Question and Answer (Board Exam)

1. (CBSE 2019) If a matrix has 18 elements, what are the possible orders it can have?

Explanation:

1. The order of a matrix is (number of rows, $m$) $\times$ (number of columns, $n$).

2. The total number of elements is the product $m \times n$.

3. We need to find all pairs of natural numbers ($m, n$) whose product is 18.

4. The factor pairs of 18 are: (1, 18), (2, 9), (3, 6).

5. We must also include the 'reverse' of these pairs, as the number of rows and columns can be swapped: (18, 1), (9, 2), (6, 3).

• Answer: The possible orders are 1x18, 18x1, 2x9, 9x2, 3x6, and 6x3.

2. (CBSE 2020) Construct a 2x2 matrix $A = [a_{ij}]$ whose elements are given by $a_{ij} = (i+2j)^2/2$.

Explanation:

1. A 2x2 matrix has the form $\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$.

2. We calculate each element using the formula $a_{ij} = (i+2j)^2/2$:

   • $a_{11}$ (i=1, j=1): $(1+2(1))^2/2 = (3)^2/2 = 9/2$.
   • $a_{12}$ (i=1, j=2): $(1+2(2))^2/2 = (5)^2/2 = 25/2$.
   • $a_{21}$ (i=2, j=1): $(2+2(1))^2/2 = (4)^2/2 = 16/2 = 8$.
   • $a_{22}$ (i=2, j=2): $(2+2(2))^2/2 = (6)^2/2 = 36/2 = 18$.

3. Construct the matrix with these values.

• Answer: The matrix is $$\begin{bmatrix} 9/2 & 25/2 \\ 8 & 18 \end{bmatrix}$$.

3. (CBSE 2018) Find the values of x and y if $$\begin{bmatrix} 2x+y \\ x-y \end{bmatrix} = \begin{bmatrix} 10 \\ 4 \end{bmatrix}$$

Explanation:

1. Use Equality of Matrices: For two matrices to be equal, their corresponding elements must be equal.

2. Set up Equations: This gives us a system of two linear equations:

   • (Row 1): $2x+y=10 \quad \cdots(1)$
   • (Row 2): $x-y=4 \quad \cdots(2)$

3. Solve the System: We can use the elimination method. Add equation (1) to equation (2): $$(2x+y) + (x-y) = 10 + 4$$ $$3x = 14 \implies x=14/3$$

4. Find y: Substitute the value of $x$ back into equation (2): $$y = x-4 = 14/3 - 4$$ $$= 14/3 - 12/3 = 2/3$$

• Answer: The solution is x=14/3, y=2/3.

4. (CBSE 2017) If $2\begin{bmatrix} 3 & 4 \\ 5 & x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix}$, find the values of x and y.

Explanation:

1. Perform Scalar Multiplication: $$2\begin{bmatrix} 3 & 4 \\ 5 & x \end{bmatrix} = \begin{bmatrix} 6 & 8 \\ 10 & 2x \end{bmatrix}$$
2. Perform Matrix Addition: $$\begin{bmatrix} 6 & 8 \\ 10 & 2x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix}$$ $$= \begin{bmatrix} 6+1 & 8+y \\ 10+0 & 2x+1 \end{bmatrix} = \begin{bmatrix} 7 & 8+y \\ 10 & 2x+1 \end{bmatrix}$$
3. Equate Matrices: $$\begin{bmatrix} 7 & 8+y \\ 10 & 2x+1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix}$$
4. Solve for x and y:
   • From element (1,2): $8+y=0 \implies y=-8$.
   • From element (2,2): $2x+1=5 \implies 2x=4 \implies x=2$.

• Answer: x=2, y=-8.

5. (CBSE 2019) Compute the product: $\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \begin{bmatrix} 2 & 3 & 4 \end{bmatrix}$.

Explanation:

1. Check Order: The first matrix is (3 x 1). The second matrix is (1 x 3). The inner dimensions (1 and 1) match, so the product is defined. The resulting matrix will have the outer dimensions: (3 x 3).

