Matrices and Determinants jee mains questions | JEE Main Maths Matrices and Determinants Previous Year Question

Q 1 :
Let $α \in (0, \infty)$ and $A = [\begin{matrix} 1 & 2 & α \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{matrix}] .$ If $\det (adj (2 A - A^{T})) \cdot adj (A - 2 A^{T}) = 2^{8},$ then $(\det {(A))}^{2}$ is equal to [2024]

1
16
36
49

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(B)

We have, $| adj (2 A - A^{T}) \cdot adj (A - 2 A^{T}) | = 2^{8}$

$\Rightarrow | adj (A - 2 A^{T}) \cdot (2 A - A^{T}) | = 2^{8}$ $(∵ adj (A B) = adj B \cdot adj A)$

$\Rightarrow | adj {- ((2 A^{T} - A) (2 A - A^{T}))} | = 2^{8}$

$\Rightarrow | adj {- ({(2 A - A^{T})}^{T} (2 A - A^{T}))} | = 2^{8}$

$\Rightarrow | adj {- (P^{T} P)} | = 2^{8}, where P = 2 A - A^{T}$

$\Rightarrow | - P^{T} \cdot P |^{2} = 2^{8} \Rightarrow | - P^{T} |^{2} | P |^{2} = 2^{8}$

$\Rightarrow | P |^{4} = 4^{4}$ $(∵ | P^{T} | = | P |)$

$\Rightarrow | P | = \pm 4 \Rightarrow | 2 A - A^{T} | = \pm 4 \Rightarrow$ $| \begin{matrix} 1 & 3 & 2 α \\ 0 & 1 & 1 \\ - α & 1 & 2 \end{matrix} |$ $= \pm 4$

$\Rightarrow 1 (0 - 1) - 3 (0 + α) + 2 α (0) = \pm 4 \Rightarrow - 1 - 3 α = \pm 4$

$\Rightarrow α = 1, \frac{- 5}{3} \Rightarrow α = 1$ $[∵ α \in (0, \infty)]$

Now, $| A | = 1 (0 - 1) - 2 (2 - 0) + α (1 - 0) = α - 5$

$∴ | A |^{2} = 16$

Q 2 :
Let $A = [\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}]$ and $B = I + adj (A) + {(adj A)}^{2} + \dots + {(adj A)}^{10} .$ Then, the sum of all the elements of the matrix $B$ is: [2024]

$- 110$
$- 124$
$22$
$- 88$

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(D)

Given that $A = [\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}], adj (A) = [\begin{matrix} 1 & - 2 \\ 0 & 1 \end{matrix}],$ ${(adj A)}^{2} = [\begin{matrix} 1 & - 4 \\ 0 & 1 \end{matrix}], {(adj A)}^{4} = [\begin{matrix} 1 & - 8 \\ 0 & 1 \end{matrix}]$

$∴$ ${(adj A)}^{r} = [\begin{matrix} 1 & - 2 r \\ 0 & 1 \end{matrix}]$

Now, $B = I + adj (A) + {(adj A)}^{2} + \dots + {(adj A)}^{10} = [\begin{matrix} 11 & \sum_{r = 0}^{10} (- 2 r) \\ 0 & 11 \end{matrix}] = [\begin{matrix} 11 & - 110 \\ 0 & 11 \end{matrix}]$

$∴$ $Sum of elements of B = 11 + (- 110) + 0 + 11 = - 88 .$

Q 3 :
Let $A$ and $B$ be two square matrices of order 3 such that $| A | = 3$ and $| B | = 2 .$ Then $| A^{T} A (adj {(2 A))}^{- 1} (adj (4 B)) (adj {(A B))}^{- 1} A A^{T} |$ is equal to : [2024]

108
32
64
81

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(C)

We have $| A^{T} A (adj {(2 A))}^{- 1} (adj (4 B)) (adj {(A B))}^{- 1} A A^{T} |$

$=$ $| A^{T} | | A | \frac{1}{| adj 2 A |} | adj 4 B | \frac{1}{| adj A B |} | A | | A^{T} |$

