, we add the two resulting matrices together. By matching the parts of this new matrix to the answer matrix, we get two simple equations. We then solve these equations to find x and y.

Exam Tip: When solving simultaneous equations, look for opportunities to eliminate a variable, as done by adding the two equations in this problem. Double-check your arithmetic, especially with negative signs.

Question 12. Given \( 3 \left[\begin{array}{ll} x & y \\ z & w \end{array}\right] = \left[\begin{array}{ll} x & 6 \\ -1 & 2w \end{array}\right] + \left[\begin{array}{ll} 4 & x+y \\ z+w & 3 \end{array}\right], \) find the values of x, y, z and w.
Answer: We are given the matrix equation:
\[ 3 \left[\begin{array}{ll} x & y \\ z & w \end{array}\right] = \left[\begin{array}{ll} x & 6 \\ -1 & 2w \end{array}\right] + \left[\begin{array}{ll} 4 & x+y \\ z+w & 3 \end{array}\right] \]
First, perform the scalar multiplication on the left side:
\[ \left[\begin{array}{ll} 3x & 3y \\ 3z & 3w \end{array}\right] = \left[\begin{array}{ll} x & 6 \\ -1 & 2w \end{array}\right] + \left[\begin{array}{ll} 4 & x+y \\ z+w & 3 \end{array}\right] \]
Next, add the two matrices on the right side:
\[ \left[\begin{array}{ll} 3x & 3y \\ 3z & 3w \end{array}\right] = \left[\begin{array}{cc} x+4 & 6+x+y \\ -1+z+w & 2w+3 \end{array}\right] \]
Now, equate the corresponding elements to form a system of equations:
1. \( 3x = x + 4 \)
2. \( 3y = 6 + x + y \)
3. \( 3z = -1 + z + w \)
4. \( 3w = 2w + 3 \)
Let's solve each equation:
From (1): \( 3x = x + 4 \)
\( \implies 3x - x = 4 \)
\( \implies 2x = 4 \)
\( \implies x = 2 \)
From (4): \( 3w = 2w + 3 \)
\( \implies 3w - 2w = 3 \)
\( \implies w = 3 \)
Now, substitute \( x = 2 \) into equation (2):
\( 3y = 6 + 2 + y \)
\( \implies 3y = 8 + y \)
\( \implies 3y - y = 8 \)
\( \implies 2y = 8 \)
\( \implies y = 4 \)
Finally, substitute \( w = 3 \) into equation (3):
\( 3z = -1 + z + 3 \)
\( \implies 3z - z = 2 \)
\( \implies 2z = 2 \)
\( \implies z = 1 \)
Thus, the values are \( x = 2, y = 4, z = 1, \) and \( w = 3 \).
In simple words: First, multiply the matrix on the left by 3. Then, add the two matrices on the right side. After that, make each number in the left matrix equal to the corresponding number in the right matrix. This gives us four equations, which we solve to find x, y, z, and w.

Exam Tip: Always solve for variables that appear in only one equation first (like x and w here) to simplify the process before moving to equations with multiple unknowns.