2. Calculate Each Element: We find the element $c_{ij}$ by multiplying the i-th row of the first matrix (a single number) by the j-th column of the second matrix (a single number).

   • $c_{11} = (1)(2) = 2$
   • $c_{12} = (1)(3) = 3$
   • $c_{13} = (1)(4) = 4$
   • $c_{21} = (2)(2) = 4$
   • $c_{22} = (2)(3) = 6$
   • $c_{23} = (2)(4) = 8$
   • $c_{31} = (3)(2) = 6$
   • $c_{32} = (3)(3) = 9$
   • $c_{33} = (3)(4) = 12$

• Answer: $$\begin{bmatrix} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{bmatrix}$$

6. (CBSE 2020) If $A = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix}$, find AB.

• Explanation:

1. Check Order: A is 2x2, B is 2x2. The result AB will be 2x2.
2. Calculate Element (1,1): (Row 1 of A) $\times$ (Col 1 of B) = $$(2)(1) + (1)(-1) = 2 - 1 = 1$$
3. Calculate Element (1,2): (Row 1 of A) $\times$ (Col 2 of B) = $$(2)(0) + (1)(2) = 0 + 2 = 2$$
4. Calculate Element (2,1): (Row 2 of A) $\times$ (Col 1 of B) = $$(3)(1) + (2)(-1) = 3 - 2 = 1$$
5. Calculate Element (2,2): (Row 2 of A) $\times$ (Col 2 of B) = $$(3)(0) + (2)(2) = 0 + 4 = 4$$

• Answer: $$AB = \begin{bmatrix} 1 & 2 \\ 1 & 4 \end{bmatrix}$$

7. (CBSE 2015) If $A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, find $A^2$.

Explanation:

1. Set up: $$A^2 = A \cdot A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$
2. Calculate Element (1,1): (Row 1) $\times$ (Col 1) = $$(0)(0) + (1)(1) = 1$$
3. Calculate Element (1,2): (Row 1) $\times$ (Col 2) = $$(0)(1) + (1)(0) = 0$$
4. Calculate Element (2,1): (Row 2) $\times$ (Col 1) = $$(1)(0) + (0)(1) = 0$$
5. Calculate Element (2,2): (Row 2) $\times$ (Col 2) = $$(1)(1) + (0)(0) = 1$$

• Answer: $$A^2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$$

8. (CBSE 2018) Find the number of all possible matrices of order 3 x 3 with each entry 0 or 1.

Explanation:

1. Count Elements: A 3x3 matrix has $3 \times 3 = 9$ total positions.

2. Count Choices: Each of these 9 positions can be filled in one of two ways (either 0 or 1).

3. Apply Multiplication Principle: The total number of distinct matrices is the product of the number of choices for each position. $$Total = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^9$$

• Answer: $2^9 = 512$.

9. (CBSE 2021) If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, find $A+A^T$.

Explanation:

1. Find the Transpose ($A^T$): Interchange rows and columns of A. $$A^T = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$$
2. Add A and $A^T$: $$A+A^T = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$$ $$= \begin{bmatrix} 1+1 & 2+3 \\ 3+2 & 4+4 \end{bmatrix}$$
3. Calculate Result: $$A+A^T = \begin{bmatrix} 2 & 5 \\ 5 & 8 \end{bmatrix}$$ .

• Answer: $\begin{bmatrix} 2 & 5 \\ 5 & 8 \end{bmatrix}$.

10. (CBSE 2016) If $f(x)=x^2-5x+6$, find $f(A)$ if $A=\begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}$.

