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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(2)

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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(4)

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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(3)

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

1

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(4)

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

(4)

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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3

4

(3)

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

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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(4)

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]

(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]

(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]

(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.

Q 11 :

Let $a \in R$ and A be a matrix of order $3 \times 3$ such that $\det (A) = - 4 and A + I = [\begin{matrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{matrix}]$ , where $I$ is the identity matrix of order $3 \times 3$ . If det((a + 1) adj((a – 1A)) is $2^{m} 3^{n},$ $m, n \in {0, 1, 2, . . ., 20}$ , then m + n is equal to : [2025]

16
17
15
14

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(1)

We have, $A = [\begin{matrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{matrix}] - I = [\begin{matrix} 0 & a & 1 \\ 2 & 0 & 0 \\ a & 1 & 1 \end{matrix}]$

Also, $| A |$ $= - 4 \Rightarrow [\begin{matrix} 0 & a & 1 \\ 2 & 0 & 0 \\ a & 1 & 1 \end{matrix}] = - 4 \Rightarrow - 2 (a - 1) = - 4 \Rightarrow a = 3$

Now, det((a + 1) adj((a – 1)A)) = |4 adj(2A)|

$= 4^{3}$ $| a d j (2 A) |$ $= 4^{3}$ $| 2 A |^{3 - 1} = 4^{3} | 2 A |^{2} = 4^{3} {(2)}^{6} | A |^{2}$

$= 4^{3} \times 2^{6} \times {(- 4)}^{2} = 4^{5} \times 2^{6} = {(2^{2})}^{5} \times 2^{6} = 2^{16} = 2^{m} \times 3^{n}$

$\Rightarrow$ m = 16 and n = 0

$∴$ m + n = 16 + 0 = 16.

Q 12 :

Let A be a matrix of order $3 \times 3$ and |A| = 5. If $| 2 a d j (3 A a d j (2 A)) | = 2^{α} \cdot 3^{β} \cdot 5^{γ}, α, β, γ \in N$ , then $α + β + γ$ is equal to [2025]

28
25
27
26

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4

(3)

We have, |2 adj(3A adj(2A))|

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

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

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

$= 2^{3} \cdot 3^{6} \cdot | A |^{2} [{(2)}^{6} \cdot | A |^{2}]^{2}$

$= 2^{3} \cdot 3^{6} \cdot | A |^{2} \cdot 2^{12} \cdot | A |^{4} = 2^{15} \cdot 3^{6} \cdot | A |^{6}$

$= 2^{15} \cdot 3^{6} \cdot 5^{6} = 2^{α} \cdot 3^{β} \cdot 5^{γ}$ [ $∵$ $| A |$ = 5]

By comparing, we get

$α = 15, β = 6, γ = 6$

$∴ α + β + γ = 27$ .

Q 13 :

Let A be a $3 \times 3$ matrix such that $| a d j (a d j (a d j A)) |$ = 81. If $S = {n \in Z : (| adj {(adj A) |)}^{\frac{{(n - 1)}^{2}}{2}} = | A |^{(3 n^{2} - 5 n - 4)}}$ , then $\sum_{n \in S}$ $| A^{(n^{2} + n)} |$ is equal to [2025]

820
750
866
732

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(4)

We have, $| a d j (a d j (a d j A)) |$ = 81

$\Rightarrow | adj A |^{4} = 81 \Rightarrow | adj A | = 3$

$\Rightarrow | A |^{2} = 3 \Rightarrow | A | = \sqrt{3}$

Now, ${(| A |^{4})}^{\frac{{(n - 1)}^{2}}{2}} =$ $| A |^{(3 n^{2} - 5 n - 4)}$

$\Rightarrow 2 {(n - 1)}^{2} = 3 n^{2} - 5 n - 4$

$\Rightarrow 2 n^{2} - 4 n + 2 = 3 n^{2} - 5 n - 4$

$\Rightarrow n^{2} - n - 6 = 0 \Rightarrow (n - 3) (n + 2) = 0$

$\Rightarrow n = 3, - 2$

So, $\sum_{n \in S}$ $| A^{(n^{2} + n)} |$ $=$ $| A^{2} |$ $+$ $| A^{12} |$ $= 3 + 729 = 732$ .

Q 14 :

Let $A = [\begin{matrix} 2 & 2 + p & 2 + p + q \\ 4 & 6 + 2 p & 8 + 3 p + 2 q \\ 6 & 12 + 3 p & 20 + 6 p + 3 q \end{matrix}]$ . If det (adj (adj (3A))) = $2^{m} \cdot 3^{n}$ , m, n $\in$ N, then m + n is equal to [2025]

20
24
26
22

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(2)

$A = [\begin{matrix} 2 & 2 + p & 2 + p + q \\ 4 & 6 + 2 p & 8 + 3 p + 2 q \\ 6 & 12 + 3 p & 20 + 6 p + 3 q \end{matrix}]$

$R_{2} \to R_{2} - 2 R_{1} and R_{3} \to R_{3} - 3 R_{1}$

$A = [\begin{matrix} 2 & 2 + p & 2 + p + q \\ 0 & 2 & 4 + p \\ 0 & 6 & 14 + 3 p \end{matrix}]$

$∴ | A | = 2 (28 + 6 p - 24 - 6 p) = 8 = 2^{3}$

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

$= | 3 A |^{4} = {(3^{3} | A |)}^{4} = 3^{12} {(2^{3})}^{4} = 2^{12} 3^{12}$

$∴ m = n = 12 \Rightarrow m + n = 24$ .