Question 13. If \( F(x) = \left[\begin{array}{ccc} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right], \) then show that \( F(x)F(y) = F(x + y). \)
Answer: We are given the matrix function \( F(x) = \left[\begin{array}{ccc} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right]. \)
So, \( F(y) \) would be:
\[ F(y) = \left[\begin{array}{ccc} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{array}\right] \]
Now, let's compute the Left Hand Side (L.H.S.), which is \( F(x)F(y) \):
\[ F(x)F(y) = \left[\begin{array}{ccc} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right] \left[\begin{array}{ccc} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{array}\right] \]
To multiply these matrices, we'll perform dot products of rows from \( F(x) \) with columns from \( F(y) \):
The (1,1) element: \( (\cos x)(\cos y) + (-\sin x)(\sin y) + (0)(0) = \cos x \cos y - \sin x \sin y \)
The (1,2) element: \( (\cos x)(-\sin y) + (-\sin x)(\cos y) + (0)(0) = -\cos x \sin y - \sin x \cos y \)
The (1,3) element: \( (\cos x)(0) + (-\sin x)(0) + (0)(1) = 0 \)
The (2,1) element: \( (\sin x)(\cos y) + (\cos x)(\sin y) + (0)(0) = \sin x \cos y + \cos x \sin y \)
The (2,2) element: \( (\sin x)(-\sin y) + (\cos x)(\cos y) + (0)(0) = -\sin x \sin y + \cos x \cos y \)
The (2,3) element: \( (\sin x)(0) + (\cos x)(0) + (0)(1) = 0 \)
The (3,1) element: \( (0)(\cos y) + (0)(\sin y) + (1)(0) = 0 \)
The (3,2) element: \( (0)(-\sin y) + (0)(\cos y) + (1)(0) = 0 \)
The (3,3) element: \( (0)(0) + (0)(0) + (1)(1) = 1 \)
So, the product matrix is:
\[ F(x)F(y) = \left[\begin{array}{ccc} \cos x \cos y - \sin x \sin y & -\cos x \sin y - \sin x \cos y & 0 \\ \sin x \cos y + \cos x \sin y & \cos x \cos y - \sin x \sin y & 0 \\ 0 & 0 & 1 \end{array}\right] \]
Now, we use the trigonometric addition formulas:
\( \cos(x+y) = \cos x \cos y - \sin x \sin y \)
\( \sin(x+y) = \sin x \cos y + \cos x \sin y \)
Substitute these identities into the product matrix:
\[ F(x)F(y) = \left[\begin{array}{ccc} \cos(x+y) & -(\sin x \cos y + \cos x \sin y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{array}\right] \]
\[ F(x)F(y) = \left[\begin{array}{ccc} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{array}\right] \]
This result is exactly \( F(x+y) \).
Therefore, \( F(x)F(y) = F(x+y) \) is shown.
In simple words: We take the two F matrices, one for x and one for y, and multiply them together. When we do the multiplication, we use special math rules called trigonometric identities to combine the sine and cosine terms. The final answer looks just like an F matrix for (x+y), proving the statement.

Exam Tip: This type of question tests your knowledge of both matrix multiplication and trigonometric identities. Be careful with signs when applying the trigonometric formulas. A common mistake is misremembering the minus sign in \( \cos(A+B) \) or \( \sin(A-B) \).

Question 14. Show that:
(i) \( \left[\begin{array}{cc} 5 & -1 \\ 6 & 7 \end{array}\right] \left[\begin{array}{cc} 2 & 1 \\ 3 & 4 \end{array}\right] \neq \left[\begin{array}{cc} 2 & 1 \\ 3 & 4 \end{array}\right] \left[\begin{array}{cc} 5 & -1 \\ 6 & 7 \end{array}\right] \)
(ii) \( \left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right] \left[\begin{array}{ccc} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right] \neq \left[\begin{array}{ccc} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right] \left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right] \)
Answer:
(i) Let's compute the Left Hand Side (L.H.S.):
\[ L.H.S. = \left[\begin{array}{cc} 5 & -1 \\ 6 & 7 \end{array}\right] \left[\begin{array}{cc} 2 & 1 \\ 3 & 4 \end{array}\right] \]
\[ = \left[\begin{array}{cc} (5)(2)+(-1)(3) & (5)(1)+(-1)(4) \\ (6)(2)+(7)(3) & (6)(1)+(7)(4) \end{array}\right] \]
\[ = \left[\begin{array}{cc} 10-3 & 5-4 \\ 12+21 & 6+28 \end{array}\right] \]
\[ = \left[\begin{array}{cc} 7 & 1 \\ 33 & 34 \end{array}\right] \]
Now, let's compute the Right Hand Side (R.H.S.):
\[ R.H.S. = \left[\begin{array}{cc} 2 & 1 \\ 3 & 4 \end{array}\right] \left[\begin{array}{cc} 5 & -1 \\ 6 & 7 \end{array}\right] \]
\[ = \left[\begin{array}{cc} (2)(5)+(1)(6) & (2)(-1)+(1)(7) \\ (3)(5)+(4)(6) & (3)(-1)+(4)(7) \end{array}\right] \]
\[ = \left[\begin{array}{cc} 10+6 & -2+7 \\ 15+24 & -3+28 \end{array}\right] \]
\[ = \left[\begin{array}{cc} 16 & 5 \\ 39 & 25 \end{array}\right] \]
Since \( \left[\begin{array}{cc} 7 & 1 \\ 33 & 34 \end{array}\right] \neq \left[\begin{array}{cc} 16 & 5 \\ 39 & 25 \end{array}\right] \), we have shown that L.H.S. \( \neq \) R.H.S.
(ii) Let's compute the Left Hand Side (L.H.S.):
\[ L.H.S. = \left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right] \left[\begin{array}{ccc} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right] \]
The product matrix elements are:
(1,1): \( (1)(-1)+(2)(0)+(3)(2) = -1+0+6 = 5 \)
(1,2): \( (1)(1)+(2)(-1)+(3)(3) = 1-2+9 = 8 \)
(1,3): \( (1)(0)+(2)(1)+(3)(4) = 0+2+12 = 14 \)
(2,1): \( (0)(-1)+(1)(0)+(0)(2) = 0+0+0 = 0 \)
(2,2): \( (0)(1)+(1)(-1)+(0)(3) = 0-1+0 = -1 \)
(2,3): \( (0)(0)+(1)(1)+(0)(4) = 0+1+0 = 1 \)
(3,1): \( (1)(-1)+(1)(0)+(0)(2) = -1+0+0 = -1 \)
(3,2): \( (1)(1)+(1)(-1)+(0)(3) = 1-1+0 = 0 \)
(3,3): \( (1)(0)+(1)(1)+(0)(4) = 0+1+0 = 1 \)
So, the L.H.S. is:
\[ L.H.S. = \left[\begin{array}{ccc} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{array}\right] \]
Now, let's compute the Right Hand Side (R.H.S.):
\[ R.H.S. = \left[\begin{array}{ccc} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right] \left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right] \]
The product matrix elements are:
(1,1): \( (-1)(1)+(1)(0)+(0)(1) = -1+0+0 = -1 \)
(1,2): \( (-1)(2)+(1)(1)+(0)(1) = -2+1+0 = -1 \)
(1,3): \( (-1)(3)+(1)(0)+(0)(0) = -3+0+0 = -3 \)
(2,1): \( (0)(1)+(-1)(0)+(1)(1) = 0+0+1 = 1 \)
(2,2): \( (0)(2)+(-1)(1)+(1)(1) = 0-1+1 = 0 \)
(2,3): \( (0)(3)+(-1)(0)+(1)(0) = 0+0+0 = 0 \)
(3,1): \( (2)(1)+(3)(0)+(4)(1) = 2+0+4 = 6 \)
(3,2): \( (2)(2)+(3)(1)+(4)(1) = 4+3+4 = 11 \)
(3,3): \( (2)(3)+(3)(0)+(4)(0) = 6+0+0 = 6 \)
So, the R.H.S. is:
\[ R.H.S. = \left[\begin{array}{ccc} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{array}\right] \]
Since \( \left[\begin{array}{ccc} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{array}\right] \neq \left[\begin{array}{ccc} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{array}\right] \), we have shown that L.H.S. \( \neq \) R.H.S.
In simple words: For both parts, we calculate the product of the two matrices in the given order (Left Hand Side) and then in the reversed order (Right Hand Side). We find that the resulting matrices are different, which confirms that matrix multiplication is not commutative.

