CBSE Class 12 Mathematics HOTs Determinants Set 01

Question. If \(\begin{vmatrix} \alpha & 3 & 4 \\ 1 & 2 & 1 \\ 1 & 4 & 1 \end{vmatrix} = 0\), then the value of \(\alpha\) is
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (d) 4

Question. If \(A\) and \(B\) are two square matrices of order 2 and \(|A| = 2\) and \(|B| = 5\), then \(|-3AB|\) is
(a) -90
(b) -30
(c) 30
(d) 90
Answer: (d) 90

Question. \(\begin{vmatrix} 2 & 3 & 4 \\ 5 & 6 & 8 \\ 6x & 9x & 12x \end{vmatrix}\) is equal to
(a) 0
(b) 3x
(c) \(9x^2\)
(d) None of the options
Answer: (a) 0

Question. If \(A\) is a square matrix of order 2 and \(|A| = -2\), then value of \(|5A'|\) is
(a) -50
(b) -10
(c) 10
(d) 50
Answer: (a) -50

Question. The value of \(|A|\), if \(A = \begin{bmatrix} 0 & 2x-1 & \sqrt{x} \\ 1-2x & 0 & 2\sqrt{x} \\ -\sqrt{x} & -2\sqrt{x} & 0 \end{bmatrix}\), where \(x \in \mathbb{R}^+\), is
(a) \((2x+1)^2\)
(b) 0
(c) \((2x+1)^3\)
(d) None of the options
Answer: (b) 0

Question. Let \(P\) be a skew-symmetric matrix of order 3. If \(\det(P) = \alpha\), then \((2025)^{\alpha}\) is
(a) 0
(b) 1
(c) 2025
(d) \((2025)^3\)
Answer: (b) 1

Question. Let \(A\) be the area of a triangle having vertices \((x_1, y_1)\), \((x_2, y_2)\) and \((x_3, y_3)\). Which of the following is correct?
(a) \(\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \pm A\)
(b) \(\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \pm 2A\)
(c) \(\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \pm \frac{A}{2}\)
(d) \(\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = A^2\)
Answer: (b) \(\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \pm 2A\)

Question. The area of the triangle with vertices \((-1, 2)\), \((4, 0)\) and \((3, 9)\) is
(a) \(\frac{43}{2}\) sq units
(b) \(\frac{43}{4}\) sq units
(c) 21 sq units
(d) 42 sq units
Answer: (a) \(\frac{43}{2}\) sq units

Question. Given that \(A = [a_{ij}]\) is a square matrix of order \(3 \times 3\) and \(|A| = -7\), then the value of \(\sum_{i=1}^3 a_{i2} A_{i2}\), where \(A_{ij}\) denotes the cofactor of element \(a_{ij}\) is
(a) 7
(b) -7
(c) 0
(d) 49
Answer: (b) -7

Question. The adjoint of the matrix \(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\) is
(a) \(\begin{bmatrix} 4 & 2 \\ 3 & 1 \end{bmatrix}\)
(b) \(\begin{bmatrix} -4 & 2 \\ 3 & -1 \end{bmatrix}\)
(c) \(\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}\)
(d) \(\begin{bmatrix} 1 & -2 \\ -3 & 4 \end{bmatrix}\)
Answer: (c) \(\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}\)

Assertion-Reason Based Questions

Question. Assertion (A): \(\begin{vmatrix} 0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0 \end{vmatrix} = 0\).
Reason (R): Determinant of odd ordered skew-symmetric matrix is zero.
(a) Both A and R are correct; R is the correct explanation of A
(b) Both A and R are correct; R is not the correct explanation of A
(c) A is correct; R is incorrect
(d) R is correct; A is incorrect
Answer: (a) Both A and R are correct; R is the correct explanation of A

