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

Q 21 :
If $A$ is a $3 \times 3$ matrix and $| A | = 2$ , then $| 3 adj (| 3 A | A^{2}) |$ is equal to [2023]

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

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(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]

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

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(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

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(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

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(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

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(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
$2 \sqrt{3}$

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(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]

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

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(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]

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

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(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
$| Adj P |$ $= 1$

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(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]

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(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 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 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 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]