Explanation: To find $f(A)$, we replace $x$ with $A$ and the constant 6 with $6I$, where $I$ is the 2x2 identity matrix. $$f(A)=A^2-5A+6I$$

1. Calculate $A^2$: $$A^2 = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}$$ $$= \begin{bmatrix} 4+0 & 0+0 \\ 2+1 & 0+1 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 3 & 1 \end{bmatrix}$$
2. Calculate $5A$: $$5A = 5\begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 10 & 0 \\ 5 & 5 \end{bmatrix}$$
3. Calculate $6I$: $$6I = 6\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$$
4. Combine: $$f(A) = \begin{bmatrix} 4 & 0 \\ 3 & 1 \end{bmatrix} - \begin{bmatrix} 10 & 0 \\ 5 & 5 \end{bmatrix} + \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$$ $$= \begin{bmatrix} 4-10+6 & 0-0+0 \\ 3-5+0 & 1-5+6 \end{bmatrix}$$ $$= \begin{bmatrix} 0 & 0 \\ -2 & 2 \end{bmatrix}$$

• Answer: $f(A) = \begin{bmatrix} 0 & 0 \\ -2 & 2 \end{bmatrix}$.

11. (CBSE 2019) If $A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$, and $A+A^T=I$, find the value of $\alpha$.

Explanation:

1. Find the Transpose ($A^T$): $$A^T = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$$
2. Calculate $A+A^T$: $$A+A^T = \begin{bmatrix} \cos\alpha+\cos\alpha & -\sin\alpha+\sin\alpha \\ \sin\alpha-\sin\alpha & \cos\alpha+\cos\alpha \end{bmatrix}$$ $$= \begin{bmatrix} 2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{bmatrix}$$
3. Set up the Equation: $$A+A^T=I$$ $$\implies \begin{bmatrix} 2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
4. Solve for $\alpha$: By equating the diagonal elements, we get $2\cos\alpha=1$, or $\cos\alpha=1/2$. A valid principal value for $\alpha$ is $\pi/3$.

• Answer: $\alpha = \pi/3$ (or $2n\pi \pm \pi/3, n \in \mathbb{Z}$).

12. (CBSE 2020) For what value of x is the matrix $$A = \begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ x & -3 & 0 \end{bmatrix}$$ a skew-symmetric matrix?

Explanation:

1. Definition: A matrix is skew-symmetric if $A^T = -A$, which implies $a_{ij} = -a_{ji}$ for all $i,j$. Also, all diagonal elements $a_{ii}$ must be 0, which is already true for matrix A.

2. Apply Condition: Let's check the $a_{31}$ element.

   • $a_{31} = x$
   • $a_{13} = -2$
   • The condition is $a_{31} = -a_{13}$.

3. Solve: $x = -(-2) \implies x = 2$.

   (We can check other elements: $a_{21} = -1$ and $a_{12} = 1$. $a_{21} = -a_{12}$ is true. $a_{32} = -3$ and $a_{23} = 3$. $a_{32} = -a_{23}$ is true.)

• Answer: x=2.

13. (CBSE 2018) If $A = \begin{bmatrix} 3 & x-1 \\ 2x+3 & x+2 \end{bmatrix}$ is a symmetric matrix, find the value of x.

Explanation:

1. Definition: A matrix A is symmetric if $A^T = A$, which implies $a_{ij} = a_{ji}$ for all $i,j$. The diagonal elements do not matter for this check.

2. Apply Condition: We must have $a_{12} = a_{21}$.

   • $a_{12} = x-1$
   • $a_{21} = 2x+3$

3. Set up and Solve: $$x-1 = 2x+3$$ $$\implies -1-3 = 2x-x$$ $$\implies -4 = x$$

• Answer: x = -4.

14. (CBSE 2017) If A is a square matrix, show that $A-A^T$ is a skew-symmetric matrix.

Explanation:

1. Define: Let $P = A-A^T$.

2. Goal: To prove $P$ is skew-symmetric, we must show that $P^T = -P$.

3. Take the Transpose: $P^T = (A-A^T)^T$.

4. Apply Transpose Properties: Using $(X-Y)^T = X^T-Y^T$ and $(X^T)^T=X$: $$P^T = (A)^T - (A^T)^T = A^T - A$$

5. Factor and Conclude: Factor out -1: $$P^T = -(A - A^T)$$ Since $P = A-A^T$, we have $P^T = -P$.

• Answer: Since its transpose is equal to its negative, $A-A^T$ is skew-symmetric.