Question. Assertion (A): Minor of the element 9 of \(\begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}\) is 1.
Reason (R): Minor of element \(a_{ij}\) is the value of determinant after deleting ith row and jth column.
(a) Both A and R are correct; R is the correct explanation of A
(b) Both A and R are correct; R is not the correct explanation of A
(c) A is correct; R is incorrect
(d) R is correct; A is incorrect
Answer: (d) R is correct; A is incorrect

Question. Assertion (A): \(\text{adj}(\text{adj } A) = |A|^{n-2} \cdot A\).
Reason (R): \(|\text{adj } A| = |A|^{n-1}\).
(a) Both A and R are correct; R is the correct explanation of A
(b) Both A and R are correct; R is not the correct explanation of A
(c) A is correct; R is incorrect
(d) R is correct; A is incorrect
Answer: (a) Both A and R are correct; R is the correct explanation of A

Question. Assertion (A): \(\det(kU) = k^n \times \det(U)\).
Reason (R): If \(W\) is a matrix obtained by multiplying any one row or column of \(V\) by the scalar \(k\), then \(\det(W) = k \times \det(V)\).
(a) Both A and R are correct; R is the correct explanation of A
(b) Both A and R are correct; R is not the correct explanation of A
(c) A is correct; R is incorrect
(d) R is correct; A is incorrect
Answer: (a) Both A and R are correct; R is the correct explanation of A

Case Study Based Questions

Gautam buys 5 pens, 3 bags and 1 instrument box and pays a sum of ₹160. From the same shop, Vikram buys 2 pens, 1 bag and 3 instrument boxes and pays a sum of ₹190. Also, Ankur buys 1 pen, 2 bags and 4 instrument boxes and pays a sum of ₹250.

Question. Convert the given situation into a matrix equation of the form \(AX=B\).
Answer: Let the price of a pen, a bag, and an instrument box be \(x, y,\) and \(z\) respectively. Based on the given conditions, we can formulate the system of equations:
\(5x + 3y + z = 160\)
\(2x + y + 3z = 190\)
\(x + 2y + 4z = 250\)
In matrix form \(AX = B\), this is written as:
\(A = \begin{bmatrix} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{bmatrix}\), \(X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\), and \(B = \begin{bmatrix} 160 \\ 190 \\ 250 \end{bmatrix}\).

Question. Find \(|A|\).
Answer: Expanding \(|A|\) along the first row:
\(|A| = 5(1(4) - 3(2)) - 3(2(4) - 3(1)) + 1(2(2) - 1(1))\)
\(= 5(4 - 6) - 3(8 - 3) + 1(4 - 1)\)
\(= 5(-2) - 3(5) + 1(3)\)
\(= -10 - 15 + 3 = -22 \neq 0\).

Question. Find \(A^{-1}\).
Answer: First, find the cofactors of elements of matrix \(A\):
\(A_{11} = (4-6) = -2\)
\(A_{12} = -(8-3) = -5\)
\(A_{13} = 4-1 = 3\)
\(A_{21} = -(12-2) = -10\)
\(A_{22} = 20-1 = 19\)
\(A_{23} = -(10-3) = -7\)
\(A_{31} = 9-1 = 8\)
\(A_{32} = -(15-2) = -13\)
\(A_{33} = 5-6 = -1\)
The adjoint of \(A\) is:
\(\text{adj } A = \begin{bmatrix} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{bmatrix}\).
Thus, the inverse of matrix \(A\) is:
\(A^{-1} = \frac{\text{adj } A}{|A|} = -\frac{1}{22} \begin{bmatrix} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{bmatrix}\).