$= | A |^{2} \frac{1}{{(2^{2})}^{3} | adj A |} {(4^{2})}^{3} | adj B | \frac{1}{| adjA B |} | A |^{2}$ $[∵ | A | = | A^{T} | and adj kA = k^{n - 1} adj A, where n is order of A]$

$= | A |^{2} \frac{1}{2^{6} | A |^{2}} 4^{6} | B |^{2} \frac{1}{| A |^{2} | B |^{2}} | A |^{2}$ $[∵ | adj A | = | A |^{n - 1}, where n is order of A]$

$= \frac{4^{6}}{2^{6}} = 64$

Q 4 :
Let $α β \neq 0$ and $A = [\begin{matrix} β & α & 3 \\ α & α & β \\ - β & α & 2 α \end{matrix}] .$ If $B = [\begin{matrix} 3 α & - 9 & 3 α \\ - α & 7 & - 2 α \\ - 2 α & 5 & - 2 β \end{matrix}]$ is the matrix of cofactors of the elements of $A,$ then $d e t (A B)$ is equal to : [2024]

125
64
343
216

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(D)

Clearly $B = {(adj A)}^{T}$

Now, $| A B |$ $= | A |$ $| {(adj A)}^{T} |$

$= | A | | adj A |$

$= | A | | A |^{2}$

$= | A |^{3}$

Now, $A = [\begin{matrix} β & α & 3 \\ α & α & β \\ - β & α & 2 α \end{matrix}]$ and ${(adj A)}^{T} = [\begin{matrix} 3 α & - 9 & 3 α \\ - α & 7 & - 2 α \\ - 2 α & 5 & - 2 β \end{matrix}]$

$⇒Cofactor of a_{11} = (2 α^{2} - α β) = 3 α$

$Cofactor of a_{21} = (2 α^{2} - 3 α) = {(- 1)}^{2 + 1} (- α)$

$\Rightarrow 2 α^{2} - 4 α = 0 \Rightarrow α = 0, 2$

$\Rightarrow β = 1 for α = 2$

$∴$ $| A |$ $= [\begin{matrix} 1 & 2 & 3 \\ 2 & 2 & 1 \\ - 1 & 2 & 4 \end{matrix}] = 6 - 18 + 18 = 6$

$\Rightarrow | A B | = 6^{3} = 216$

Q 5 :
If $A$ is a square matrix of order 3 such that $\det (A) = 3$ and $\det (adj (- 4 adj (- 3 adj (3 adj ({(2 A)}^{- 1}))))) = 2^{m} 3^{n},$ then $m + 2 n$ is equal to : [2024]

6
2
3
4

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4

(D)

We have, $| A | = 3$

Now, $|adj(- 4 adj (- 3 adj (3 adj {(2 A)}^{- 1}))) |$

$=$ $| - 4 adj (- 3 adj (3 adj {(2 A)}^{- 1})) |^{2}$ $[∵ | adj A | = | A |^{n - 1}, n is order of A]$

$= ({{(- 4)}^{3})}^{2}$ $| adj (- 3 adj (3 adj {(2 A)}^{- 1})) |^{2}$

$= 4^{6}$ $| - 3 adj (3 adj {(2 A)}^{- 1}) |^{4}$

$= 4^{6} 3^{12}$ $| adj (3 adj {(2 A)}^{- 1}) |^{4}$

$= 4^{6} 3^{12}$ $| 3 adj {(2 A)}^{- 1} |^{8}$

$= 4^{6} 3^{12} 3^{24}$ $| {(2 A)}^{- 1} |^{16}$

$= 4^{6} 3^{36} 2^{- 48}$ $| A^{- 1} |^{16}$

$= \frac{4^{6} 3^{36}}{2^{48} \cdot 3^{16}} = 2^{- 36} 3^{20}$

$\Rightarrow m = - 36, n = 20$

$\Rightarrow m + 2 n = - 36 + 40 = 4$

Q 6 :
Let $B = [\begin{matrix} 1 & 3 \\ 1 & 5 \end{matrix}]$ and $A$ be a $2 \times 2$ matrix such that $A B^{- 1} = A^{- 1} .$ If $B C B^{- 1} = A$ and $C^{4} + α C^{2} + β I = O,$ then $2 β - α$ is equal to [2024]