15. (CBSE 2021) Express $A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$ as the sum of a symmetric and a skew-symmetric matrix.

Explanation: We use the theorem $A = P+Q$, where $P = \frac{1}{2}(A+A^T)$ (symmetric part) and $Q = \frac{1}{2}(A-A^T)$ (skew-symmetric part).

1. Find $A^T$: $$A^T = \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}$$
2. Find P: $$A+A^T = \begin{bmatrix} 3+3 & 5+1 \\ 1+5 & -1-1 \end{bmatrix} = \begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix}$$ $$P = \frac{1}{2}(A+A^T) = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}$$
3. Find Q: $$A-A^T = \begin{bmatrix} 3-3 & 5-1 \\ 1-5 & -1-(-1) \end{bmatrix} = \begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix}$$ $$Q = \frac{1}{2}(A-A^T) = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$$

• Answer: $$A = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$$

16. (CBSE 2019) If a matrix is both symmetric and skew-symmetric, show that it is a zero matrix.

• Explanation:

1. Given: Let A be the matrix.

   • Since A is symmetric, by definition $A^T = A$.
   • Since A is skew-symmetric, by definition $A^T = -A$.

2. Combine Conditions: For both to be true, we must have $A = -A$.

3. Solve for A: $$A + A = -A + A \implies 2A = O$$ (where O is the zero matrix).

   This implies $A = O$.

• Answer: A must be the zero matrix.

17. (CBSE 2022) For the matrix $A = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}$, verify that $A+A^T$ is a symmetric matrix.

Explanation:

1. Find $A^T$: $$A^T = \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}$$
2. Calculate $P = A+A^T$: $$P = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix} + \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}$$ $$= \begin{bmatrix} 1+1 & 5+6 \\ 6+5 & 7+7 \end{bmatrix}$$ $$= \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}$$
3. Check if P is Symmetric: We find the transpose of P. $$P^T = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}$$
4. Conclusion: Since $P = P^T$, the matrix $A+A^T$ is symmetric.

• Answer: Verified.

18. (CBSE 2020) If A and B are symmetric matrices of the same order, prove that AB is symmetric if and only if AB = BA.

Explanation: Forward Proof (if AB=BA, then AB is symmetric):

1. Given: $A^T=A$, $B^T=B$, and $AB=BA$.

2. We must prove that $(AB)^T = AB$.

3. $(AB)^T = B^TA^T$ (Reversal Law)

4. $(AB)^T = BA$ (Substituting A=A^T, B=B^T)

5. $(AB)^T = AB$ (Substituting the given AB=BA)

6. Thus, AB is symmetric.

   Backward Proof (if AB is symmetric, then AB=BA):

7. Given: $A^T=A$, $B^T=B$, and $(AB)^T=AB$.

8. We must prove $AB=BA$.

9. From the reversal law, $(AB)^T = B^TA^T$.

10. Substituting $A=A^T, B=B^T$, we get $(AB)^T = BA$.

11. From the given, we have $(AB)^T=AB$.

12. Therefore, by transitivity, $AB = BA$.

• Answer: Proven.

19. (CBSE 2019) Find the inverse of the matrix $A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$ using the shortcut formula.

Explanation:

1. Recall the 2x2 Inverse Formula: For $$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ $$A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$
2. Calculate the Determinant: $$\det(A) = ad-bc = (2)(2) - (3)(1) = 4 - 3 = 1$$
3. Apply the Formula: $$A^{-1} = \frac{1}{1}\begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$$

• Answer: $A^{-1} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$.

20. (CBSE 2020) Find the inverse of $A = \begin{bmatrix} 1 & 3 \\ 2 & 7 \end{bmatrix}$ using elementary row operations.