Q 15 :

For a $3 \times 3$ matrix M, let trace (M) denote the sum of all the diagonal elements of M. Let A be a $3 \times 3$ matrix such that $| A |$ $= \frac{1}{2}$ and trace (A) = 3. If B = adj (adj (2A)), then the value of $| B |$ + trace (B) equals : [2025]

56
280
132
174

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(2)

Given, $| A |$ $= \frac{1}{2}$ , trace (A) = 3

B = adj (adj (2A)) [ $∵ adj (adj A) = | A |^{n - 2} A$ ]

$= | 2 A |^{3 - 2} (2 A) = 2^{3} | A | (2 A) = 8 A$

$\Rightarrow trace (B) = 8 trace (A) = 8 \times 3 = 24$

Hence, $| B |$ + trace (B)

$= 8^{3}$ $| A |$ $+ 24 = \frac{8^{3}}{2} + 24 = 280$ .

Q 16 :

If A, B and $(adj (A^{- 1}) + adj (B^{- 1}))$ are non-singular matrices of same order, then the inverse of $A (adj (A^{- 1}) + adj {(B^{- 1}))}^{- 1} B$ , is equal to [2025]

$A B^{- 1} + A^{- 1} B$
$\frac{A B^{- 1}}{| A |} + \frac{B A^{- 1}}{| B |}$
$adj (B^{- 1}) + adj (A^{- 1})$
$\frac{1}{| A B |} (adj (B) + adj (A))$

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(4)

${[A (adj (A^{- 1}) + adj {(B^{- 1}))}^{- 1} B]}^{- 1}$

$= B^{- 1} (adj (A^{- 1}) + adj (B^{- 1})) A^{- 1}$

$= B^{- 1} (adj (A^{- 1})) A^{- 1} + B^{- 1} (adj (B^{- 1})) A^{- 1}$

$= B^{- 1}$ $| A^{- 1} |$ $+$ $| B^{- 1} |$ $A^{- 1}$

$= \frac{B^{- 1}}{| A |} + \frac{A^{- 1}}{| B |} = \frac{adj (B)}{| B | | A |} + \frac{adj (A)}{| A | | B |}$

$= \frac{1}{| A | | B |} (adj (B) + adj (A))$

$= \frac{1}{| A B |} (adj (B) + adj (A))$ .

Q 17 :

Let A be a square matrix of order 3 such that det(A) = – 2 and det(3 adj(– 6 adj(3A))) = $2^{m + n} \cdot 3^{m n}$ , m > n. Then 4m + 2n is equal to __________. [2025]

(34)

We have, $| A |$ = – 2 and det(3 adj(– 6 adj(3A))) = $2^{m + n} \cdot 3^{m n}$

$\Rightarrow 3^{3} \det (adj (- 6 adj (3 A))) = 2^{m + n} \cdot 3^{m n}$

$\Rightarrow 3^{3} | - 6 adj (3 A) |^{2} = 2^{m + n} \cdot 3^{m n}$

$\Rightarrow 3^{3} ({{(- 6)}^{3})}^{2}$ $| adj (3 A) |^{2}$ $= 2^{m + n} \cdot 3^{m n}$

$\Rightarrow 3^{3} {(6)}^{6}$ $| {(3 A)}^{2} |^{2}$ $= 2^{m + n} \cdot 3^{m n}$

$\Rightarrow 3^{3} \times 6^{6} \times {(3^{3})}^{4} | A |^{4} = 2^{m + n} \cdot 3^{m n}$

$\Rightarrow 3^{21} \times 2^{6} \times {(- 2)}^{4} = 2^{m + n} \cdot 3^{m n}$ $\Rightarrow 3^{21} \times 2^{10} = 2^{m + n} \cdot 3^{m n}$

On comparing the powers, we get m + n = 10 and mn = 21

$∴$ m = 7 and n = 3

$∴$ Value of $4 m + 2 n = 4 \times 7 + 2 \times 3 = 34$ .

Q 18 :

Let A be a $3 \times 3$ matrix such that $X^{T} A X =' O'$ for all nonzero $3 \times 1 matrices X = [\begin{matrix} x \\ y \\ z \end{matrix}]$ . If $A [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] = [\begin{matrix} 1 \\ 4 \\ - 5 \end{matrix}], A [\begin{matrix} 1 \\ 2 \\ 1 \end{matrix}] = [\begin{matrix} 0 \\ 4 \\ - 8 \end{matrix}]$ , and $\det (adj (2 (A + I))) = 2^{α} 3^{β} 5^{γ}, α, β, γ \in ℕ$ , then $α^{2} + β^{2} + γ^{2}$ is __________. [2025]

(44)

Given, $X^{T} A X = O$

$∴ [\begin{matrix} X & Y & Z \end{matrix}] [\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}] [\begin{matrix} X \\ Y \\ Z \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}], where A = [\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}]$

$\Rightarrow X (a_{1} X + a_{2} Y + a_{3} Z) + Y (b_{1} X + b_{2} Y + b_{3} Z) + Z (c_{1} X + c_{2} Y + c_{3} Z) = 0$

On comparing cofficients, we get

$\Rightarrow a_{1} = 0, b_{2} = 0, c_{3} = 0 and a_{2} + b_{1} = 0, a_{3} + c_{1} = 0, b_{3} + c_{2} = 0$

$∴ A = [\begin{matrix} 0 & a_{2} & a_{3} \\ - a_{2} & 0 & b_{3} \\ - a_{3} & - b_{3} & 0 \end{matrix}] = [\begin{matrix} 0 & x & y \\ - x & 0 & z \\ - y & - z & 0 \end{matrix}] (Let)$

