PYQs Guides Notes Mocks Board Questions MHT-CET Maths Determinants and Matrices 102 MHT-CET Maths previous year questions (PYQs) from 68 papers, 2021 to 2026, with answers and worked solutions.
About this chapter Difficulty: 14 easy · 52 moderate · 36 hard. Most-asked subtopics: Inverse of a Matrix — Adjoint Formula, Products and Verification (44), Determinants, Cofactors and the Adjoint Identities (27), Cayley–Hamilton, Matrix Polynomials and Powers (17). Updated 13 October 2026. #1 · Determinants, Cofactors and the Adjoint Identities
If w = − 1 2 + i 3 2 w = -\dfrac{1}{2} + \dfrac{i\sqrt{3}}{2} w = − 2 1 + 2 i 3 , then the value of ∣ 1 1 1 1 − 1 − w 2 w 2 1 w 2 w 4 ∣ \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 - w^2 & w^2 \\ 1 & w^2 & w^4 \end{vmatrix} 1 1 1 1 − 1 − w 2 w 2 1 w 2 w 4 is... A B C − 3 w ( w − 1 ) -3w(w - 1) − 3 w ( w − 1 ) D − 3 w ( w + 1 ) -3w(w + 1) − 3 w ( w + 1 ) #2 · Determinants, Cofactors and the Adjoint Identities
If A = [ − 1 2 − 3 2 3 2 − 1 2 ] A = \begin{bmatrix} -\dfrac{1}{2} & -\dfrac{\sqrt{3}}{2} \\ \dfrac{\sqrt{3}}{2} & -\dfrac{1}{2} \end{bmatrix} A = − 2 1 2 3 − 2 3 − 2 1 , the value of ∣ A d j ( A 2026 ) ∣ |\mathrm{Adj}(A^{2026})| ∣ Adj ( A 2026 ) ∣ is equal to: #3 · Determinants, Cofactors and the Adjoint Identities
If A = [ 1 − 1 1 0 2 − 3 2 1 0 ] A = \begin{bmatrix} 1 & -1 & 1 \\ 0 & 2 & -3 \\ 2 & 1 & 0 \end{bmatrix} A = 1 0 2 − 1 2 1 1 − 3 0 , B = a d j A B = \mathrm{adj}\ A B = adj A and C = 5 A C = 5A C = 5 A , then the value of ∣ a d j B ∣ ∣ C ∣ \dfrac{|\mathrm{adj}\ B|}{|C|} ∣ C ∣ ∣ adj B ∣ is: #4 · Determinants, Cofactors and the Adjoint Identities
If matrix ( x 1 2 2 4 x − 3 3 2 ) \begin{pmatrix} x & 1 & 2 \\ 2 & 4 & x \\ -3 & 3 & 2 \end{pmatrix} x 2 − 3 1 4 3 2 x 2 is non-invertible matrix and distinct values of x x x are a a a and b b b then 9 ( a 2 + b 2 ) = ⋅ s ⋅ s ⋅ s 9(a^2 + b^2) = \cdot s\cdot s\cdot s 9 ( a 2 + b 2 ) = ⋅ s ⋅ s ⋅ s MHT-CET · 2026 · 12 May Shift I · Q68 Hard Add #5 · Determinants, Cofactors and the Adjoint Identities
Let A A A be an invertible matrix. Identify the correct statement(s) from the following.