Exam Tip: This question demonstrates that matrix multiplication is generally not commutative (i.e., \( AB \neq BA \)). Always perform matrix multiplication carefully, row by column, to avoid arithmetic errors. Double-check each element calculation.

Question 15. Find \( A^2 - 5A + 6I, \) if \( A = \left[\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right] \)
Answer: We are given the matrix \( A = \left[\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]. \)
First, we need to calculate \( A^2 = A \times A \):
\[ A^2 = \left[\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right] \left[\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right] \]
The elements of \( A^2 \) are calculated as follows:
(1,1): \( (2)(2)+(0)(2)+(1)(1) = 4+0+1 = 5 \)
(1,2): \( (2)(0)+(0)(1)+(1)(-1) = 0+0-1 = -1 \)
(1,3): \( (2)(1)+(0)(3)+(1)(0) = 2+0+0 = 2 \)
(2,1): \( (2)(2)+(1)(2)+(3)(1) = 4+2+3 = 9 \)
(2,2): \( (2)(0)+(1)(1)+(3)(-1) = 0+1-3 = -2 \)
(2,3): \( (2)(1)+(1)(3)+(3)(0) = 2+3+0 = 5 \)
(3,1): \( (1)(2)+(-1)(2)+(0)(1) = 2-2+0 = 0 \)
(3,2): \( (1)(0)+(-1)(1)+(0)(-1) = 0-1+0 = -1 \)
(3,3): \( (1)(1)+(-1)(3)+(0)(0) = 1-3+0 = -2 \)
So, \( A^2 = \left[\begin{array}{ccc} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{array}\right]. \)
Next, we calculate \( 5A \):
\[ 5A = 5\left[\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right] = \left[\begin{array}{ccc} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{array}\right] \]
Now, we need the identity matrix \( I \) of order 3, which is \( I = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]. \)
So, \( 6I \):
\[ 6I = 6\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] = \left[\begin{array}{ccc} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}\right] \]
Finally, we compute \( A^2 - 5A + 6I \):
\[ A^2 - 5A + 6I = \left[\begin{array}{ccc} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{array}\right] - \left[\begin{array}{ccc} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{array}\right] + \left[\begin{array}{ccc} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}\right] \]
Perform the subtraction first:
\[ = \left[\begin{array}{ccc} 5-10 & -1-0 & 2-5 \\ 9-10 & -2-5 & 5-15 \\ 0-5 & -1-(-5) & -2-0 \end{array}\right] + \left[\begin{array}{ccc} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}\right] \]
\[ = \left[\begin{array}{ccc} -5 & -1 & -3 \\ -1 & -7 & -10 \\ -5 & 4 & -2 \end{array}\right] + \left[\begin{array}{ccc} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}\right] \]
Now, perform the addition:
\[ = \left[\begin{array}{ccc} -5+6 & -1+0 & -3+0 \\ -1+0 & -7+6 & -10+0 \\ -5+0 & 4+0 & -2+6 \end{array}\right] \]
\[ = \left[\begin{array}{ccc} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{array}\right] \]
In simple words: First, we multiply matrix A by itself to find \( A^2 \). Then, we multiply A by 5 and the identity matrix (I) by 6. Finally, we combine these results by subtracting \( 5A \) from \( A^2 \) and then adding \( 6I \), following the order of operations for matrices.

