1. ⇒ (MHT CET 2023 12th May Evening Shift )

If $A = [\begin{array}{cc} 2 a & - 3 b \\ 3 & 2 \end{array}]$ and $A \cdot adj A = A A^{T}$ , then $2 a + 3 b$ is

A. -1

B. 1

C. 5

D. -5

Correct answer option is (C)

$\begin{aligned} A & = [\begin{array}{cc} 2 a & - 3 b \\ 3 & 2 \end{array}] \\ A & \cdot adj A = [\begin{array}{cc} 2 a & - 3 b \\ 3 & 2 \end{array}] [\begin{array}{cc} 2 & 3 b \\ - 3 & 2 a \end{array}] \\ = [\begin{array}{cc} 4 a + 9 b & 0 \\ 0 & 9 b + 4 a \end{array}] \\ A \cdot A^{T} = [\begin{array}{cc} 2 a & - 3 b \\ 3 & 2 \end{array}] [\begin{array}{cc} 2 a & 3 \\ - 3 b & 2 \end{array}] \\ = [\begin{array}{cc} 4 a^{2} + 9 b^{2} & 6 a - 6 b \\ 6 a - 6 b & 13 \end{array}] \\ ∴ & A \cdot adj A = A \cdot A^{T} \\ \Rightarrow [\begin{array}{cc} 4 a + 9 b & 0 \\ 0 & 4 a + 9 b \end{array}] = [\begin{array}{c} 4 a^{2} + 9 b^{2} \\ 6 a - 6 b \end{array}] \\ \Rightarrow a = b and 4 a + 9 b = 13 \\ \Rightarrow a = b = 1 \\ \Rightarrow 2 a + 3 b = 5 \end{aligned}$

2. ⇒ (MHT CET 2023 11th May Morning Shift )

If $P = [\begin{array}{lll} 1 & α & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{array}]$ is the adjoint of a $3 \times 3$ matrix $A$ and $| A | = 4$ , then value of $α$ is

A. 4

B. 11

C. 5

D. 0

Correct answer option is (B)

$\begin{aligned} P = adj A .... [Given] \\ ∴ & | P | = | A |^{2} . . . . [∵ | adjA | = | A |^{n - 1}] \\ \Rightarrow | \begin{array}{lll} 1 & α & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{array} | = 4^{2} \\ \Rightarrow 2 α - 6 = 16 \\ \Rightarrow α = 11 \end{aligned}$

3. ⇒ (MHT CET 2023 10th May Evening Shift )

If $B = [\begin{array}{ccc} 3 & α & - 1 \\ 1 & 3 & 1 \\ - 1 & 1 & 3 \end{array}]$ is the adjoint of a $3 \times 3$ matrix $A$ and $| A | = 4$ , then $α$ is equal to

A. 1

B. 0

C. -1

D. -2

Correct answer option is (A)

$\begin{aligned} \begin{array}{ll} Using | adj A | = | A |^{n - 1} \\ B u t B = Adj (A) . . . . [Given] \\ ∴ & | B | = | A |^{2} \\ \Rightarrow & | \begin{array}{ccc} 3 & α & - 1 \\ 1 & 3 & 1 \\ - 1 & 1 & 3 \end{array} | = | A |^{2} \\ \Rightarrow & 24 - 4 α - 4 = 4^{2} \\ \Rightarrow & 20 - 4 α = 16 \\ \Rightarrow & α = 1 \end{array} \end{aligned}$

4. ⇒ (MHT CET 2023 10th May Morning Shift )

If $B = [\begin{array}{lll} 1 & α & 2 \\ 1 & 2 & 2 \\ 2 & 3 & 3 \end{array}]$ is the adjoint of a $3 \times 3$ matrix A and $| A | = 5$ , then $α$ is equal to

A. 25

B. 27

C. 3 $\sqrt{3}$

D. 5

Correct answer option is (B)

Using $| Adj A | = | A |^{n - 1}$

But $B = Adj (A)$ .... [Given]

$∴ | B | = | A |^{2}$

$\begin{aligned} \Rightarrow | \begin{array}{lll} 1 & α & 2 \\ 1 & 2 & 2 \\ 2 & 3 & 3 \end{array} | = | A |^{2} \\ \Rightarrow α - 2 = 5^{2} \\ \Rightarrow α - 2 = 25 \\ \Rightarrow α = 27 \end{aligned}$

5. ⇒ (MHT CET 2023 9th May Evening Shift )

If $| \begin{array}{ccc} \cos (A + B) & - \sin (A + B) & \cos (2 B) \\ \sin A & \cos A & \sin B \\ - \cos A & \sin A & \cos B \end{array} | = 0$ , then the value of $B$ is

A. $n π, n \in Z$

B. $(2 n + 1) \frac{π}{2}, n \in Z$

C. $(2 n + 1) \frac{π}{4}, n \in Z$

D. $2 n \frac{π}{3}, n \in Z$

Correct answer option is (B)

$| \begin{array}{ccc} \cos (A + B) & - \sin (A + B) & \cos (2 B) \\ \sin A & \cos A & \sin B \\ - \cos A & \sin A & \cos B \end{array} | = 0$

$∴ \cos (A + B) [(\cos A \cos B - \sin A \sin B)] + \sin (A + B) [\sin A \cos B + \sin B \cos A] + \cos 2 B [\sin^{2} A + \cos^{2} A] = 0$

$\begin{aligned} ∴ & \cos (A + B) \cdot \cos (A + B) \\ + \sin (A + B) \cdot \sin (A + B) + \cos 2 B = 0 \\ \cos^{2} (A + B) + \sin^{2} (A + B) + \cos 2 B = 0 \\ 1 + \cos 2 B = 0 \end{aligned}$

$\begin{aligned} 2 \cos^{2} B = 0 \\ ∴ \cos B = 0 \end{aligned}$

$∴ B = (2 n + 1) \frac{π}{2} for (n \in Z)$

⇒Matrices