(I) A ( a d j A − 1 ) = 1 ∣ A ∣ A 2 A(\mathrm{adj}\,A^{-1}) = \dfrac{1}{|A|}A^2 A ( adj A − 1 ) = ∣ A ∣ 1 A 2 (II) ( a d j A − 1 ) − 1 = 1 ∣ A ∣ A − 1 (\mathrm{adj}\,A^{-1})^{-1} = \dfrac{1}{|A|}A^{-1} ( adj A − 1 ) − 1 = ∣ A ∣ 1 A − 1 (III) ( a d j A ) = 1 ∣ A ∣ A − 1 (\mathrm{adj}\,A) = \dfrac{1}{|A|}A^{-1} ( adj A ) = ∣ A ∣ 1 A − 1 Done a few? Here’s your next move
#6 · Determinants, Cofactors and the Adjoint Identities
Let A = [ 3 1 2 1 2 0 1 1 4 ] A = \begin{bmatrix} 3 & 1 & 2 \\ 1 & 2 & 0 \\ 1 & 1 & 4 \end{bmatrix} A = 3 1 1 1 2 1 2 0 4 and p C 11 + 4 C 21 − 5 C 32 = − 2 pC_{11} + 4C_{21} - 5C_{32} = -2 p C 11 + 4 C 21 − 5 C 32 = − 2 , where C i j C_{ij} C ij denotes the cofactor of an element a i j a_{ij} a ij of matrix A A A , then the value of p p p is : #7 · Determinants, Cofactors and the Adjoint Identities
If matrix A = [ − 1 2025 2026 0 2 2027 0 0 − 1 ] A = \begin{bmatrix} -1 & 2025 & 2026 \\ 0 & 2 & 2027 \\ 0 & 0 & -1 \end{bmatrix} A = − 1 0 0 2025 2 0 2026 2027 − 1 , then the sum of all elements in a d j ( A − 1 ) \mathrm{adj}(A^{-1}) adj ( A − 1 ) is equal to... #8 · Determinants, Cofactors and the Adjoint Identities
If A = [ a i j ] 3 × 3 A = [a_{ij}]_{3 \times 3} A = [ a ij ] 3 × 3 , where a i j = { 1 , if i + j is even 0 , if i + j is odd a_{ij} = \begin{cases} 1, & \text{if } i + j \text{ is even} \\ 0, & \text{if } i + j \text{ is odd} \end{cases} a ij = { 1 , 0 , if i + j is even if i + j is odd , then a d j ( A ) = … \mathrm{adj}(A) = \ldots adj ( A ) = … A [ 0 1 0 1 0 1 0 1 0 ] \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} 0 1 0 1 0 1 0 1 0 B [ 1 0 − 1 0 0 0 − 1 0 1 ] \begin{bmatrix} 1 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & 1 \end{bmatrix} 1 0 − 1 0 0 0 − 1 0 1 C [ 0 1 0 1 1 1 0 1 0 ] \begin{bmatrix} 0 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix} 0 1 0 1 1 1 0 1 0 D [ 0 0 1 0 1 1 1 1 1 ] \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} 0 0 1 0 1 1 1 1 1 MHT-CET · 2026 · 18 April Shift I · Q65 Moderate Add #9 · Determinants, Cofactors and the Adjoint Identities
Let A = [ cos α − sin α 0 sin α cos α 0 0 0 1 ] A = \begin{bmatrix} \cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} A = cos α sin α 0 − sin α cos α 0 0 0 1 . If B = a d j A B = \mathrm{adj}\,A B = adj A , then the matrix B − 1 B^{-1} B − 1 is equal to... #10 · Determinants, Cofactors and the Adjoint Identities
If A = [ 5 a − b 3 2 ] A = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix} A = [ 5 a 3 − b 2 ] and A ( a d j A ) = A A T A\,(\mathrm{adj}\,A) = AA^T A ( adj A ) = A A T , then 5 a + b = 5a + b = 5 a + b = #11 · Determinants, Cofactors and the Adjoint Identities