Question. Determine \(P = A^2 - 5A\).
Answer: We have:
\(A^2 = \begin{bmatrix} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{bmatrix} \begin{bmatrix} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 25+6+1 & 15+3+2 & 5+9+4 \\ 10+2+3 & 6+1+6 & 2+3+12 \\ 5+4+4 & 3+2+8 & 1+6+16 \end{bmatrix} = \begin{bmatrix} 32 & 20 & 18 \\ 15 & 13 & 17 \\ 13 & 13 & 23 \end{bmatrix}\).
\(5A = \begin{bmatrix} 25 & 15 & 5 \\ 10 & 5 & 15 \\ 5 & 10 & 20 \end{bmatrix}\).
Therefore, \(P = A^2 - 5A = \begin{bmatrix} 32 & 20 & 18 \\ 15 & 13 & 17 \\ 13 & 13 & 23 \end{bmatrix} - \begin{bmatrix} 25 & 15 & 5 \\ 10 & 5 & 15 \\ 5 & 10 & 20 \end{bmatrix} = \begin{bmatrix} 7 & 5 & 13 \\ 5 & 8 & 2 \\ 8 & 3 & 3 \end{bmatrix}\).

Case Study Based Questions

In an army camp, three teams Alpha, Beta and Charlie are located at the three corners of a triangular plot. The area of the triangular plot is 37 sq units and the coordinates of team Charlie lie on the X-axis (positive direction). Team Charlie received a message from the camp head about a secret room as follows. The coordinates of the secret room are \((a_{12}, a_{21})\) of the adjoint of the matrix \(\begin{bmatrix} 3 & 8 \\ 11 & 10 \end{bmatrix}\).

Question. Find the x-coordinate of the location of team Charlie. Use determinant method and show your steps.
Answer: Let the coordinates of team Charlie be \((x, 0)\), as it lies on the X-axis. The vertices of the triangle are Alpha \((3, 8)\), Beta \((11, 10)\), and Charlie \((x, 0)\).
Since the area of the triangular plot is 37 sq units:
\(\pm 37 = \frac{1}{2} \begin{vmatrix} 3 & 8 & 1 \\ 11 & 10 & 1 \\ x & 0 & 1 \end{vmatrix}\)
\(\implies \pm 74 = 3(10 - 0) - 8(11 - x) + 1(0 - 10x)\)
\(\implies \pm 74 = 30 - 88 + 8x - 10x\)
\(\implies \pm 74 = -58 - 2x\)
Case 1: \(74 = -58 - 2x \implies 2x = -132 \implies x = -66\).
Case 2: \(-74 = -58 - 2x \implies 2x = 16 \implies x = 8\).
Since the coordinates of Charlie lie on the positive direction of the X-axis, we choose \(x = 8\). Thus, the x-coordinate of the location of team Charlie is 8.

Question. The head of the camp plans to have a medical centre on the line joining the coordinates of teams Alpha and Beta such that its x-coordinate is 5. Find the y-coordinate of the medical centre using the determinant method. Show your steps and give a valid reason.
Answer: Let the coordinates of the medical centre be \((5, y)\). Since it lies on the line joining Alpha \((3, 8)\) and Beta \((11, 10)\), these three points are collinear. Consequently, the area of the triangle formed by them must be zero.
Using the determinant method:
\(\frac{1}{2} \begin{vmatrix} 3 & 8 & 1 \\ 11 & 10 & 1 \\ 5 & y & 1 \end{vmatrix} = 0\)
\(\implies 3(10 - y) - 8(11 - 5) + 1(11y - 50) = 0\)
\(\implies 30 - 3y - 48 + 11y - 50 = 0\)
\(\implies 8y - 68 = 0\)
\(\implies y = \frac{68}{8} = 8.5\).
Thus, the y-coordinate of the medical centre is 8.5.

Case Study Based Questions

A student in class XII is studying the concept of matrices. He comes across matrices \(A\) and \(B\), defined as \(A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}\) and \(B = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}\). Later on, he uses the product of \(AB\) in solving the system of equations \(x + 3z = 9\), \(-x + 2y - 2z = 4\) and \(2x - 3y + 4z = -3\).