$∴ A = [\begin{matrix} 0 & x & y \\ - x & 0 & z \\ - y & - z & 0 \end{matrix}]$ , which is skew-symmetric matrix

Given, $A [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] = [\begin{matrix} 1 \\ 4 \\ - 5 \end{matrix}] \Rightarrow A = [\begin{matrix} 0 & x & y \\ - x & 0 & z \\ - y & - z & 0 \end{matrix}] [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}] = [\begin{matrix} 1 \\ 4 \\ - 5 \end{matrix}]$

x + y = 1, – x + z = 4, y + z = – 5

$[\begin{matrix} 0 & x & y \\ - z & 0 & z \\ - y & - x & 0 \end{matrix}] [\begin{matrix} 1 \\ 2 \\ 1 \end{matrix}] = [\begin{matrix} 0 \\ 4 \\ - 8 \end{matrix}]$

2x + y = 0, – x + z = 4, – y – 2z = – 8

$\Rightarrow x = - 1, y = 2, z = 3$

$∴ A = [\begin{matrix} 0 & - 1 & 2 \\ 1 & 0 & 3 \\ - 2 & - 3 & 0 \end{matrix}]$

$∴ 2 (A + I) = [\begin{matrix} 2 & - 2 & 4 \\ 2 & 2 & 6 \\ - 4 & - 6 & 2 \end{matrix}]$

$∴ \det (adj (2 (A + I))) = | 2 (A + I) |^{2} = {(120)}^{2}$

$2^{α} 3^{β} 5^{γ} = 2^{6} \times 3^{2} \times 5^{2}$

$∴ α = 6, β = 2, γ = 2$

$∴ α^{2} + β^{2} + γ^{2} = 36 + 4 + 4 = 44$ .

Q 19 :

Let $A =$ $[\begin{matrix} 2 & 1 & 0 \\ 1 & 2 & - 1 \\ 0 & - 1 & 2 \end{matrix}]$ . If $| adj (adj (adj (2 A))) |$ $= {(16)}^{n},$ then $n$ is equal to [2023]

9
10
12
8

1

2

3

4

(2)

We have,
$A = [\begin{matrix} 2 & 1 & 0 \\ 1 & 2 & - 1 \\ 0 & - 1 & 2 \end{matrix}] \Rightarrow | A | = 4$

Now, $| adj (adj (adj (2 A))) | = | 2 A |^{{(n - 1)}^{3}} = | 2 A |^{2^{3}} = | 2 A |^{8}$

$= {[2^{3} | A |]}^{8} = 8^{8} \cdot 4^{8} = 32^{8} = 2^{8} \cdot 16^{8} = 16^{2} \cdot 16^{8} = 16^{n}$

$\Rightarrow n = 10$

Q 20 :

If $A = [\begin{matrix} 1 & 5 \\ λ & 10 \end{matrix}], A^{- 1} = α A + β I and α + β = - 2,$ then $4 α^{2} + β^{2} + λ^{2}$ is equal to [2023]

10
14
19
12

1

2

3

4

(2)

Given, $A = [\begin{matrix} 1 & 5 \\ λ & 10 \end{matrix}] \Rightarrow | A | = 10 - 5 λ$

Then, $A^{- 1} = [\begin{matrix} \frac{10}{10 - 5 λ} & \frac{- 5}{10 - 5 λ} \\ \frac{- λ}{10 - 5 λ} & \frac{1}{10 - 5 λ} \end{matrix}] = α [\begin{matrix} 1 & 5 \\ λ & 10 \end{matrix}] + β [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} α + β & 5 α \\ α λ & 10 α + β \end{matrix}]$

So, $α + β = \frac{10}{10 - 5 λ}; Given α + β = - 2$

$\Rightarrow - 2 = \frac{10}{10 - 5 λ} \Rightarrow - 20 + 10 λ = 10 \Rightarrow λ = 3$

Also, $5 α = \frac{- 5}{10 - 5 λ} \Rightarrow 5 α = \frac{- 5}{10 - 15}$

$\Rightarrow α = \frac{1}{5}$ then $β = \frac{- 11}{5}$

Now, $4 α^{2} + β^{2} + λ^{2} = 4 {(\frac{1}{5})}^{2} + {(\frac{- 11}{5})}^{2} + 3^{2}$

$= \frac{4}{25} + \frac{121}{25} + 9 = 14$

Q 21 :

If $A$ is a $3 \times 3$ matrix and $| A | = 2$ , then $| 3 adj (| 3 A | A^{2}) |$ is equal to [2023]

$3^{11} \cdot 6^{10}$
$3^{12} \cdot 6^{11}$
$3^{10} \cdot 6^{11}$
$3^{12} \cdot 6^{10}$

1

2

3

4

(1)

We have, $| A |$ $= 2 .$

$∴$ $| 3 A |$ $= 3^{3} | A | = 27 \times 2 = 54$

Now, $| 3 adj (| 3 A | A^{2}) |$ $=$ $| 3 adj (54 A^{2}) |$ $=$ $| 3 \times 54^{2} adj (A^{2}) |$

$= {(3 \times 54^{2})}^{3} | adj A |^{2}$

$= 3^{3} \times {(54^{2})}^{3} {(| A |^{2})}^{2} = 3^{3} \cdot 2^{4} \cdot 2^{6} \cdot 27^{6} = 3^{11} 6^{10}$

Q 22 :