If A = [ sec θ − tan θ − tan θ sec θ ] , θ ∈ ( 0 , π 2 ) A = \begin{bmatrix} \sec\theta & -\tan\theta \\ -\tan\theta & \sec\theta \end{bmatrix}, \theta \in \left(0, \dfrac{\pi}{2}\right) A = [ sec θ − tan θ − tan θ sec θ ] , θ ∈ ( 0 , 2 π ) such that A + a d j A = 4 I A + \mathrm{adj}A = 4I A + adj A = 4 I , then θ = \theta = θ = #12 · Determinants, Cofactors and the Adjoint Identities
Let A = [ 2 k − 1 1 1 0 2 k − 1 1 0 0 2 k − 1 ] A = \begin{bmatrix} 2k-1 & 1 & 1 \\ 0 & 2k-1 & 1 \\ 0 & 0 & 2k-1 \end{bmatrix} A = 2 k − 1 0 0 1 2 k − 1 0 1 1 2 k − 1 and B = [ 0 2 k − 1 1 1 − 2 k 0 k − 1 − k 0 ] B = \begin{bmatrix} 0 & 2k-1 & 1 \\ 1-2k & 0 & k \\ -1 & -k & 0 \end{bmatrix} B = 0 1 − 2 k − 1 2 k − 1 0 − k 1 k 0 where k is a real number. If det ( a d j A ) + det ( a d j B ) = 11 6 \det(\mathrm{adj}A) + \det(\mathrm{adj}B) = 11^6 det ( adj A ) + det ( adj B ) = 1 1 6 , then the value of k − 5 k - 5 k − 5 is equal to... #13 · Determinants, Cofactors and the Adjoint Identities
If A = [ 5 a − b 3 2 ] A=\begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix} A = [ 5 a 3 − b 2 ] and A ⋅ a d j A = A A T A\cdot adjA=AA^{T} A ⋅ a d j A = A A T , then 5 a + b = 5a+b= 5 a + b = #14 · Determinants, Cofactors and the Adjoint Identities
Let A be a non-singular matrix of order n and ∣ A ∣ = k |A| =k ∣ A ∣ = k , then ( a d j A ) − 1 (adjA)^{- 1} ( a d j A ) − 1 is A B k n − 1 ( a d j A ) k^{n - 1}(adj\text{ }A) k n − 1 ( a d j A ) C k n − 2 A k^{n - 2}\text{ }A k n − 2 A D #15 · Determinants, Cofactors and the Adjoint Identities
The third element in the second row of adjoint of a matrix A = [ a i j ] 3 × 3 A =\left\lbrack a_{ij} \right\rbrack_{3 \times 3} A = [ a ij ] 3 × 3 (where a i j = 2 i + j a_{ij}= 2i + j a ij = 2 i + j ) is MHT-CET · 2025 · 19 April Shift I · Q149 Moderate Add #16 · Determinants, Cofactors and the Adjoint Identities
If A = [ cos θ sin θ 0 − sin θ cos θ 0 0 0 1 ] A=\begin{bmatrix} \cos\theta & \sin\theta & 0 \\ - \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} A = cos θ − sin θ 0 sin θ cos θ 0 0 0 1 , where A 21 , A 22 , A 23 A_{21},A_{22},A_{23} A 21 , A 22 , A 23 are cofactors of a 21 , a 22 , a 23 a_{21},a_{22},a_{23} a 21 , a 22 , a 23 respectively, then the value of a 21 A 21 + a 22 A 22 + a 23 A 23 = a_{21}A_{21}+a_{22}A_{22}+a_{23}A_{23}= a 21 A 21 + a 22 A 22 + a 23 A 23 = #17 · Determinants, Cofactors and the Adjoint Identities
If A = ( 5 a − b 3 2 ) A = \begin{pmatrix} 5a & -b \\ 3 & 2 \end{pmatrix} A = ( 5 a 3 − b 2 ) and A ⋅ adj A = A A T A \cdot \text{adj}\,A = AA^T A ⋅ adj A = A A T , then 5 a + b 5a+b 5 a + b is equal to MHT-CET · 2023 · 9th May Shift 2 · Q113 Hard Add #18 · Determinants, Cofactors and the Adjoint Identities