Question. Find the product \(AB\).
Answer: We multiply \(A\) and \(B\):
\(AB = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix} \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}\)
\(= \begin{bmatrix} -2 - 9 + 12 & 0 - 2 + 2 & 1 + 3 - 4 \\ 0 + 18 - 18 & 0 + 4 - 3 & 0 - 6 + 6 \\ -6 - 18 + 24 & 0 - 4 + 4 & 3 + 6 - 8 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I\).
Thus, the product \(AB\) is the identity matrix \(I\).

Question. Find the value of \(A^{-1}\).
Answer: Since \(AB = I\), we have \(A^{-1} = B\). Therefore:
\(A^{-1} = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}\).

Question. What is the solution of given system of equation?
Answer: The given system of equations is:
\(x + 3z = 9\)
\(-x + 2y - 2z = 4\)
\(2x - 3y + 4z = -3\)
This can be represented as \(CX = D\), where:
\(C = \begin{bmatrix} 1 & 0 & 3 \\ -1 & 2 & -2 \\ 2 & -3 & 4 \end{bmatrix}\), \(X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\), and \(D = \begin{bmatrix} 9 \\ 4 \\ -3 \end{bmatrix}\).
Notice that the matrix \(C\) is the transpose of matrix \(A\) (i.e., \(C = A^T\)).
Hence, \(C^{-1} = (A^T)^{-1} = (A^{-1})^T = B^T\).
\(C^{-1} = \begin{bmatrix} -2 & 9 & 6 \\ 0 & 2 & 1 \\ 1 & -3 & -2 \end{bmatrix}\).
The solution is given by \(X = C^{-1}D\):
\(X = \begin{bmatrix} -2 & 9 & 6 \\ 0 & 2 & 1 \\ 1 & -3 & -2 \end{bmatrix} \begin{bmatrix} 9 \\ 4 \\ -3 \end{bmatrix} = \begin{bmatrix} -18 + 36 - 18 \\ 0 + 8 - 3 \\ 9 - 12 + 6 \end{bmatrix} = \begin{bmatrix} 0 \\ 5 \\ 3 \end{bmatrix}\).
So, the solution is \(x = 0, y = 5, z = 3\).

Question. Find the value of \(x+y+z\).
Answer: Using the solution from the system of equations, \(x = 0\), \(y = 5\), and \(z = 3\).
Thus, \(x + y + z = 0 + 5 + 3 = 8\).

Very Short Answer Type Questions

Question. If \(\begin{vmatrix} 3x & 7 \\ -2 & 4 \end{vmatrix} = \begin{vmatrix} 8 & 7 \\ 6 & 4 \end{vmatrix}\), then find the value of \(x\).
Answer: We have, \(\begin{vmatrix} 3x & 7 \\ -2 & 4 \end{vmatrix} = \begin{vmatrix} 8 & 7 \\ 6 & 4 \end{vmatrix}\)
\(\Rightarrow 12x - (-14) = 32 - 42\)
\(\Rightarrow 12x + 14 = -10\)
\(\Rightarrow 12x = -24\)
\(\Rightarrow x = -2\)

Question. If \(x \in \mathbb{N}\) and \(\begin{vmatrix} x + 3 & -2 \\ -3x & 2x \end{vmatrix} = 8\), then find the value of \(x\).
Answer: We have, \(\begin{vmatrix} x + 3 & -2 \\ -3x & 2x \end{vmatrix} = 8\)
\(\Rightarrow 2x(x + 3) - (-2)(-3x) = 8\)
\(\Rightarrow 2x^2 + 6x - 6x = 8\)
\(\Rightarrow 2x^2 = 8\)
\(\Rightarrow x^2 = 4\)
\(\Rightarrow x = \pm 2\)
Since \(x \in \mathbb{N}\), we have \(x = 2\).