If $A = \frac{1}{5! 6! 7!} [\begin{matrix} 5! & 6! & 7! \\ 6! & 7! & 8! \\ 7! & 8! & 9! \end{matrix}],$ then $| adj (adj (2 A)) |$ is equal to [2023]

$2^{12}$
$2^{20}$
$2^{8}$
$2^{16}$

1

2

3

4

(4)

Given $A = \frac{1}{5! 6! 7!} [\begin{matrix} 5! & 6! & 7! \\ 6! & 7! & 8! \\ 7! & 8! & 9! \end{matrix}]$

$| A |$ $= \frac{1}{5! 6! 7!} 5! 6! 7!$ $| \begin{matrix} 1 & 6 & 42 \\ 1 & 7 & 56 \\ 1 & 8 & 72 \end{matrix} |$

Applying $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{2}$ , we get

$| A |$ $=$ $| \begin{matrix} 1 & 6 & 42 \\ 0 & 1 & 14 \\ 0 & 1 & 16 \end{matrix} |$

Now, $| adj (adj (2 A)) |$ $= | 2 A |^{{(n - 1)}^{2}} = | 2 A |^{4} = {(2^{3} | A |)}^{4} = 2^{12} | A |^{4} = 2^{12} 2^{4} = 2^{16}$

Q 23 :

Let $B = [\begin{matrix} 1 & 3 & α \\ 1 & 2 & 3 \\ α & α & 4 \end{matrix}], α > 2$ be the adjoint of a matrix $A$ and $| A |$ $= 2$ . Then $[α - 2 α α] B [\begin{matrix} α \\ - 2 α \\ α \end{matrix}] is equal to ?$ [2023]

0
16
32
- 16

1

2

3

4

(4)

We have, $adj A = B = [\begin{matrix} 1 & 3 & α \\ 1 & 2 & 3 \\ α & α & 4 \end{matrix}], α > 2$ and $| A |$ $= 2$

$Now, [α - 2 α α] B [\begin{matrix} α \\ - 2 α \\ α \end{matrix}]$

$= [α - 2 α α] [\begin{matrix} 1 & 3 & α \\ 1 & 2 & 3 \\ α & α & 4 \end{matrix}] [\begin{matrix} α \\ - 2 α \\ α \end{matrix}]$

$= [α - 2 α α] [\begin{matrix} α - 6 α + α^{2} \\ α - 4 α + 3 α \\ α^{2} - 2 α^{2} + 4 α \end{matrix}]$

$= α (α - 6 α + α^{2}) - 2 α (α - 4 α + 3 α) + α (α^{2} - 2 α^{2} + 4 α)$

$= α^{2} - 6 α^{2} + α^{3} - 2 α^{2} + 8 α^{2} - 6 α^{2} + α^{3} - 2 α^{3} + 4 α^{2} = - α^{2}$

$Now,$ $| B |$ $=$ $| adj A |$ $= {(| A |)}^{2} = 2^{2} = 4$

$∴$ $8 - 3 α - 12 + 9 α - α^{2} = 4$

$\Rightarrow α^{2} - 6 α + 8 = 0 \Rightarrow (α - 2) (α - 4) = 0 \Rightarrow α = 2, 4$

$But α > 2$ $∴$ $α = 4$

$∴$ $[α - 2 α α] B [\begin{matrix} α \\ - 2 α \\ α \end{matrix}] = - {(4)}^{2} = - 16$

Q 24 :

Let $A = [\begin{matrix} 1 & 2 & 3 \\ α & 3 & 1 \\ 1 & 1 & 2 \end{matrix}],$ $| A |$ $= 2 .$ If $| 2 adj (2 adj (2 A)) | = 32^{n},$ then $3 n + α$ is equal to [2023]

10
11
9
12

1

2

3

4

(2)

We have, $| 2 (adj (2 adj 2 A)) |$

$=$ $| 2 (adj (8 adj A)) |$ $(∵ adj (k A) = k^{n - 1} adj (A))$

$=$ $| 2 (64 adj (adj A)) |$ $=$ $| 128 adj (adj A)) |$

$= {(128)}^{3}$ $| adj (adj A) |$ $= {(128)}^{3} | A |^{4}$

$\Rightarrow {(128)}^{3} {(2)}^{4} = 32^{n} \Rightarrow {(2^{7})}^{3} {(2)}^{4} = {(2^{5})}^{n}$

$\Rightarrow 2^{25} = 2^{5 n} \Rightarrow 25 = 5 n \Rightarrow n = 5$

Also, $| A |$ $= 2 \Rightarrow$ $| \begin{matrix} 1 & 2 & 3 \\ α & 3 & 1 \\ 1 & 1 & 2 \end{matrix} |$ $= 2$

$\Rightarrow 1 (6 - 1) - 2 (2 α - 1) + 3 (α - 3) = 2$

$\Rightarrow 5 - 4 α + 2 + 3 α - 9 = 2 \Rightarrow - α - 4 = 0 \Rightarrow α = - 4$

$So, 3 n + α = 3 (5) - 4 = 11$

Q 25 :

Let the determinant of a square matrix $A$ of order $m$ be $m - n$ , where $m$ and $n$ satisfy $4 m + n = 22 and 17 m + 4 n = 93 .$ If $det(n adj(adj(mA))) = 3^{a} 5^{b} 6^{c},$ then $a + b + c$ is equal to [2023]

109
101
84
96

1

2

3

4

(4)

Given $4 m + n = 22$

$17 m + 4 n = 93$

Solving the above two equations, we get $m = 5$ and $n = 2$ .