For the matrix A = ( 2 0 − 1 3 1 2 − 1 1 2 ) A = \begin{pmatrix} 2 & 0 & -1 \\ 3 & 1 & 2 \\ -1 & 1 & 2 \end{pmatrix} A = 2 3 − 1 0 1 1 − 1 2 2 , the matrix of cofactors is A ( 0 8 − 4 − 1 3 2 1 − 7 2 ) \begin{pmatrix} 0 & 8 & -4 \\ -1 & 3 & 2 \\ 1 & -7 & 2 \end{pmatrix} 0 − 1 1 8 3 − 7 − 4 2 2 B ( 0 − 8 4 − 1 3 − 2 1 − 7 2 ) \begin{pmatrix} 0 & -8 & 4 \\ -1 & 3 & -2 \\ 1 & -7 & 2 \end{pmatrix} 0 − 1 1 − 8 3 − 7 4 − 2 2 C ( 0 8 − 4 1 − 3 2 − 1 7 − 2 ) \begin{pmatrix} 0 & 8 & -4 \\ 1 & -3 & 2 \\ -1 & 7 & -2 \end{pmatrix} 0 1 − 1 8 − 3 7 − 4 2 − 2 D ( 0 − 8 4 1 3 2 − 1 − 7 2 ) \begin{pmatrix} 0 & -8 & 4 \\ 1 & 3 & 2 \\ -1 & -7 & 2 \end{pmatrix} 0 1 − 1 − 8 3 − 7 4 2 2 #19 · Determinants, Cofactors and the Adjoint Identities
Let A = ( 2 − 1 0 2 ) A=\begin{pmatrix}2&-1\\0&2\end{pmatrix} A = ( 2 0 − 1 2 ) . If B = I − 3 C 1 ( adj A ) + 3 C 2 ( adj A ) 2 − 3 C 3 ( adj A ) 3 B=I-{}^3C_1(\text{adj}\,A)+{}^3C_2(\text{adj}\,A)^2-{}^3C_3(\text{adj}\,A)^3 B = I − 3 C 1 ( adj A ) + 3 C 2 ( adj A ) 2 − 3 C 3 ( adj A ) 3 , then the sum of all elements of the matrix B B B is #20 · Determinants, Cofactors and the Adjoint Identities
If A = ( 2 a − 3 b 3 2 ) A = \begin{pmatrix}2a & -3b \\ 3 & 2\end{pmatrix} A = ( 2 a 3 − 3 b 2 ) and A ⋅ adj A = A A T A \cdot \text{adj}\, A = A A^T A ⋅ adj A = A A T , then 2 a + 3 b 2a+3b 2 a + 3 b is #21 · Determinants, Cofactors and the Adjoint Identities
If P = ( 1 α 3 1 3 3 2 4 4 ) P = \begin{pmatrix}1&\alpha&3\\1&3&3\\2&4&4\end{pmatrix} P = 1 1 2 α 3 4 3 3 4 is the adjoint of a 3 × 3 3\times3 3 × 3 matrix A and ∣ A ∣ = 4 |A|=4 ∣ A ∣ = 4 , then value of α \alpha α is #22 · Determinants, Cofactors and the Adjoint Identities
Let ω ≠ 1 \omega \neq 1 ω = 1 be a cube root of unity and S be the set of all non-singular matrices of the form ( 1 a b ω 1 c ω 2 ω 1 ) \begin{pmatrix}1&a&b\\\omega&1&c\\\omega^2&\omega&1\end{pmatrix} 1 ω ω 2 a 1 ω b c 1 where each of a, b and c is either ω \omega ω or ω 2 \omega^2 ω 2 , then the number of distinct matrices in the set S is #23 · Determinants, Cofactors and the Adjoint Identities
If B = ( 3 a − 1 1 3 1 − 1 1 3 ) B=\begin{pmatrix}3&a&-1\\1&3&1\\-1&1&3\end{pmatrix} B = 3 1 − 1 a 3 1 − 1 1 3 is the adjoint of a 3 × 3 3\times3 3 × 3 matrix A A A and ∣ A ∣ = 4 |A|=4 ∣ A ∣ = 4 , then a a a is equal to #24 · Determinants, Cofactors and the Adjoint Identities
B= ( 1 α 2 1 2 2 2 3 3 ) \begin{pmatrix}1&\alpha&2\\1&2&2\\2&3&3\end{pmatrix} 1 1 2 α 2 3 2 2 3 is adjoint of 3 × 3 3\times 3 3 × 3 matrix A, ∣ A ∣ = 5 |A|=5 ∣ A ∣ = 5 . Value of α \alpha α is #25 · Determinants, Cofactors and the Adjoint Identities