Explanation:

1. Set up: Start with the augmented matrix $[A|I]$. Goal is to get $[I|A^{-1}]$. $$[\begin{array}{cc|cc} 1 & 3 & 1 & 0 \\ 2 & 7 & 0 & 1 \end{array}]$$
2. Step 1: Create 0 in (2,1). Apply $R_2 \to R_2 - 2R_1$. $$R_2 = [2-2(1) \quad 7-2(3) \quad 0-2(1) \quad 1-2(0)] = [0 \quad 1 \quad -2 \quad 1]$$ Matrix becomes: $$[\begin{array}{cc|cc} 1 & 3 & 1 & 0 \\ 0 & 1 & -2 & 1 \end{array}]$$
3. Step 2: Create 0 in (1,2). Apply $R_1 \to R_1 - 3R_2$. $$R_1 = [1-3(0) \quad 3-3(1) \quad 1-3(-2) \quad 0-3(1)] = [1 \quad 0 \quad 7 \quad -3]$$ Matrix becomes: $$[\begin{array}{cc|cc} 1 & 0 & 7 & -3 \\ 0 & 1 & -2 & 1 \end{array}]$$
4. Conclusion: The LHS is now $I$, so the RHS is $A^{-1}$.

• Answer: $A^{-1} = \begin{bmatrix} 7 & -3 \\ -2 & 1 \end{bmatrix}$.

21. (CBSE 2018) A matrix $A=\begin{bmatrix} 6 & x \\ 2 & 1 \end{bmatrix}$ is singular (has no inverse). Find the value of x.

Explanation: A matrix is singular if its determinant is zero.

1. Calculate the Determinant: $$\det(A) = ad-bc = (6)(1) - (x)(2) = 6-2x$$

2. Set Determinant to Zero: $$6-2x=0 \implies 6=2x \implies x=3$$

• Answer: x=3.

22. (CBSE 2017) If $A = \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix}$, find x and y such that $$A^2+xI=yA$$

Explanation:

1. Calculate $A^2$: $$A^2 = \begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix}$$ $$= \begin{bmatrix} 9+7 & 3+5 \\ 21+35 & 7+25 \end{bmatrix}$$ $$= \begin{bmatrix} 16 & 8 \\ 56 & 32 \end{bmatrix}$$

2. Write out the equation: $$LHS: A^2+xI = \begin{bmatrix} 16 & 8 \\ 56 & 32 \end{bmatrix} + \begin{bmatrix} x & 0 \\ 0 & x \end{bmatrix}$$ $$= \begin{bmatrix} 16+x & 8 \\ 56 & 32+x \end{bmatrix}$$ $$RHS: yA = y\begin{bmatrix} 3 & 1 \\ 7 & 5 \end{bmatrix} = \begin{bmatrix} 3y & y \\ 7y & 5y \end{bmatrix}$$

3. Equate Matrices and Solve: $$\begin{bmatrix} 16+x & 8 \\ 56 & 32+x \end{bmatrix} = \begin{bmatrix} 3y & y \\ 7y & 5y \end{bmatrix}$$

   • $a_{12}$: $8 = y \implies y=8$.
   • $a_{21}$: $56 = 7y \implies y=8$.
   • $a_{11}$: $16+x = 3y \implies 16+x = 3(8) \implies 16+x=24 \implies x=8$.
   • $a_{22}$: $32+x = 5y \implies 32+x = 5(8) \implies 32+x=40 \implies x=8$.

   All equations are consistent.

• Answer: x=8, y=8.

23. (CBSE 2021) If $A^2 - A + I = O$, prove that the inverse of A is $I-A$.

Explanation:

1. Goal: To find $A^{-1}$, we must rearrange the equation into the form $A \cdot B = I$ or $B \cdot A = I$, where $B$ will be the inverse.

2. Rearrange the Equation: Start with $A^2-A+I=O$. Isolate the identity matrix $I$.

   $I = A - A^2$.