$∴$ $A is a square matrix of order 5 and$ $| A |$ $= 5 - 2 = 3$

Now, we know that $adj (k A) = k^{n - 1} adj (A),$ where $A$ is a matrix of order $n$

$∴$ $adj (m A) = adj (5 A) = 5^{5 - 1} (adj A) = 5^{4} (adj A)$

Again, $adj (5^{4} adj A) = {(5^{4})}^{4} adj A (adj A) = 5^{16} | A |^{5 - 2} \cdot A = 5^{16} 3^{3} A$

Now, $\det (n adj (adj m A)) = \det (2 \cdot 5^{16} \cdot 3^{3} \cdot A)$

$= (2 \cdot 5^{16} \cdot 3^{5}) \det A = 2^{5} \cdot 5^{80} \cdot 3^{15} \cdot 3 = 2^{5} \cdot 5^{80} \cdot 3^{16} = 6^{5} \cdot 5^{80} \cdot 3^{11}$

$∴$ $a = 11, b = 80, c = 5$

$\Rightarrow a + b + c = 80 + 11 + 5 = 96$

Q 26 :

$Let A be a 3 \times 3 matrix such that$ $| adj (adj (adj A)) |$ $= 12^{4} . Then | A^{- 1} adj A | is equal to$ [2023]

1
$\sqrt{6}$
12
$2 \sqrt{3}$

1

2

3

4

(4)

Given, $| adj (adj (adj A)) |$ $= 12^{4}$

$\Rightarrow$ $| A |^{{(n - 1)}^{3}}$ $= 12^{4}$

At $n = 3,$ $| A |^{8} = 12^{4} \Rightarrow | A |^{2} = 12 \Rightarrow$ $| A |$ $= 2 \sqrt{3}$

$∴$ $| A^{- 1} adj A |$ $=$ $| A^{- 1} |$ $\cdot$ $| adj A |$ $= \frac{1}{| A |} \cdot | A |^{2} =$ $| A |$ $= 2 \sqrt{3}$

Q 27 :

$Let x, y, z > 1 and A = [\begin{matrix} 1 & \log_{x} y & \log_{x} z \\ \log_{y} x & 2 & \log_{y} z \\ \log_{z} x & \log_{z} y & 3 \end{matrix}] .$ Then $| adj (adj A^{2}) |$ is equal to [2023]

$4^{8}$
$2^{8}$
$2^{4}$
$6^{4}$

1

2

3

4

(2)

$| A |$ $=$ $| \begin{matrix} \log_{x} x & \log_{x} y & \log_{x} z \\ \log_{y} x & 2 \log_{y} y & \log_{y} z \\ \log_{z} x & \log_{z} y & 3 \log_{z} z \end{matrix} |$

$= \frac{1}{\log x \cdot \log y \cdot \log z}$ $| \begin{matrix} \log x & \log y & \log z \\ \log x & 2 \log y & \log z \\ \log x & \log y & 3 \log z \end{matrix} |$ $[∵ \log_{a} b = \frac{\log b}{\log a}]$

$= \frac{\log x \log y \log z}{\log x \log y \log z} \times$ $| \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 3 \end{matrix} |$ $= 2$

$∴$ $| adj (adj (A^{2})) |$ $= | adj (A^{2}) |^{2}$ $= {(| A^{2} |^{2})}^{2} = | A |^{8} = 2^{8}$

Q 28 :

$Let A = (\begin{matrix} m & n \\ p & q \end{matrix}), d = | A | \neq 0 and$ $| A - d (Adj A) |$ $= 0 .$ Then [2023]

$1 + d^{2} = m^{2} + q^{2}$
$1 + d^{2} = {(m + q)}^{2}$
${(1 + d)}^{2} = m^{2} + q^{2}$
${(1 + d)}^{2} = {(m + q)}^{2}$

1

2

3

4

(4)

We have, A = $[\begin{matrix} m & n \\ p & q \end{matrix}]$ , $d =$ $| A |$ $\neq 0 .$

$and$ $| A - d (Adj A) | = 0$ $∴$ $Adj (A) =$ $[\begin{matrix} q & - n \\ - p & m \end{matrix}]$

$\Rightarrow$ $| A - d (Adj A) |$ $=$ $| [\begin{matrix} m & n \\ p & q \end{matrix}] - d [\begin{matrix} q & - n \\ - p & m \end{matrix}] |$
$=$ $| \begin{matrix} m - d q & n + d n \\ p + d p & q - d m \end{matrix} |$

$\Rightarrow (m - d q) (q - d m) - (n + d n) (p + d p) = 0$

$\Rightarrow m q - d m^{2} - d q^{2} + d^{2} m q - n p - d p n - d n p - d^{2} n p = 0$

$\Rightarrow (m q - n p) - d (m^{2} + q^{2} + 2 p n) + d^{2} (m q - n p) = 0$

$\Rightarrow (1 + d^{2}) (m q - n p) - d (m^{2} + q^{2} + 2 p n) = 0$

$\Rightarrow (1 + d^{2}) (m q - n p) - d (m^{2} + q^{2} + 2 m q - 2 m q + 2 p n) = 0$

$\Rightarrow (1 + d^{2}) (m q - n p) - d {(m + q)}^{2} + 2 d (m q - p n) = 0$

$\Rightarrow (1 + d^{2} + 2 d) (m q - p n) - d {(m + q)}^{2} = 0$

$\Rightarrow (1 + d^{2}) (m q - p n) = d {(m + q)}^{2}$

$\Rightarrow (1 + d^{2}) = {(m + q)}^{2} ∴ d = m q - p n .$

Q 29 :