If ∣ cos ( A + B ) − sin ( A + B ) cos 2 B sin A cos A sin B − cos A sin A cos B ∣ = 0 \begin{vmatrix}\cos(A+B) & -\sin(A+B) & \cos 2B \\ \sin A & \cos A & \sin B \\ -\cos A & \sin A & \cos B\end{vmatrix} = 0 cos ( A + B ) sin A − cos A − sin ( A + B ) cos A sin A cos 2 B sin B cos B = 0 , then the value of B B B is A n π , n ∈ Z n\pi,\;n\in\mathbb{Z} nπ , n ∈ Z B ( 2 n + 1 ) π 2 , n ∈ Z (2n+1)\frac{\pi}{2},\;n\in\mathbb{Z} ( 2 n + 1 ) 2 π , n ∈ Z C ( 2 n + 1 ) π 4 , n ∈ Z (2n+1)\frac{\pi}{4},\;n\in\mathbb{Z} ( 2 n + 1 ) 4 π , n ∈ Z D 2 n π 3 , n ∈ Z \frac{2n\pi}{3},\;n\in\mathbb{Z} 3 2 nπ , n ∈ Z #26 · Determinants, Cofactors and the Adjoint Identities
For an invertible matrix A A A , if A ( adj A ) = ( 20 0 0 20 ) A(\text{adj}\,A) = \begin{pmatrix}20 & 0\\0 & 20\end{pmatrix} A ( adj A ) = ( 20 0 0 20 ) , then ∣ A ∣ = |A| = ∣ A ∣ = MHT-CET · 2022 · Shift 1 · Q139 Hard Add #27 · Determinants, Cofactors and the Adjoint Identities
Given A = ( x 3 2 1 y 4 2 2 z ) A = \begin{pmatrix} x & 3 & 2 \\ 1 & y & 4 \\ 2 & 2 & z \end{pmatrix} A = x 1 2 3 y 2 2 4 z , if x y z = 60 xyz = 60 x y z = 60 and 8 x + 4 y + 3 z = 20 8x + 4y + 3z = 20 8 x + 4 y + 3 z = 20 , then A ( adj A ) A(\text{adj}\,A) A ( adj A ) A ( 60 0 0 0 60 0 0 0 60 ) \begin{pmatrix} 60 & 0 & 0 \\ 0 & 60 & 0 \\ 0 & 0 & 60 \end{pmatrix} 60 0 0 0 60 0 0 0 60 B ( 108 0 0 0 108 0 0 0 108 ) \begin{pmatrix} 108 & 0 & 0 \\ 0 & 108 & 0 \\ 0 & 0 & 108 \end{pmatrix} 108 0 0 0 108 0 0 0 108 C ( 20 0 0 0 20 0 0 0 20 ) \begin{pmatrix} 20 & 0 & 0 \\ 0 & 20 & 0 \\ 0 & 0 & 20 \end{pmatrix} 20 0 0 0 20 0 0 0 20 D ( 68 0 0 0 68 0 0 0 68 ) \begin{pmatrix} 68 & 0 & 0 \\ 0 & 68 & 0 \\ 0 & 0 & 68 \end{pmatrix} 68 0 0 0 68 0 0 0 68 #28 · Inverse of a Matrix — Adjoint Formula, Products and Verification
If A = [ 2 − 3 5 − 7 ] A = \begin{bmatrix} 2 & -3 \\ 5 & -7 \end{bmatrix} A = [ 2 5 − 3 − 7 ] , then A − A − 1 = A - A^{-1} = A − A − 1 = A [ − 9 6 − 10 9 ] \begin{bmatrix} -9 & 6 \\ -10 & 9 \end{bmatrix} [ − 9 − 10 6 9 ] B [ 9 − 6 10 − 9 ] \begin{bmatrix} 9 & -6 \\ 10 & -9 \end{bmatrix} [ 9 10 − 6 − 9 ] C [ − 9 10 − 6 9 ] \begin{bmatrix} -9 & 10 \\ -6 & 9 \end{bmatrix} [ − 9 − 6 10 9 ] D [ 9 − 10 6 − 9 ] \begin{bmatrix} 9 & -10 \\ 6 & -9 \end{bmatrix} [ 9 6 − 10 − 9 ] #29 · Inverse of a Matrix — Adjoint Formula, Products and Verification
If A + I = ( 3 − 2 1 − 1 ) A + I = \begin{pmatrix} 3 & -2 \\ 1 & -1 \end{pmatrix} A + I = ( 3 1 − 2 − 1 ) , then the inverse of the matrix ( A − I ) (A - I) ( A − I ) is: A ( 1 − 2 1 − 3 ) \begin{pmatrix} 1 & -2 \\ 1 & -3 \end{pmatrix} ( 1 1 − 2 − 3 ) B ( 3 − 2 1 − 1 ) \begin{pmatrix} 3 & -2 \\ 1 & -1 \end{pmatrix} ( 3 1 − 2 − 1 ) C ( − 3 2 − 1 1 ) \begin{pmatrix} -3 & 2 \\ -1 & 1 \end{pmatrix} ( − 3 − 1 2 1 ) D ( − 3 2 − 1 − 1 ) \begin{pmatrix} -3 & 2 \\ -1 & -1 \end{pmatrix} ( − 3 − 1 2 − 1 ) #30 · Inverse of a Matrix — Adjoint Formula, Products and Verification