2
8
10
16

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4

(C)

$B = [\begin{matrix} 1 & 3 \\ 1 & 5 \end{matrix}]$

We have, $B C B^{- 1} = A$

$\Rightarrow (B C B^{- 1}) (B C B^{- 1}) = A^{2} \Rightarrow B C B^{- 1} B C B^{- 1} = A^{2}$

$\Rightarrow B C^{2} B^{- 1} = A^{2} \Rightarrow B^{- 1} B C^{2} B^{- 1} = B^{- 1} A^{2}$

$\Rightarrow C^{2} B^{- 1} B = B^{- 1} A^{2} B$

$\Rightarrow C^{2} = B^{- 1} A^{2} B$

$\Rightarrow C^{2} = B^{- 1} A A B$ ...(i)

$\Rightarrow C^{2} \cdot C^{2} = B^{- 1} A^{2} B B^{- 1} A^{2} B$

$\Rightarrow C^{4} = B^{- 1} A^{4} B$ ...(ii)

Now, $A B^{- 1} = A^{- 1}$

$\Rightarrow A^{2} B^{- 1} = I$

$\Rightarrow A^{2} = B$ ...(iii)

Using (i), (ii) and (iii), we get

$C^{2} = B^{- 1} B B = B$ and $C^{4} = B^{- 1} B^{2} B = B^{2}$

So, $C^{4} + α C^{2} + β I = 0 \Rightarrow B^{2} + α B + β I = 0$

$= [\begin{matrix} 1 & 3 \\ 1 & 5 \end{matrix}] [\begin{matrix} 1 & 3 \\ 1 & 5 \end{matrix}] + [\begin{matrix} α & 3 α \\ α & 5 α \end{matrix}] + [\begin{matrix} β & 0 \\ 0 & β \end{matrix}] = 0$

$\Rightarrow [\begin{matrix} 4 & 18 \\ 6 & 28 \end{matrix}] + [\begin{matrix} α & 3 α \\ α & 5 α \end{matrix}] + [\begin{matrix} β & 0 \\ 0 & β \end{matrix}] = 0$

$\Rightarrow 4 + α + β = 0$ and $6 + α = 0$

$\Rightarrow α = - 6, β = 2 \Rightarrow 2 β - α = 10$

Q 7 :
Let $R = (\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix})$ be a non-zero $3 \times 3$ matrix, where $x \sin θ = y \sin (θ + \frac{2 π}{3}) = z \sin (θ + \frac{4 π}{3}) \neq 0, θ \in (0, 2 π) .$ For a square matrix $M,$ let trace $(M)$ denote the sum of all the diagonal entries of $M .$ Then, among the statements:

(I) Trace(R) = 0

(II) If trace $(adj(adj (R))) = 0$ , then $R$ has exactly one non-zero entry.

Only (I) is true.
Both (I) and (II) are true.
Neither (I) nor (II) is true.
Only (II) is true.

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(D)

We have, $x \sin θ = y \sin (θ + \frac{2 π}{3}) = z \sin (θ + \frac{4 π}{3})$

$\Rightarrow y = \frac{x \sin θ}{\sin (θ + \frac{2 π}{3})}$ and $z = \frac{x \sin θ}{\sin (θ + \frac{4 π}{3})}$

$∴$ $x + y + z = x + \frac{x \sin θ}{\sin (θ + \frac{2 π}{3})} + \frac{x \sin θ}{\sin (θ + \frac{4 π}{3})}$ $= \frac{- 3 x}{4 \sin (θ + \frac{2 π}{3}) \sin (θ + \frac{4 π}{3})} \neq 0$

Also, $R = (\begin{matrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{matrix}) or adj R = (\begin{matrix} y z & 0 & 0 \\ 0 & x z & 0 \\ 0 & 0 & x y \end{matrix})$

$\Rightarrow adj(adj R)$ $= (\begin{matrix} x^{2} y z & 0 & 0 \\ 0 & y^{2} x z & 0 \\ 0 & 0 & z^{2} x y \end{matrix})$

$∴$ $Tr(adj(adj R)) = x y z (x + y + z) \neq 0$

Only (II) is true.