Question. Find the area of triangle, whose vertices are \((2, 7)\), \((1, 1)\) and \((10, 8)\).
Answer: Given, vertices of a triangle are \((2, 7)\), \((1, 1)\) and \((10, 8)\).
Let \((x_1, y_1) = (2, 7)\), \((x_2, y_2) = (1, 1)\) and \((x_3, y_3) = (10, 8)\).
Then, area of triangle \(= \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \frac{1}{2} \begin{vmatrix} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{vmatrix}\)
\(= \frac{1}{2} [2(1 - 8) - 7(1 - 10) + 1(8 - 10)]\)
\(= \frac{1}{2} [-14 + 63 - 2] = \frac{1}{2} (47) = 23.5\) sq units.

Short Answer Type Questions

Question. If \(f(x) = \begin{vmatrix} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{vmatrix}\), then show that \(f(0)=0\).
Answer: Given, \(f(x) = \begin{vmatrix} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{vmatrix}\).
\(\therefore f(0) = \begin{vmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{vmatrix}\)
Expanding along \(R_1\):
\(f(0) = -(-a)(0 - (-bc)) + (-b)(ac - 0)\)
\(= a(bc) - b(ac) = abc - abc = 0\)
Hence proved.

Question. If \(A\) is skew-symmetric matrix of order 3, then prove that \(\det A = 0\).
Answer: Given, \(A\) is a skew-symmetric matrix of order 3.
\(\Rightarrow A^T = -A\)
Taking determinants on both sides:
\(\Rightarrow |A^T| = |-A|\)
\(\Rightarrow |A| = (-1)^3 |A|\) [Since \(|A| = |A^T|\) and \(|-A| = (-1)^n |A|\) for order \(n\)]
\(\Rightarrow |A| = -|A|\)
\(\Rightarrow 2|A| = 0\)
\(\Rightarrow |A| = 0\)

Question. Find the values of \(k\), if the area of triangle is 4 sq units and vertices are \((-2, 0)\), \((0, 4)\) and \((0, k)\).
Answer: Given area of a triangle with vertices \((-2, 0)\), \((0, 4)\) and \((0, k)\) is 4 sq units.
We have, \(\frac{1}{2} \begin{vmatrix} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{vmatrix} = \pm 4\)
\(\Rightarrow \begin{vmatrix} -2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1 \end{vmatrix} = \pm 8\)
Expanding along \(C_1\):
\(\Rightarrow -2(4 - k) = \pm 8\)
\(\Rightarrow -8 + 2k = \pm 8\)
\(\Rightarrow 2k = \pm 8 + 8\)
\(\Rightarrow 2k = 16\) or \(2k = 0\)
\(\therefore k = 8\) or \(k = 0\)

Question. Show that the points \((a + 5, a - 4)\), \((a - 2, a + 3)\) and \((a, a)\) do not lie on a straight line for any value of \(a\).
Answer: To prove that the points do not lie on a straight line, we show that the area of the triangle formed by these three points is non-zero for all values of \(a\).
\(\text{Area} = \frac{1}{2} \begin{vmatrix} a+5 & a-4 & 1 \\ a-2 & a+3 & 1 \\ a & a & 1 \end{vmatrix}\)
Applying row operations \(R_1 \rightarrow R_1 - R_3\) and \(R_2 \rightarrow R_2 - R_3\):
\(\text{Area} = \frac{1}{2} \begin{vmatrix} 5 & -4 & 0 \\ -2 & 3 & 0 \\ a & a & 1 \end{vmatrix}\)
Expanding along \(C_3\):
\(\text{Area} = \frac{1}{2} [1 \cdot (15 - 8)] = \frac{7}{2} = 3.5 \neq 0\)
Since the area of the triangle is \(3.5\) square units, which is a constant and non-zero for any value of \(a\), the given points are never collinear and thus do not lie on a straight line.