The sum of the matrix [ 2 − 3 5 − 7 ] \begin{bmatrix} 2 & -3 \\ 5 & -7 \end{bmatrix} [ 2 5 − 3 − 7 ] and its multiplicative inverse is A [ 2 − 3 5 − 7 ] \begin{bmatrix} 2 & -3 \\ 5 & -7 \end{bmatrix} [ 2 5 − 3 − 7 ] B [ − 7 3 − 5 2 ] \begin{bmatrix} -7 & 3 \\ -5 & 2 \end{bmatrix} [ − 7 − 5 3 2 ] C [ − 5 0 0 − 5 ] \begin{bmatrix} -5 & 0 \\ 0 & -5 \end{bmatrix} [ − 5 0 0 − 5 ] D [ 1 0 0 1 ] \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} [ 1 0 0 1 ] #31 · Inverse of a Matrix — Adjoint Formula, Products and Verification
If A − 1 = [ 1 0 − 4 0 − 1 2 − 1 2 1 ] = a d j ( A ) k A^{-1} = \begin{bmatrix} 1 & 0 & -4 \\ 0 & -1 & 2 \\ -1 & 2 & 1 \end{bmatrix} = \dfrac{\mathrm{adj}(A)}{k} A − 1 = 1 0 − 1 0 − 1 2 − 4 2 1 = k adj ( A ) , then k k k is a root of the equation... A B C D x 2 − x + 1 = 0 x^2 - x + 1 = 0 x 2 − x + 1 = 0 #32 · Inverse of a Matrix — Adjoint Formula, Products and Verification
If A = [ 0 1 2 1 2 3 3 a 1 ] A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & a & 1 \end{bmatrix} A = 0 1 3 1 2 a 2 3 1 and A − 1 = [ 1 2 − 1 2 1 2 − 4 3 c 5 2 − 3 2 1 2 ] A^{-1} = \begin{bmatrix} \dfrac{1}{2} & -\dfrac{1}{2} & \dfrac{1}{2} \\ -4 & 3 & c \\ \dfrac{5}{2} & -\dfrac{3}{2} & \dfrac{1}{2} \end{bmatrix} A − 1 = 2 1 − 4 2 5 − 2 1 3 − 2 3 2 1 c 2 1 , then ... A a = 1 , c = 1 a = 1,\ c = 1 a = 1 , c = 1 B a = 1 , c = − 1 a = 1,\ c = -1 a = 1 , c = − 1 C a = 1 , c = 2 a = 1,\ c = 2 a = 1 , c = 2 D a = − 1 , c = 2 a = -1,\ c = 2 a = − 1 , c = 2 #33 · Inverse of a Matrix — Adjoint Formula, Products and Verification
If the matrix A = [ 1 2 3 1 1 5 2 4 7 ] A = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7 \end{bmatrix} A = 1 1 2 2 1 4 3 5 7 and its inverse is A − 1 = [ a c − 7 b − 1 c − 2 0 1 ] A^{-1} = \begin{bmatrix} a & c & -7 \\ b & -1 & c \\ -2 & 0 & 1 \end{bmatrix} A − 1 = a b − 2 c − 1 0 − 7 c 1 , then .... A a = 13 , b = − 3 , c = 0 a = 13,\ b = -3,\ c = 0 a = 13 , b = − 3 , c = 0 B a = 13 , b = 3 , c = − 2 a = 13,\ b = 3,\ c = -2 a = 13 , b = 3 , c = − 2 C a = 13 , b = − 7 , c = 2 a = 13,\ b = -7,\ c = 2 a = 13 , b = − 7 , c = 2 D a = 13 , b = − 3 , c = 2 a = 13,\ b = -3,\ c = 2 a = 13 , b = − 3 , c = 2 #34 · Inverse of a Matrix — Adjoint Formula, Products and Verification