If P is a $3 \times 3$ real matrix such that $P^{T} = a P + (a - 1) I,$ where $a > 1$ , then [2023]

$| Adj P |$ $> 1$
$| Adj P |$ $= \frac{1}{2}$
P is a singular matrix
$| Adj P |$ $= 1$

1

2

3

4

(4)

$P is a 3 \times 3 real matrix$

$∴ P^{T} = a P + (a - 1) I, a > 1 \Rightarrow {(P^{T})}^{T} = a P^{T} + (a - 1) I$

$\Rightarrow P = a P^{T} + (a - 1) I \Rightarrow P = a (a P + (a - 1) I) + (a - 1) I$

$\Rightarrow P = a^{2} P + a (a - 1) I + (a - 1) I$

$\Rightarrow P - a^{2} P = (a - 1) I (a + 1) = (a^{2} - 1) I$

$\Rightarrow P (1 - a^{2}) = (a^{2} - 1) I$

$\Rightarrow P (1 - a^{2}) - (a^{2} - 1) I = 0 \Rightarrow (1 - a^{2}) (P + I) = 0$

$\Rightarrow P = - I \Rightarrow$ $| P |$ $=$ $- 1$

$∴$ $| Adj P |$ $= {(- 1)}^{2} = 1$

Q 30 :

Let A be a $n \times n$ matrix such that $| A |$ = 2. If the determinant of the matrix $Adj (2 \cdot Adj (2 A^{- 1}))$ is $2^{84}$ , then $n$ is equal to ______. [2023]

(5)

$| Adj (2 Adj (2 A^{- 1})) |$

$=$ $| 2 (Adj (2 A^{- 1})) |^{n - 1}$ $= 2^{n (n - 1)}$ $| Adj (2 A^{- 1}) |^{n - 1}$

$= 2^{n (n - 1)}$ $| (2 A^{- 1}) |^{(n - 1) (n - 1)}$ $= 2^{n (n - 1) + n (n - 1) (n - 1)} \cdot \frac{1}{| A |^{{(n - 1)}^{2}}}$

$= \frac{2^{n (n - 1) + n (n - 1) (n - 1)}}{2^{{(n - 1)}^{2}}} 2^{n (n - 1) + n {(n - 1)}^{2} - {(n - 1)}^{2}}$

$= 2^{(n - 1) (n^{2} - n + 1)} = 2^{84}$

$\Rightarrow (n - 1) (n^{2} - n + 1) = 84$

$\Rightarrow (n - 5) (n^{2} + 3 n + 17) = 0 \Rightarrow n = 5$

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]

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]

Q 12 :

Let A be a matrix of order $3 \times 3$ and |A| = 5. If $| 2 a d j (3 A a d j (2 A)) | = 2^{α} \cdot 3^{β} \cdot 5^{γ}, α, β, γ \in N$ , then $α + β + γ$ is equal to [2025]

Q 13 :

Let A be a $3 \times 3$ matrix such that $| a d j (a d j (a d j A)) |$ = 81. If $S = {n \in Z : (| adj {(adj A) |)}^{\frac{{(n - 1)}^{2}}{2}} = | A |^{(3 n^{2} - 5 n - 4)}}$ , then $\sum_{n \in S}$ $| A^{(n^{2} + n)} |$ is equal to [2025]

Q 14 :

Let $A = [\begin{matrix} 2 & 2 + p & 2 + p + q \\ 4 & 6 + 2 p & 8 + 3 p + 2 q \\ 6 & 12 + 3 p & 20 + 6 p + 3 q \end{matrix}]$ . If det (adj (adj (3A))) = $2^{m} \cdot 3^{n}$ , m, n $\in$ N, then m + n is equal to [2025]

Q 15 :

For a $3 \times 3$ matrix M, let trace (M) denote the sum of all the diagonal elements of M. Let A be a $3 \times 3$ matrix such that $| A |$ $= \frac{1}{2}$ and trace (A) = 3. If B = adj (adj (2A)), then the value of $| B |$ + trace (B) equals : [2025]

Q 16 :

If A, B and $(adj (A^{- 1}) + adj (B^{- 1}))$ are non-singular matrices of same order, then the inverse of $A (adj (A^{- 1}) + adj {(B^{- 1}))}^{- 1} B$ , is equal to [2025]

Q 17 :

Let A be a square matrix of order 3 such that det(A) = – 2 and det(3 adj(– 6 adj(3A))) = $2^{m + n} \cdot 3^{m n}$ , m > n. Then 4m + 2n is equal to __________. [2025]

Q 19 :

Let $A =$ $[\begin{matrix} 2 & 1 & 0 \\ 1 & 2 & - 1 \\ 0 & - 1 & 2 \end{matrix}]$ . If $| adj (adj (adj (2 A))) |$ $= {(16)}^{n},$ then $n$ is equal to [2023]

Q 20 :

If $A = [\begin{matrix} 1 & 5 \\ λ & 10 \end{matrix}], A^{- 1} = α A + β I and α + β = - 2,$ then $4 α^{2} + β^{2} + λ^{2}$ is equal to [2023]

Q 21 :

If $A$ is a $3 \times 3$ matrix and $| A | = 2$ , then $| 3 adj (| 3 A | A^{2}) |$ is equal to [2023]

Q 22 :

If $A = \frac{1}{5! 6! 7!} [\begin{matrix} 5! & 6! & 7! \\ 6! & 7! & 8! \\ 7! & 8! & 9! \end{matrix}],$ then $| adj (adj (2 A)) |$ is equal to [2023]