Long Answer Type Questions

Question. If \(A = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}\), then find \(\text{adj } A\) and verify that \(A(\text{adj } A) = (\text{adj } A)A = |A|I_3\).
Answer: We have, \(A = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}\).
First, find the determinant of \(A\):
\(|A| = 1 \cdot (\cos^2 \alpha - (-\sin^2 \alpha)) = \cos^2 \alpha + \sin^2 \alpha = 1\).
The cofactors of the elements of \(|A|\) are:
\(A_{11} = \cos \alpha\), \(A_{12} = -\sin \alpha\), \(A_{13} = 0\)
\(A_{21} = \sin \alpha\), \(A_{22} = \cos \alpha\), \(A_{23} = 0\)
\(A_{31} = 0\), \(A_{32} = 0\), \(A_{33} = 1\).
\(\therefore \text{adj } A = \begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{bmatrix} = \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}\).
Now, we check \(A(\text{adj } A)\):
\(A(\text{adj } A) = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & 0 & 0 \\ 0 & \sin^2 \alpha + \cos^2 \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I_3\).
Similarly, \((\text{adj } A)A = \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I_3\).
Since \(|A| = 1\), we have \(A(\text{adj } A) = (\text{adj } A)A = |A|I_3\). Hence verified.

Question. Given, \(A = \begin{bmatrix} 5 & 3 \\ -2 & 4 \end{bmatrix}\) and \(B = \begin{bmatrix} 2 & -3 \\ 4 & -8 \end{bmatrix}\). Show that \((\text{adj } A) \times (\text{adj } B) = \text{adj}(BA)\).
Answer: We have, \(A = \begin{bmatrix} 5 & 3 \\ -2 & 4 \end{bmatrix}\) and \(B = \begin{bmatrix} 2 & -3 \\ 4 & -8 \end{bmatrix}\).
The adjoint of \(A\) and \(B\) are given by:
\(\text{adj } A = \begin{bmatrix} 4 & -3 \\ 2 & 5 \end{bmatrix}\) and \(\text{adj } B = \begin{bmatrix} -8 & 3 \\ -4 & 2 \end{bmatrix}\).
Now, \((\text{adj } A) \times (\text{adj } B) = \begin{bmatrix} 4 & -3 \\ 2 & 5 \end{bmatrix} \begin{bmatrix} -8 & 3 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} -32 + 12 & 12 - 6 \\ -16 - 20 & 6 + 10 \end{bmatrix} = \begin{bmatrix} -20 & 6 \\ -36 & 16 \end{bmatrix}\)     ...(i)
Next, compute the product \(BA\):
\(BA = \begin{bmatrix} 2 & -3 \\ 4 & -8 \end{bmatrix} \begin{bmatrix} 5 & 3 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} 10 + 6 & 6 - 12 \\ 20 + 16 & 12 - 32 \end{bmatrix} = \begin{bmatrix} 16 & -6 \\ 36 & -20 \end{bmatrix}\).
Thus, the adjoint of \(BA\) is:
\(\text{adj}(BA) = \begin{bmatrix} -20 & 6 \\ -36 & 16 \end{bmatrix}\).     ...(ii)
From (i) and (ii), we get:
\((\text{adj } A) \times (\text{adj } B) = \text{adj}(BA)\). Hence proved.