Exam Tip: When performing multiple matrix operations, break them down into smaller, manageable steps (e.g., calculate \( A^2 \) first, then \( 5A \), then \( 6I \), and finally combine them). Pay close attention to signs and individual element calculations.

Question 16. If \( A = \left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right], \) prove that \( A^3 - 6A^2 + 7A + 2I = 0. \)
Answer: We are given the matrix \( A = \left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]. \)
First, we calculate \( A^2 = A \times A \):
\[ A^2 = \left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right] \left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right] \]
(1,1): \( (1)(1)+(0)(0)+(2)(2) = 1+0+4 = 5 \)
(1,2): \( (1)(0)+(0)(2)+(2)(0) = 0+0+0 = 0 \)
(1,3): \( (1)(2)+(0)(1)+(2)(3) = 2+0+6 = 8 \)
(2,1): \( (0)(1)+(2)(0)+(1)(2) = 0+0+2 = 2 \)
(2,2): \( (0)(0)+(2)(2)+(1)(0) = 0+4+0 = 4 \)
(2,3): \( (0)(2)+(2)(1)+(1)(3) = 0+2+3 = 5 \)
(3,1): \( (2)(1)+(0)(0)+(3)(2) = 2+0+6 = 8 \)
(3,2): \( (2)(0)+(0)(2)+(3)(0) = 0+0+0 = 0 \)
(3,3): \( (2)(2)+(0)(1)+(3)(3) = 4+0+9 = 13 \)
So, \( A^2 = \left[\begin{array}{ccc} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right]. \)
Next, we calculate \( A^3 = A^2 \times A \):
\[ A^3 = \left[\begin{array}{ccc} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right] \left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right] \]
(1,1): \( (5)(1)+(0)(0)+(8)(2) = 5+0+16 = 21 \)
(1,2): \( (5)(0)+(0)(2)+(8)(0) = 0+0+0 = 0 \)
(1,3): \( (5)(2)+(0)(1)+(8)(3) = 10+0+24 = 34 \)
(2,1): \( (2)(1)+(4)(0)+(5)(2) = 2+0+10 = 12 \)
(2,2): \( (2)(0)+(4)(2)+(5)(0) = 0+8+0 = 8 \)
(2,3): \( (2)(2)+(4)(1)+(5)(3) = 4+4+15 = 23 \)
(3,1): \( (8)(1)+(0)(0)+(13)(2) = 8+0+26 = 34 \)
(3,2): \( (8)(0)+(0)(2)+(13)(0) = 0+0+0 = 0 \)
(3,3): \( (8)(2)+(0)(1)+(13)(3) = 16+0+39 = 55 \)
So, \( A^3 = \left[\begin{array}{ccc} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{array}\right]. \)
Now, we need to find \( A^3 - 6A^2 + 7A + 2I \). The identity matrix \( I \) of order 3 is \( \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]. \)
Calculate \( 6A^2 \), \( 7A \), and \( 2I \):
\[ 6A^2 = 6\left[\begin{array}{ccc} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right] = \left[\begin{array}{ccc} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{array}\right] \]
\[ 7A = 7\left[\begin{array}{ccc} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right] = \left[\begin{array}{ccc} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{array}\right] \]
\[ 2I = 2\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] = \left[\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right] \]
Now, combine these matrices:
\[ A^3 - 6A^2 + 7A + 2I = \left[\begin{array}{ccc} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{array}\right] - \left[\begin{array}{ccc} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{array}\right] + \left[\begin{array}{ccc} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{array}\right] + \left[\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right] \]
Add/subtract element by element:
(1,1): \( 21 - 30 + 7 + 2 = -9 + 9 = 0 \)
(1,2): \( 0 - 0 + 0 + 0 = 0 \)
(1,3): \( 34 - 48 + 14 + 0 = -14 + 14 = 0 \)
(2,1): \( 12 - 12 + 0 + 0 = 0 \)
(2,2): \( 8 - 24 + 14 + 2 = -16 + 16 = 0 \)
(2,3): \( 23 - 30 + 7 + 0 = -7 + 7 = 0 \)
(3,1): \( 34 - 48 + 14 + 0 = -14 + 14 = 0 \)
(3,2): \( 0 - 0 + 0 + 0 = 0 \)
(3,3): \( 55 - 78 + 21 + 2 = -23 + 23 = 0 \)
So, the result is the zero matrix:
\[ \left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] = O \]
Thus, \( A^3 - 6A^2 + 7A + 2I = O \) is proven.
In simple words: We first calculate \( A^2 \) by multiplying A by itself, then \( A^3 \) by multiplying \( A^2 \) by A. Next, we multiply \( A^2 \) by 6, A by 7, and the identity matrix (I) by 2. Finally, we add and subtract all these resulting matrices together. When we do this, every element in the final matrix turns out to be zero, showing that the expression equals the zero matrix.