Q 23 :

Let $B = [\begin{matrix} 1 & 3 & α \\ 1 & 2 & 3 \\ α & α & 4 \end{matrix}], α > 2$ be the adjoint of a matrix $A$ and $| A |$ $= 2$ . Then $[α - 2 α α] B [\begin{matrix} α \\ - 2 α \\ α \end{matrix}] is equal to ?$ [2023]

Q 24 :

Let $A = [\begin{matrix} 1 & 2 & 3 \\ α & 3 & 1 \\ 1 & 1 & 2 \end{matrix}],$ $| A |$ $= 2 .$ If $| 2 adj (2 adj (2 A)) | = 32^{n},$ then $3 n + α$ is equal to [2023]

Q 25 :

Let the determinant of a square matrix $A$ of order $m$ be $m - n$ , where $m$ and $n$ satisfy $4 m + n = 22 and 17 m + 4 n = 93 .$ If $det(n adj(adj(mA))) = 3^{a} 5^{b} 6^{c},$ then $a + b + c$ is equal to [2023]

Q 26 :

$Let A be a 3 \times 3 matrix such that$ $| adj (adj (adj A)) |$ $= 12^{4} . Then | A^{- 1} adj A | is equal to$ [2023]

Q 27 :

$Let x, y, z > 1 and A = [\begin{matrix} 1 & \log_{x} y & \log_{x} z \\ \log_{y} x & 2 & \log_{y} z \\ \log_{z} x & \log_{z} y & 3 \end{matrix}] .$ Then $| adj (adj A^{2}) |$ is equal to [2023]

Q 28 :

$Let A = (\begin{matrix} m & n \\ p & q \end{matrix}), d = | A | \neq 0 and$ $| A - d (Adj A) |$ $= 0 .$ Then [2023]

Q 29 :

If P is a $3 \times 3$ real matrix such that $P^{T} = a P + (a - 1) I,$ where $a > 1$ , then [2023]

Q 30 :

Let A be a $n \times n$ matrix such that $| A |$ = 2. If the determinant of the matrix $Adj (2 \cdot Adj (2 A^{- 1}))$ is $2^{84}$ , then $n$ is equal to ______. [2023]

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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]

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]

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]

Q 11 : Let a∈R and A be a matrix of order 3×3 such that det(A)=–4 and A+I=[1a1210a12], where I is the identity matrix of order 3×3. If det((a + 1) adj((a – 1A)) is 2m3n,m, n∈{0,1,2,...,20}, then m + n is equal to : [2025]

Q 12 : Let A be a matrix of order 3×3 and |A| = 5. If |2 adj(3A adj(2A))|=2α·3β·5γ, α,β,γ∈N, then α+β+γ is equal to [2025]

Q 13 : Let A be a 3×3 matrix such that |adj (adj (adj A))| = 81. If S={n∈Z : (|adj (adj A)|)(n–1)22=|A|(3n2–5n–4)}, then ∑n∈S|A(n2+n)| is equal to [2025]

Q 14 : Let A=[22+p2+p+q46+2p8+3p+2q612+3p20+6p+3q]. If det (adj (adj (3A))) = 2m·3n, m, n ∈ N, then m + n is equal to [2025]

Q 15 : For a 3×3 matrix M, let trace (M) denote the sum of all the diagonal elements of M. Let A be a 3×3 matrix such that |A|=12 and trace (A) = 3. If B = adj (adj (2A)), then the value of |B| + trace (B) equals : [2025]

Q 16 : If A, B and (adj(A–1)+adj(B–1)) are non-singular matrices of same order, then the inverse of A(adj(A–1)+adj(B–1))–1B, is equal to [2025]

Q 17 : Let A be a square matrix of order 3 such that det(A) = – 2 and det(3 adj(– 6 adj(3A))) = 2m+n·3mn, m > n. Then 4m + 2n is equal to __________. [2025]

Q 18 : Let A be a 3×3 matrix such that XTAX='O' for all nonzero 3×1 matrices X=[xyz]. If A[111]=[14–5], A[121]=[04–8], and det(adj(2(A+I)))=2α3β5γ, α, β, γ∈ℕ, then α2+β2+γ2 is __________. [2025]

Q 19 : Let A=[21012-10-12]. If |adj(adj(adj(2A)))|=(16)n, then n is equal to [2023]

Q 20 : If A=[15λ10], A-1=αA+βI and α+β=-2, then 4α2+β2+λ2 is equal to [2023]

Q 21 : If A is a 3×3 matrix and |A|=2, then |3 adj(|3A|A2)| is equal to [2023]

Q 22 : If A=15!6!7![5!6!7!6!7!8!7!8!9!], then |adj(adj(2A))| is equal to [2023]

Q 23 : Let B=[13α123αα4], α>2 be the adjoint of a matrix A and |A|=2. Then [α -2α α] B[α-2αα] is equal to ? [2023]

Q 24 : Let A=[123α31112], |A|=2. If |2 adj(2 adj(2A))|=32n, then 3n+α is equal to [2023]

Q 25 : Let the determinant of a square matrix A of order m be m-n, where m and n satisfy 4m+n=22 and 17m+4n=93. If det(n adj(adj(mA)))=3a5b6c, then a+b+c is equal to [2023]

Q 26 : Let A be a 3×3 matrix such that |adj(adj(adjA))|=124. Then |A-1adjA| is equal to [2023]

Q 27 : Let x,y,z>1 and A=[1logxylogxzlogyx2logyzlogzxlogzy3]. Then |adj(adjA2)| is equal to [2023]