Q 8 :
Let $A$ be a $2 \times 2$ symmetric matrix such that $A [\begin{matrix} 1 \\ 1 \end{matrix}] = [\begin{matrix} 3 \\ 7 \end{matrix}]$ and the determinant of $A$ be 1. If $A^{- 1} = α A + β I,$ where $I$ is an identity matrix of order $2 \times 2,$ then $α + β$ equals _______ . [2024]

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(5)

Let $A = [\begin{matrix} a & b \\ b & c \end{matrix}]$

$| A | = 1 \Rightarrow a c - b^{2} = 1$ ...(i)

Given, $A [\begin{matrix} 1 \\ 1 \end{matrix}] = [\begin{matrix} 3 \\ 7 \end{matrix}] \Rightarrow [\begin{matrix} a & b \\ b & c \end{matrix}] [\begin{matrix} 1 \\ 1 \end{matrix}] = [\begin{matrix} 3 \\ 7 \end{matrix}]$

$\Rightarrow a + b = 3$ ...(ii)

and $b + c = 7$

On solving (i), (ii) and (iii), we get

$a = 1, b = 2, c = 5$

$∴$ $A = [\begin{matrix} 1 & 2 \\ 2 & 5 \end{matrix}] \Rightarrow A^{- 1} = [\begin{matrix} 5 & - 2 \\ - 2 & 1 \end{matrix}]$

Given, $A^{- 1} = α A + β I$

$\Rightarrow [\begin{matrix} 5 & - 2 \\ - 2 & 1 \end{matrix}] = α [\begin{matrix} 1 & 2 \\ 2 & 5 \end{matrix}] + β [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

$\Rightarrow [\begin{matrix} α + β & 2 α \\ 2 α & 5 α + β \end{matrix}] = [\begin{matrix} 5 & - 2 \\ - 2 & 1 \end{matrix}]$

On comparing, we get

$α = - 1 and β = 6$

Hence, $α + β = - 1 + 6 = 5$

Q 9 :
Let $A$ be a non-singular matrix of order 3. If $\det (3 adj(2 adj((\det A) A))) = 3^{- 13} \cdot 2^{- 10}$ and $\det (3 adj (2 A)) = 2^{m} \cdot 3^{n},$ then $| 3 m + 2 n |$ is equal to ______. [2024]

0%

(14)

We have, $| 3 adj(2 adj (| A | A)) |$ $= 3^{- 13} \cdot 2^{- 10}$

Let $| A |$ $A = B$

$\Rightarrow | B | = | A | | A | = | A |^{3} | A | = | A |^{4}$ ...(i)

$\Rightarrow a d j (| A | A) = (adj B)$

$\Rightarrow 2 a d j (| A | A) = (2 adj B) = C (say)$

Now, $| 3 adj (C) |$ $= 3^{3} | C |^{2}$ ...(ii)

$∴$ $| C |$ $=$ $| (2 adj B) |$ $= 2^{3} | B |^{2} = 2^{3} {(| A |^{4})}^{2}$ (using (i))

$= 2^{3} \cdot | A |^{8}$ ...(iii)

From (ii) and (iii), we get

$| 3 adj C |$ $= 3^{3} \cdot {(2^{3} | A |^{8})}^{2} = 3^{3} 2^{6} | A |^{16} = 3^{- 13} 2^{- 10}$

$\Rightarrow | A |^{16} = {(3 \cdot 2)}^{- 16} = {(\frac{1}{6})}^{16} \Rightarrow | A | = \pm \frac{1}{6}$

So, $| 3 adj 2 A |$ $= 3^{3} | 2 A |^{2} = 3^{3} \cdot {(2^{3} | A |)}^{2} = 3^{3} \cdot 2^{6} | A |^{2}$

$3^{3} \cdot 2^{6} \cdot \frac{1}{36} = 2^{m} \cdot 3^{n}$ $(∵ Given)$

$\Rightarrow 2^{m} \cdot 3^{n} = 48$

$\Rightarrow 2^{m} \cdot 3^{n} = 2^{4} \cdot 3^{1} \Rightarrow m = 4$ and $n = 1$