Exam Tip: For problems involving polynomial expressions of matrices, meticulous calculation of each matrix term (\( A^2 \), \( A^3 \), \( cA \), \( kI \)) is crucial. Organize your work clearly and double-check every arithmetic operation to avoid errors.

Question 17. If \( A = \left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right] \) and \( I = \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right], \) then find k such that \( A^2 = kA - 2I. \)
Answer: We are given the matrix \( A = \left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right] \) and the identity matrix \( I = \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]. \)
First, we calculate \( A^2 = A \times A \):
\[ A^2 = \left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right] \left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right] \]
\[ = \left[\begin{array}{ll} (3)(3)+(-2)(4) & (3)(-2)+(-2)(-2) \\ (4)(3)+(-2)(4) & (4)(-2)+(-2)(-2) \end{array}\right] \]
\[ = \left[\begin{array}{ll} 9-8 & -6+4 \\ 12-8 & -8+4 \end{array}\right] \]
\[ = \left[\begin{array}{ll} 1 & -2 \\ 4 & -4 \end{array}\right] \]
Next, we calculate \( kA \):
\[ kA = k\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right] = \left[\begin{array}{ll} 3k & -2k \\ 4k & -2k \end{array}\right] \]
Then, we calculate \( 2I \):
\[ 2I = 2\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] = \left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right] \]
Now, substitute these into the given equation \( A^2 = kA - 2I \):
\[ \left[\begin{array}{ll} 1 & -2 \\ 4 & -4 \end{array}\right] = \left[\begin{array}{ll} 3k & -2k \\ 4k & -2k \end{array}\right] - \left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right] \]
Perform the subtraction on the right side:
\[ \left[\begin{array}{ll} 1 & -2 \\ 4 & -4 \end{array}\right] = \left[\begin{array}{ll} 3k-2 & -2k-0 \\ 4k-0 & -2k-2 \end{array}\right] \]
\[ \left[\begin{array}{ll} 1 & -2 \\ 4 & -4 \end{array}\right] = \left[\begin{array}{ll} 3k-2 & -2k \\ 4k & -2k-2 \end{array}\right] \]
By equating the corresponding elements, we can find the value of k. Let's pick an element and solve:
From the (1,1) element: \( 1 = 3k - 2 \)
\( \implies 3k = 1 + 2 \)
\( \implies 3k = 3 \)
\( \implies k = 1 \)
Let's check this value with other elements to ensure consistency:
From the (1,2) element: \( -2 = -2k \)
\( \implies k = 1 \)
From the (2,1) element: \( 4 = 4k \)
\( \implies k = 1 \)
From the (2,2) element: \( -4 = -2k - 2 \)
\( \implies -2k = -4 + 2 \)
\( \implies -2k = -2 \)
\( \implies k = 1 \)
Since all equations yield \( k = 1 \), the value of k is 1.
In simple words: First, we calculate \( A^2 \) by multiplying matrix A by itself. Then, we write \( kA \) and \( 2I \). We plug these into the given equation \( A^2 = kA - 2I \). By matching the numbers in the matrices on both sides, we can solve for the value of k.