Q 28 : Let A=(mnpq), d=|A|≠0 and |A-d(AdjA)|=0. Then [2023]

Q 29 : If P is a 3×3 real matrix such that PT=aP+(a-1)I, where a>1, then [2023]

Q 30 : Let A be a n×n matrix such that |A| = 2. If the determinant of the matrix Adj(2·Adj(2A-1)) is 284, then n is equal to ______. [2023]

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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]

Q 12 :

Let A be a matrix of order $3 \times 3$ and |A| = 5. If $| 2 a d j (3 A a d j (2 A)) | = 2^{α} \cdot 3^{β} \cdot 5^{γ}, α, β, γ \in N$ , then $α + β + γ$ is equal to [2025]

Q 13 :

Let A be a $3 \times 3$ matrix such that $| a d j (a d j (a d j A)) |$ = 81. If $S = {n \in Z : (| adj {(adj A) |)}^{\frac{{(n - 1)}^{2}}{2}} = | A |^{(3 n^{2} - 5 n - 4)}}$ , then $\sum_{n \in S}$ $| A^{(n^{2} + n)} |$ is equal to [2025]

Q 14 :

Let $A = [\begin{matrix} 2 & 2 + p & 2 + p + q \\ 4 & 6 + 2 p & 8 + 3 p + 2 q \\ 6 & 12 + 3 p & 20 + 6 p + 3 q \end{matrix}]$ . If det (adj (adj (3A))) = $2^{m} \cdot 3^{n}$ , m, n $\in$ N, then m + n is equal to [2025]

Q 15 :

For a $3 \times 3$ matrix M, let trace (M) denote the sum of all the diagonal elements of M. Let A be a $3 \times 3$ matrix such that $| A |$ $= \frac{1}{2}$ and trace (A) = 3. If B = adj (adj (2A)), then the value of $| B |$ + trace (B) equals : [2025]

Q 16 :

If A, B and $(adj (A^{- 1}) + adj (B^{- 1}))$ are non-singular matrices of same order, then the inverse of $A (adj (A^{- 1}) + adj {(B^{- 1}))}^{- 1} B$ , is equal to [2025]

Q 17 :

Let A be a square matrix of order 3 such that det(A) = – 2 and det(3 adj(– 6 adj(3A))) = $2^{m + n} \cdot 3^{m n}$ , m > n. Then 4m + 2n is equal to __________. [2025]

Q 19 :

Let $A =$ $[\begin{matrix} 2 & 1 & 0 \\ 1 & 2 & - 1 \\ 0 & - 1 & 2 \end{matrix}]$ . If $| adj (adj (adj (2 A))) |$ $= {(16)}^{n},$ then $n$ is equal to [2023]

Q 20 :

If $A = [\begin{matrix} 1 & 5 \\ λ & 10 \end{matrix}], A^{- 1} = α A + β I and α + β = - 2,$ then $4 α^{2} + β^{2} + λ^{2}$ is equal to [2023]

Q 21 :

If $A$ is a $3 \times 3$ matrix and $| A | = 2$ , then $| 3 adj (| 3 A | A^{2}) |$ is equal to [2023]

Q 22 :

If $A = \frac{1}{5! 6! 7!} [\begin{matrix} 5! & 6! & 7! \\ 6! & 7! & 8! \\ 7! & 8! & 9! \end{matrix}],$ then $| adj (adj (2 A)) |$ is equal to [2023]

Q 23 :

Let $B = [\begin{matrix} 1 & 3 & α \\ 1 & 2 & 3 \\ α & α & 4 \end{matrix}], α > 2$ be the adjoint of a matrix $A$ and $| A |$ $= 2$ . Then $[α - 2 α α] B [\begin{matrix} α \\ - 2 α \\ α \end{matrix}] is equal to ?$ [2023]

Q 24 :

Let $A = [\begin{matrix} 1 & 2 & 3 \\ α & 3 & 1 \\ 1 & 1 & 2 \end{matrix}],$ $| A |$ $= 2 .$ If $| 2 adj (2 adj (2 A)) | = 32^{n},$ then $3 n + α$ is equal to [2023]

Q 25 :

Let the determinant of a square matrix $A$ of order $m$ be $m - n$ , where $m$ and $n$ satisfy $4 m + n = 22 and 17 m + 4 n = 93 .$ If $det(n adj(adj(mA))) = 3^{a} 5^{b} 6^{c},$ then $a + b + c$ is equal to [2023]

Q 26 :

$Let A be a 3 \times 3 matrix such that$ $| adj (adj (adj A)) |$ $= 12^{4} . Then | A^{- 1} adj A | is equal to$ [2023]

Q 27 :

$Let x, y, z > 1 and A = [\begin{matrix} 1 & \log_{x} y & \log_{x} z \\ \log_{y} x & 2 & \log_{y} z \\ \log_{z} x & \log_{z} y & 3 \end{matrix}] .$ Then $| adj (adj A^{2}) |$ is equal to [2023]

Q 28 :

$Let A = (\begin{matrix} m & n \\ p & q \end{matrix}), d = | A | \neq 0 and$ $| A - d (Adj A) |$ $= 0 .$ Then [2023]

Q 29 :

If P is a $3 \times 3$ real matrix such that $P^{T} = a P + (a - 1) I,$ where $a > 1$ , then [2023]

Q 30 :

Let A be a $n \times n$ matrix such that $| A |$ = 2. If the determinant of the matrix $Adj (2 \cdot Adj (2 A^{- 1}))$ is $2^{84}$ , then $n$ is equal to ______. [2023]