Question. If \(A = \begin{bmatrix} 4 & -8 \\ 12 & 16 \end{bmatrix}\) and \(B = \begin{bmatrix} -16 & -8 \\ 12 & -4 \end{bmatrix}\), identify what type of matrix is \((\text{adj } A) \times (\text{adj } B)\). Show your work.
Answer: We have, \(A = \begin{bmatrix} 4 & -8 \\ 12 & 16 \end{bmatrix}\) and \(B = \begin{bmatrix} -16 & -8 \\ 12 & -4 \end{bmatrix}\).
First, find the cofactors of elements of \(|A|\):
\(A_{11} = 16\), \(A_{12} = -12\), \(A_{21} = 8\), \(A_{22} = 4\).
\(\therefore \text{adj } A = \begin{bmatrix} 16 & 8 \\ -12 & 4 \end{bmatrix}\).
Next, find the cofactors of elements of \(|B|\):
\(B_{11} = -4\), \(B_{12} = -12\), \(B_{21} = 8\), \(B_{22} = -16\).
\(\dots \text{adj } B = \begin{bmatrix} -4 & 8 \\ -12 & -16 \end{bmatrix}\).
Now, evaluate the product \((\text{adj } A) \times (\text{adj } B)\):
\((\text{adj } A) \times (\text{adj } B) = \begin{bmatrix} 16 & 8 \\ -12 & 4 \end{bmatrix} \begin{bmatrix} -4 & 8 \\ -12 & -16 \end{bmatrix} = \begin{bmatrix} -64 - 96 & 128 - 128 \\ 48 - 48 & -96 - 64 \end{bmatrix} = \begin{bmatrix} -160 & 0 \\ 0 & -160 \end{bmatrix}\).
Since all non-diagonal elements of the resulting matrix are zero and the diagonal elements are equal, \((\text{adj } A) \times (\text{adj } B)\) is a **scalar matrix** (or a **diagonal matrix**).

Question. If \(A = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix}\), then show that \(A^T A^{-1} = \begin{bmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \end{bmatrix}\).
Answer: We have, \(A = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix}\).
First, find the determinant of \(A\):
\(|A| = 1 - (-\tan^2 x) = 1 + \tan^2 x \neq 0\).
So, \(A\) is invertible.
The cofactors of elements in \(A\) are:
\(C_{11} = 1\), \(C_{12} = \tan x\), \(C_{21} = -\tan x\), \(C_{22} = 1\).
\(\therefore \text{adj } A = \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}\).
Now, \(A^{-1} = \frac{1}{|A|} \text{adj } A = \frac{1}{1+\tan^2 x} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}\).
Next, compute \(A^T A^{-1}\):
\(A^T A^{-1} = \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \left[ \frac{1}{1+\tan^2 x} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \right]\)
\(= \frac{1}{1+\tan^2 x} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}\)
\(= \frac{1}{1+\tan^2 x} \begin{bmatrix} 1 - \tan^2 x & -2\tan x \\ 2\tan x & 1 - \tan^2 x \end{bmatrix}\)
\(= \begin{bmatrix} \frac{1-\tan^2 x}{1+\tan^2 x} & -\frac{2\tan x}{1+\tan^2 x} \\ \frac{2\tan x}{1+\tan^2 x} & \frac{1-\tan^2 x}{1+\tan^2 x} \end{bmatrix}\)
Using trigonometric identities \(\cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x}\) and \(\sin 2x = \frac{2\tan x}{1+\tan^2 x}\), we get:
\(A^T A^{-1} = \begin{bmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \end{bmatrix}\). Hence proved.

Question. If \(A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\) and \(I\) is the identity matrix of order 2, then show that \(A^2 = 4A - 3I\). Hence, find \(A^{-1}\).
Answer: We have, \(A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\).
First, find \(A^2\):
\(A^2 = A \cdot A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 4 + 1 & -2 - 2 \\ -2 - 2 & 1 + 4 \end{bmatrix} = \begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix}\)     ...(i)
Now, find \(4A - 3I\):
\(4A - 3I = 4 \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} - 3 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 8 & -4 \\ -4 & 8 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix}\)     ...(ii)
From (i) and (ii), we get \(A^2 = 4A - 3I\). Hence proved.
Now, pre-multiplying both sides of the equation by \(A^{-1}\):
\(A^{-1} A^2 = A^{-1} (4A - 3I)\)
\(\Rightarrow (A^{-1} A)A = 4(A^{-1} A) - 3(A^{-1} I)\)
\(\Rightarrow IA = 4I - 3A^{-1}\)
\(\Rightarrow A = 4I - 3A^{-1}\)
\(\Rightarrow 3A^{-1} = 4I - A\)
\(\Rightarrow 3A^{-1} = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} - \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}\)
\(\therefore A^{-1} = \frac{1}{3} \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}\).