Exam Tip: Always verify your value of 'k' by checking it against all corresponding elements in the matrix equation. If 'k' satisfies one equation but not others, it indicates an error in your calculations or the initial equation might be inconsistent (though rarely in exam questions). Also, remember the identity matrix I for the given order.

Question 18. If \( A = \left[\begin{array}{rr} 0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0 \end{array}\right] \) and I is the identity matrix of order 2, show that \( I + A = (I - A)\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]. \)
Answer: We are given \( A = \left[\begin{array}{rr} 0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0 \end{array}\right] \) and \( I = \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]. \)
Let \( t = \tan \frac{\alpha}{2} \). Then \( A = \left[\begin{array}{cc} 0 & -t \\ t & 0 \end{array}\right]. \)
First, calculate the Left Hand Side (L.H.S.):
\[ L.H.S. = I + A = \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] + \left[\begin{array}{cc} 0 & -t \\ t & 0 \end{array}\right] = \left[\begin{array}{cc} 1 & -t \\ t & 1 \end{array}\right] \]
Now, calculate \( I - A \):
\[ I - A = \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] - \left[\begin{array}{cc} 0 & -t \\ t & 0 \end{array}\right] = \left[\begin{array}{cc} 1-0 & 0-(-t) \\ 0-t & 1-0 \end{array}\right] = \left[\begin{array}{cc} 1 & t \\ -t & 1 \end{array}\right] \]
We also need the trigonometric identities for \( \cos \alpha \) and \( \sin \alpha \) in terms of \( t = \tan \frac{\alpha}{2} \):
\( \cos \alpha = \frac{1 - \tan^2 \frac{\alpha}{2}}{1 + \tan^2 \frac{\alpha}{2}} = \frac{1 - t^2}{1 + t^2} \)
\( \sin \alpha = \frac{2 \tan \frac{\alpha}{2}}{1 + \tan^2 \frac{\alpha}{2}} = \frac{2t}{1 + t^2} \)
Now, calculate the Right Hand Side (R.H.S.):
\[ R.H.S. = (I - A)\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right] \]
\[ = \left[\begin{array}{cc} 1 & t \\ -t & 1 \end{array}\right] \left[\begin{array}{cc} \frac{1 - t^2}{1 + t^2} & -\frac{2t}{1 + t^2} \\ \frac{2t}{1 + t^2} & \frac{1 - t^2}{1 + t^2} \end{array}\right] \]
Let's perform the matrix multiplication:
(1,1) element: \( (1)\left(\frac{1 - t^2}{1 + t^2}\right) + (t)\left(\frac{2t}{1 + t^2}\right) = \frac{1 - t^2 + 2t^2}{1 + t^2} = \frac{1 + t^2}{1 + t^2} = 1 \)
(1,2) element: \( (1)\left(-\frac{2t}{1 + t^2}\right) + (t)\left(\frac{1 - t^2}{1 + t^2}\right) = \frac{-2t + t(1 - t^2)}{1 + t^2} = \frac{-2t + t - t^3}{1 + t^2} = \frac{-t - t^3}{1 + t^2} = \frac{-t(1 + t^2)}{1 + t^2} = -t \)
(2,1) element: \( (-t)\left(\frac{1 - t^2}{1 + t^2}\right) + (1)\left(\frac{2t}{1 + t^2}\right) = \frac{-t(1 - t^2) + 2t}{1 + t^2} = \frac{-t + t^3 + 2t}{1 + t^2} = \frac{t + t^3}{1 + t^2} = \frac{t(1 + t^2)}{1 + t^2} = t \)
(2,2) element: \( (-t)\left(-\frac{2t}{1 + t^2}\right) + (1)\left(\frac{1 - t^2}{1 + t^2}\right) = \frac{2t^2 + 1 - t^2}{1 + t^2} = \frac{1 + t^2}{1 + t^2} = 1 \)
So, the R.H.S. simplifies to:
\[ R.H.S. = \left[\begin{array}{cc} 1 & -t \\ t & 1 \end{array}\right] \]
Since L.H.S. \( = \left[\begin{array}{cc} 1 & -t \\ t & 1 \end{array}\right] \) and R.H.S. \( = \left[\begin{array}{cc} 1 & -t \\ t & 1 \end{array}\right] \), we have shown that L.H.S. = R.H.S.
Hence, the result is proven.
In simple words: First, we find I+A (Left Hand Side) and I-A. Then, we use special rules to write \( \cos \alpha \) and \( \sin \alpha \) using \( \tan \frac{\alpha}{2} \). We plug these into the Right Hand Side equation, multiply the matrices, and simplify. The simplified Right Hand Side matches the Left Hand Side, proving the given statement.

