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

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 $m, n \in {0, 1, 2, . . ., 20}$ , then m + n is equal to : [2025]

16
17
15
14

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2

3

4

(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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2

3

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]

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

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

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

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

1

2

3

4

(4)

We have, |adj (adj (adj 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

3

4

(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

3

4

(2)

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

= 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]

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

1

2

3

4

(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]

0%

34

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

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

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

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

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

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

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

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

$∴$ m = 7 and n = 3

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

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 $\det (adj (2 (A + I))) = a^{α} 3^{β} 5^{γ}, α, β, γ \in ℕ$ , then $α^{2} + β^{2} + γ^{2}$ is __________. [2025]

0%

44

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

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

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

On comparing cofficients, we get

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

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

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

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

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

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

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

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

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

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

$∴ α = 6, β = 2, γ = 2$

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

Topic Question Set

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 |adj (adj (adj 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]

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Topic Question Set

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 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 3×3matrix 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]

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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)))=aα3β5γ, α, β, γ∈ℕ, then α2+β2+γ2 is __________. [2025]

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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 |adj (adj (adj 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]