So, $3 m + 2 n = 12 + 2 = 14$

Q 10 :
Let $A$ be a $3 \times 3$ matrix and det(A) = 2. If $n = \det \underset{2024 - t i m e s}{\underset{⏟}{(a d j (a d j (. . . (a d j A))))}},$ then the remainder when $n$ is divided by 9 is equal to ________ . [2024]

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(7)

We know that $| adj A |$ $= | A |^{n - 1} = | A |^{2}$

$\Rightarrow | adj adj A | = | adj A |^{2} = | A |^{2^{2}}$

$n = \underset{2024 t i m e s}{\underset{⏟}{| a d j a d j . . . a d j A |}} = | A |^{2^{2024}}$

Here, $2^{2024} = 4^{1012} = {(3 + 1)}^{1012} = 3 k + 1$

$\Rightarrow n = | A |^{3 k + 1} = 2^{3 k + 1} [∵ | A | = 2]$

$= 2 \cdot 8^{k} = 2 {(9 - 1)}^{k} = 2 (9 λ - 1) = 18 λ - 2$

$= 18 (λ - 1) + 16 = 18 (λ - 1) + 9 + 7$

Hence, the required remainder is 7.

Topic Question Set

Q 1 :
Let $α \in (0, \infty)$ and $A = [\begin{matrix} 1 & 2 & α \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{matrix}] .$ If $\det (adj (2 A - A^{T})) \cdot adj (A - 2 A^{T}) = 2^{8},$ then $(\det {(A))}^{2}$ is equal to [2024]

Q 2 :
Let $A = [\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}]$ and $B = I + adj (A) + {(adj A)}^{2} + \dots + {(adj A)}^{10} .$ Then, the sum of all the elements of the matrix $B$ is: [2024]

Q 3 :
Let $A$ and $B$ be two square matrices of order 3 such that $| A | = 3$ and $| B | = 2 .$ Then $| A^{T} A (adj {(2 A))}^{- 1} (adj (4 B)) (adj {(A B))}^{- 1} A A^{T} |$ is equal to : [2024]

Q 5 :
If $A$ is a square matrix of order 3 such that $\det (A) = 3$ and $\det (adj (- 4 adj (- 3 adj (3 adj ({(2 A)}^{- 1}))))) = 2^{m} 3^{n},$ then $m + 2 n$ is equal to : [2024]

Q 6 :
Let $B = [\begin{matrix} 1 & 3 \\ 1 & 5 \end{matrix}]$ and $A$ be a $2 \times 2$ matrix such that $A B^{- 1} = A^{- 1} .$ If $B C B^{- 1} = A$ and $C^{4} + α C^{2} + β I = O,$ then $2 β - α$ is equal to [2024]

Q 8 :
Let $A$ be a $2 \times 2$ symmetric matrix such that $A [\begin{matrix} 1 \\ 1 \end{matrix}] = [\begin{matrix} 3 \\ 7 \end{matrix}]$ and the determinant of $A$ be 1. If $A^{- 1} = α A + β I,$ where $I$ is an identity matrix of order $2 \times 2,$ then $α + β$ equals _______ . [2024]

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Q 9 :
Let $A$ be a non-singular matrix of order 3. If $\det (3 adj(2 adj((\det A) A))) = 3^{- 13} \cdot 2^{- 10}$ and $\det (3 adj (2 A)) = 2^{m} \cdot 3^{n},$ then $| 3 m + 2 n |$ is equal to ______. [2024]

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Q 10 :
Let $A$ be a $3 \times 3$ matrix and det(A) = 2. If $n = \det \underset{2024 - t i m e s}{\underset{⏟}{(a d j (a d j (. . . (a d j A))))}},$ then the remainder when $n$ is divided by 9 is equal to ________ . [2024]

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Topic Question Set

Q 1 : Let α∈(0,∞) and A=[12α101012]. If det(adj(2A-AT))·adj(A-2AT)=28, then (det(A))2 is equal to [2024]