Exam Tip: For problems involving trigonometric functions in matrices, it's often helpful to use a substitution like \( t = \tan \frac{\alpha}{2} \) and apply the half-angle formulas for \( \sin \alpha \) and \( \cos \alpha \). Be very careful with algebraic simplification, especially when dealing with fractions.

Question 19. A trust has Rs. 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year and second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs. 30,000 among the two types of bond, if the trust fund obtains an annual total interest of :
(a) Rs. 1,800
(b) Rs. 2,000.
Answer: Let the total investment be Rs. 30,000. Let x be the amount invested in the first bond (5% interest) and \( y \) be the amount invested in the second bond (7% interest).
So, \( x + y = 30000 \). This means \( y = 30000 - x \).
We can represent the amounts invested as a 1x2 row matrix: \( A = [x \quad (30000-x)] \).
The annual interest rates are 5% (0.05) and 7% (0.07). These can be represented as a 2x1 column matrix for rates:
\[ R = \left[\begin{array}{l} 0.05 \\ 0.07 \end{array}\right] \]
The total interest obtained is given by the matrix product \( AR \).
(a) Total interest is Rs. 1,800.
\[ AR = [1800] \]
\[ [x \quad (30000-x)] \left[\begin{array}{l} 0.05 \\ 0.07 \end{array}\right] = [1800] \]
Perform the matrix multiplication:
\[ [x \times 0.05 + (30000 - x) \times 0.07] = [1800] \]
\( 0.05x + 0.07(30000 - x) = 1800 \)
To remove decimals, multiply the entire equation by 100:
\( 5x + 7(30000 - x) = 180000 \)
\( 5x + 210000 - 7x = 180000 \)
\( -2x + 210000 = 180000 \)
\( -2x = 180000 - 210000 \)
\( -2x = -30000 \)
\( x = 15000 \)
The amount invested in the first bond is Rs. 15,000.
The amount invested in the second bond is \( 30000 - x = 30000 - 15000 = 15000 \).
So, for an annual interest of Rs. 1,800, the trust should invest Rs. 15,000 in each bond.

(b) Total interest is Rs. 2,000.
\[ AR = [2000] \]
\[ [x \quad (30000-x)] \left[\begin{array}{l} 0.05 \\ 0.07 \end{array}\right] = [2000] \]
Perform the matrix multiplication:
\[ [x \times 0.05 + (30000 - x) \times 0.07] = [2000] \]
\( 0.05x + 0.07(30000 - x) = 2000 \)
To remove decimals, multiply the entire equation by 100:
\( 5x + 7(30000 - x) = 200000 \)
\( 5x + 210000 - 7x = 200000 \)
\( -2x + 210000 = 200000 \)
\( -2x = 200000 - 210000 \)
\( -2x = -10000 \)
\( x = 5000 \)
The amount invested in the first bond is Rs. 5,000.
The amount invested in the second bond is \( 30000 - x = 30000 - 5000 = 25000 \).
So, for an annual interest of Rs. 2,000, the trust should invest Rs. 5,000 in the first bond and Rs. 25,000 in the second bond.
In simple words: We set up a matrix for the invested amounts and another for the interest rates. We multiply these matrices to get the total interest. Then, we set this total equal to either Rs. 1,800 or Rs. 2,000 and solve the resulting equation for x (the first investment). Once x is found, we subtract it from Rs. 30,000 to get the second investment.

Exam Tip: Remember that the number of columns in the first matrix must match the number of rows in the second matrix for multiplication. In financial problems, representing investments as a row matrix and rates as a column matrix is common for calculating total returns.