3. Factor out A: Factor $A$ out of the right-hand side. We can pre-factor or post-factor.

   • Post-factor: $I = A(I - A)$.
   • Pre-factor: $I = (I - A)A$.

4. Conclusion: Both factorizations match the definition of an inverse $AB=BA=I$, where $B = I-A$.

• Answer: $A^{-1} = I-A$.

24. (CBSE 2022) For what value of x, is the matrix $$A = \begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ x & -3 & 0 \end{bmatrix}$$ skew-symmetric?

Explanation:

1. Definition: A matrix is skew-symmetric if $A^T = -A$, which implies $a_{ij} = -a_{ji}$ for all $i,j$. Also, all diagonal elements $a_{ii}$ must be 0, which is already true for matrix A.

2. Apply Condition: Let's check the $a_{31}$ element.

   • $a_{31} = x$
   • $a_{13} = -2$
   • The condition is $a_{31} = -a_{13}$.

3. Solve: $x = -(-2) \implies x = 2$.

   (We can check other elements: $a_{21} = -1$ and $a_{12} = 1$. $a_{21} = -a_{12}$ is true. $a_{32} = -3$ and $a_{23} = 3$. $a_{32} = -a_{23}$ is true.)

• Answer: x=2.

25. (CBSE Sample Paper) If A is an invertible matrix of order 3 and $|A|=5$, then find $|adj A|$.

Explanation: This question uses a property of determinants and adjoints (covered in the next chapter).

1. Recall Property: For an invertible matrix A of order $n$, the determinant of its adjoint is given by the formula $$|adj A| = |A|^{n-1}$$

2. Substitute Values: We are given $n=3$ and $|A|=5$.

3. Calculate: $$|adj A| = 5^{3-1} = 5^2=25$$

• Answer: 25.

26. (CBSE 2019) If $A = \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix}$, find $A^{-1}$.

Explanation:

1. Recall the 2x2 Inverse Formula: $$A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$
2. Calculate the Determinant: $$\det(A) = ad-bc$$ $$= (2)(7) - (-3)(-4) = 14 - 12 = 2$$
3. Apply the Formula: $$A^{-1} = \frac{1}{2}\begin{bmatrix} 7 & -(-3) \\ -(-4) & 2 \end{bmatrix}$$ $$= \frac{1}{2}\begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix}$$
4. Scalar Multiplication: $$A^{-1} = \begin{bmatrix} 7/2 & 3/2 \\ 2 & 1 \end{bmatrix}$$

• Answer: $\begin{bmatrix} 7/2 & 3/2 \\ 2 & 1 \end{bmatrix}$.

27. (CBSE 2018) If $A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$, show that $$A^2-4A-5I=O$$

Explanation: This is a verification problem. We need to compute $A^2$, $4A$, and $5I$ and show they sum to the zero matrix $O$.

1. Calculate $A^2$: $$A^2 = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$$ $$= \begin{bmatrix} 1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 4+4+1 \end{bmatrix}$$ $$= \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix}$$
2. Calculate $4A$ and $5I$: $$4A = \begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix}$$ $$5I = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$$
3. Compute the Expression: $$A^2-4A-5I = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} - \begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix} - \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$$ $$= \begin{bmatrix} 9-4-5 & 8-8-0 & 8-8-0 \\ 8-8-0 & 9-4-5 & 8-8-0 \\ 8-8-0 & 8-8-0 & 9-4-5 \end{bmatrix}$$ $$= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O$$

• Answer: Proven.