Q 2 : Let A=[1201] and B=I+adj(A)+(adjA)2+⋯+(adjA)10. Then, the sum of all the elements of the matrix B is: [2024]

Q 3 : Let A and B be two square matrices of order 3 such that |A|=3 and |B|=2. Then |ATA(adj(2A))-1(adj(4B))(adj(AB))-1AAT| is equal to : [2024]

Q 4 : Let αβ≠0 and A=[βα3ααβ-βα2α]. If B=[3α-93α-α7-2α-2α5-2β] is the matrix of cofactors of the elements of A, then det(AB) is equal to : [2024]

Q 5 : If A is a square matrix of order 3 such that det(A)=3 and det(adj(-4adj(-3adj(3adj((2A)-1)))))=2m3n, then m+2n is equal to : [2024]

Q 6 : Let B=[1315] and A be a 2×2 matrix such that AB-1=A-1. If BCB-1=A and C4+αC2+βI=O, then 2β-α is equal to [2024]

Q 8 : Let A be a 2×2 symmetric matrix such that A[11]=[37] and the determinant of A be 1. If A-1=αA+βI, where I is an identity matrix of order 2×2, then α+β equals _______ . [2024]

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Q 9 : Let A be a non-singular matrix of order 3. If det(3 adj(2 adj((detA)A)))=3-13·2-10 and det(3 adj(2A))=2m·3n, then |3m+2n| is equal to ______. [2024]

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Q 10 : Let A be a 3×3 matrix and det(A) = 2. If n=det(adj(adj(...(adj A))))⏟2024-times, then the remainder when n is divided by 9 is equal to ________ . [2024]

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Q 1 :
Let $α \in (0, \infty)$ and $A = [\begin{matrix} 1 & 2 & α \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{matrix}] .$ If $\det (adj (2 A - A^{T})) \cdot adj (A - 2 A^{T}) = 2^{8},$ then $(\det {(A))}^{2}$ is equal to [2024]

Q 2 :
Let $A = [\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}]$ and $B = I + adj (A) + {(adj A)}^{2} + \dots + {(adj A)}^{10} .$ Then, the sum of all the elements of the matrix $B$ is: [2024]

Q 3 :
Let $A$ and $B$ be two square matrices of order 3 such that $| A | = 3$ and $| B | = 2 .$ Then $| A^{T} A (adj {(2 A))}^{- 1} (adj (4 B)) (adj {(A B))}^{- 1} A A^{T} |$ is equal to : [2024]

Q 5 :
If $A$ is a square matrix of order 3 such that $\det (A) = 3$ and $\det (adj (- 4 adj (- 3 adj (3 adj ({(2 A)}^{- 1}))))) = 2^{m} 3^{n},$ then $m + 2 n$ is equal to : [2024]

Q 6 :
Let $B = [\begin{matrix} 1 & 3 \\ 1 & 5 \end{matrix}]$ and $A$ be a $2 \times 2$ matrix such that $A B^{- 1} = A^{- 1} .$ If $B C B^{- 1} = A$ and $C^{4} + α C^{2} + β I = O,$ then $2 β - α$ is equal to [2024]

Q 8 :
Let $A$ be a $2 \times 2$ symmetric matrix such that $A [\begin{matrix} 1 \\ 1 \end{matrix}] = [\begin{matrix} 3 \\ 7 \end{matrix}]$ and the determinant of $A$ be 1. If $A^{- 1} = α A + β I,$ where $I$ is an identity matrix of order $2 \times 2,$ then $α + β$ equals _______ . [2024]

Q 9 :
Let $A$ be a non-singular matrix of order 3. If $\det (3 adj(2 adj((\det A) A))) = 3^{- 13} \cdot 2^{- 10}$ and $\det (3 adj (2 A)) = 2^{m} \cdot 3^{n},$ then $| 3 m + 2 n |$ is equal to ______. [2024]

Q 10 :
Let $A$ be a $3 \times 3$ matrix and det(A) = 2. If $n = \det \underset{2024 - t i m e s}{\underset{⏟}{(a d j (a d j (. . . (a d j A))))}},$ then the remainder when $n$ is divided by 9 is equal to ________ . [2024]