Question 20. The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics ozen economics books. Their selling prices are Rs. 80, Rs. 60, Rs. 40 each respectively. Find the total amount the bookshop will receive from selling all the books, using matrix algebra.
Answer: First, let's convert the number of dozens to individual books:
1 dozen = 12 books
Chemistry books: \( 10 \times 12 = 120 \)
Physics books: \( 8 \times 12 = 96 \)
Economics books: \( 10 \times 12 = 120 \)
We can represent the number of books as a 1x3 row matrix (matrix A):
\[ A = [120 \quad 96 \quad 120] \]
The selling prices per book are Rs. 80 for chemistry, Rs. 60 for physics, and Rs. 40 for economics. These prices can be represented as a 3x1 column matrix (matrix R):
\[ R = \left[\begin{array}{l} 80 \\ 60 \\ 40 \end{array}\right] \]
To find the total amount the bookshop will receive, we perform the matrix multiplication \( AR \):
\[ AR = [120 \quad 96 \quad 120] \left[\begin{array}{l} 80 \\ 60 \\ 40 \end{array}\right] \]
\[ AR = [(120)(80) + (96)(60) + (120)(40)] \]
\[ AR = [9600 + 5760 + 4800] \]
\[ AR = [20160] \]
The total amount received by the bookshop is Rs. 20,160.
In simple words: We first change the dozens of books into actual numbers. Then, we make a matrix for the number of books and another matrix for their prices. By multiplying these two matrices together, we can figure out the total money the bookshop will get.

Exam Tip: In word problems involving matrix multiplication, correctly setting up the matrices (row vs. column) and ensuring their dimensions are compatible for multiplication is crucial. Always convert units (like dozens) before forming the matrices.

Question 21. The restriction on n, k and p so that PY = WY will be defined are
(A) k = 3, p = n
(B) k is arbitrary, p = 2
(C) p is arbitrary, k = 3
(D) k = 2, p = 3
Answer: (A) k = 3, p = n
Let's first understand the orders of the given matrices:
Matrix P has order \( p \times k \).
Matrix Y has order \( 3 \times k \).
Matrix W has order \( n \times 3 \).
For the product PY to be defined, the number of columns in P must be equal to the number of rows in Y.
Columns in P = k, Rows in Y = 3.
So, \( k = 3 \).
The resulting order of PY will be \( p \times 3 \).
For the product WY to be defined, the number of columns in W must be equal to the number of rows in Y.
Columns in W = 3, Rows in Y = 3.
So, WY is always defined and its order will be \( n \times k \). Since we found \( k=3 \), the order is \( n \times 3 \).
For PY = WY to be defined, both matrices must have the same order.
Order of PY is \( p \times 3 \).
Order of WY is \( n \times 3 \).
Therefore, for their orders to be the same, we must have \( p = n \).
Combining these conditions, we get \( k = 3 \) and \( p = n \).
In simple words: For two matrices to be multiplied, the columns of the first must match the rows of the second. For PY, this means k must be 3. For WY, the columns of W already match the rows of Y. For PY and WY to be equal, they must also have the same final shape. This means p must equal n.

Exam Tip: Always analyze matrix multiplication compatibility (inner dimensions must match) and the resulting matrix dimensions (outer dimensions). For matrix equality, both dimensions (rows and columns) must be identical.

Question 22. If n = p, then the order of the matrix 7X – 5Z is
(A) p x 2
(B) 2 x n
(C) n x 3
(D) p x n
Answer: (B) 2 x n
Let's recall the orders of the matrices X and Z:
Matrix X has order \( 2 \times n \).
Matrix Z has order \( 2 \times p \).
For the operation \( 7X - 5Z \) to be defined, matrices X and Z must have the same order. This means their number of rows must be equal, and their number of columns must be equal.
We are given that \( n = p \).
Since the number of rows of X is 2 and the number of rows of Z is 2, the row counts already match.
Since \( n = p \), the number of columns of X (which is n) is equal to the number of columns of Z (which is p). So, the column counts also match.
Because X and Z have the same order, the operation \( 7X - 5Z \) is defined.
When you multiply a matrix by a scalar (like 7X or 5Z), the order of the matrix does not change. So, \( 7X \) has order \( 2 \times n \), and \( 5Z \) has order \( 2 \times p \).
Since \( n = p \), both \( 7X \) and \( 5Z \) have order \( 2 \times n \) (or \( 2 \times p \)).
When you subtract two matrices of the same order, the resulting matrix also has that same order.
Therefore, the order of \( 7X - 5Z \) will be \( 2 \times n \).
In simple words: For us to subtract two matrices, they must be the same size. We're given that X is a \( 2 \times n \) matrix and Z is a \( 2 \times p \) matrix. Since n equals p, both X and Z are effectively \( 2 \times n \) matrices. Multiplying them by numbers (like 7 or 5) doesn't change their size. So, subtracting them results in another matrix that is also \( 2 \times n \).

Exam Tip: A fundamental rule of matrix addition and subtraction is that matrices must be of the same order. Scalar multiplication does not alter the order of a matrix. Always ensure these conditions are met before proceeding with calculations.