28. (CBSE 2020) Given $A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}, B = \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix}$, verify $$(AB)^{-1}=B^{-1}A^{-1}$$

Explanation:

1. Calculate LHS: $(AB)^{-1}$ $$AB = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix}$$ $$= \begin{bmatrix} 2-3 & -4+9 \\ 1-2 & -2+6 \end{bmatrix}$$ $$= \begin{bmatrix} -1 & 5 \\ -1 & 4 \end{bmatrix}$$ $$\det(AB) = (-1)(4) - (5)(-1) = -4+5 = 1$$ $$(AB)^{-1} = \frac{1}{1}\begin{bmatrix} 4 & -5 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 4 & -5 \\ 1 & -1 \end{bmatrix}$$

2. Calculate RHS: $B^{-1}A^{-1}$ $$A^{-1} = \frac{1}{4-3}\begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$$ $$B^{-1} = \frac{1}{3-2}\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$$ $$B^{-1}A^{-1} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$$ $$= \begin{bmatrix} 6-2 & -9+4 \\ 2-1 & -3+2 \end{bmatrix}$$ $$= \begin{bmatrix} 4 & -5 \\ 1 & -1 \end{bmatrix}$$

3. Conclusion: Since LHS = RHS, the property is verified.

• Answer: Verified.

29. (CBSE 2022) Let $A=\begin{bmatrix} 1 & -1 \\ 2 & 1 \end{bmatrix}, B=\begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix}$ and $(A+B)^2=A^2+B^2$. Find a and b.

Explanation

1. Analyze the Given Equation: We are given the equation $$(A+B)^2 = A^2+B^2$$ Let's expand the left-hand side (LHS): $$(A+B)^2 = (A+B)(A+B) = A(A+B) + B(A+B) = A^2 + AB + BA + B^2$$

2. Simplify the Condition: Substituting the expansion into the original equation: $$A^2 + AB + BA + B^2 = A^2 + B^2$$ Subtracting $A^2$ and $B^2$ from both sides, we get the simplified condition: $AB + BA = O$ (where O is the 2x2 zero matrix).

3. Calculate the product AB: $$AB = \begin{bmatrix} 1 & -1 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix}$$ $$= \begin{bmatrix} (1)(a)+(-1)(b) & (1)(1)+(-1)(-1) \\ (2)(a)+(1)(b) & (2)(1)+(1)(-1) \end{bmatrix}$$ $$= \begin{bmatrix} a-b & 2 \\ 2a+b & 1 \end{bmatrix}$$

4. Calculate the product BA: $$BA = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 2 & 1 \end{bmatrix}$$ $$= \begin{bmatrix} (a)(1)+(1)(2) & (a)(-1)+(1)(1) \\ (b)(1)+(-1)(2) & (b)(-1)+(-1)(1) \end{bmatrix}$$ $$= \begin{bmatrix} a+2 & 1-a \\ b-2 & -b-1 \end{bmatrix}$$

5. Set up the equation $AB + BA = O$: $$\begin{bmatrix} a-b & 2 \\ 2a+b & 1 \end{bmatrix} + \begin{bmatrix} a+2 & 1-a \\ b-2 & -b-1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$ Adding the corresponding elements gives us a new matrix: $$\begin{bmatrix} (a-b)+(a+2) & 2+(1-a) \\ (2a+b)+(b-2) & 1+(-b-1) \end{bmatrix}$$ $$= \begin{bmatrix} 2a-b+2 & 3-a \\ 2a+2b-2 & -b \end{bmatrix}$$ $$= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$

6. Solve the System of Equations: From the equality of matrices, we get four equations:

   (i) $2a-b+2 = 0$

   (ii) $3-a = 0$

   (iii) $2a+2b-2 = 0$

   (iv) $-b = 0$

   From (iv), we get $b=0$.

   From (ii), we get $a=3$.

7. Check for Contradiction: We must check if these values ($a=3, b=0$) satisfy the other two equations:

   • Check (i): $2(3) - 0 + 2 = 6 + 2 = 8$. The equation requires this to be 0, but $8 \neq 0$.
   • Check (iii): $2(3) + 2(0) - 2 = 6 - 2 = 4$. The equation requires this to be 0, but $4 \neq 0$.

Conclusion: The system of equations is inconsistent. There are no values of $a$ and $b$ that can satisfy all four conditions simultaneously.

Answer: There are no such values for a and b.