If A − 1 = [ 3 − 1 1 − 15 6 − 5 5 − 2 2 ] A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix} A − 1 = 3 − 15 5 − 1 6 − 2 1 − 5 2 and B = [ 1 2 − 2 − 1 3 0 0 − 2 1 ] B = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} B = 1 − 1 0 2 3 − 2 − 2 0 1 , then ( A B ) − 1 = (AB)^{-1} = ( A B ) − 1 = A [ 9 − 3 5 − 2 1 0 1 0 2 ] \begin{bmatrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix} 9 − 2 1 − 3 1 0 5 0 2 B [ 9 3 − 5 − 2 − 1 0 1 0 2 ] \begin{bmatrix} 9 & 3 & -5 \\ -2 & -1 & 0 \\ 1 & 0 & 2 \end{bmatrix} 9 − 2 1 3 − 1 0 − 5 0 2 C [ 9 − 3 5 − 2 − 1 0 1 0 − 2 ] \begin{bmatrix} 9 & -3 & 5 \\ -2 & -1 & 0 \\ 1 & 0 & -2 \end{bmatrix} 9 − 2 1 − 3 − 1 0 5 0 − 2 D [ 9 − 3 5 − 2 1 0 1 0 − 2 ] \begin{bmatrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & -2 \end{bmatrix} 9 − 2 1 − 3 1 0 5 0 − 2 #35 · Inverse of a Matrix — Adjoint Formula, Products and Verification
Let A = [ 2 1 3 2 ] A = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} A = [ 2 3 1 2 ] and B = [ − 1 − 1 1 1 ] B = \begin{bmatrix} -1 & -1 \\ 1 & 1 \end{bmatrix} B = [ − 1 1 − 1 1 ] . If the variables x x x and y y y satisfy the matrix equation ( A − 1 + B ) [ x y ] = [ 0 − 2 ] (A^{-1} + B)\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ -2 \end{bmatrix} ( A − 1 + B ) [ x y ] = [ 0 − 2 ] , then the value of x + 3 y x + 3y x + 3 y is...... #36 · Inverse of a Matrix — Adjoint Formula, Products and Verification
If A = [ 1 3 2 1 ] A = \begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix} A = [ 1 2 3 1 ] , then the determinant of ( A 2 − 2 A ) − 1 (A^2 - 2A)^{-1} ( A 2 − 2 A ) − 1 is... #37 · Inverse of a Matrix — Adjoint Formula, Products and Verification
If the matrix A = [ 3 − 5 4 − 7 ] A = \begin{bmatrix} 3 & -5 \\ 4 & -7 \end{bmatrix} A = [ 3 4 − 5 − 7 ] and A − 1 = [ 7 − 5 4 a ] A^{-1} = \begin{bmatrix} 7 & -5 \\ 4 & a \end{bmatrix} A − 1 = [ 7 4 − 5 a ] , then the value of a a a is... MHT-CET · 2026 · 12 May Shift I · Q67 Moderate Add #38 · Inverse of a Matrix — Adjoint Formula, Products and Verification
If A = [ sin α − cos α cos α sin α ] A = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix} A = [ sin α cos α − cos α sin α ] and S = { α : α ∈ ( π 2 , 3 π 2 ) , A + A − 1 = I } S = \left\{\alpha : \alpha \in \left(\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right), A + A^{-1} = I\right\} S = { α : α ∈ ( 2 π , 2 3 π ) , A + A − 1 = I } , then α = \alpha = α = ... #39 · Inverse of a Matrix — Adjoint Formula, Products and Verification
If A = [ 1 − 5 − 2 4 ] A = \begin{bmatrix} 1 & -5 \\ -2 & 4 \end{bmatrix} A = [ 1 − 2 − 5 4 ] , then A − 1 = A^{-1} = A − 1 = A − 1 6 [ − 4 5 2 − 1 ] -\dfrac{1}{6}\begin{bmatrix} -4 & 5 \\ 2 & -1 \end{bmatrix} − 6 1 [ − 4 2 5 − 1 ] B 1 14 [ − 1 5 2 − 4 ] \dfrac{1}{14}\begin{bmatrix} -1 & 5 \\ 2 & -4 \end{bmatrix} 14 1 [ − 1 2 5 − 4 ] C 1 14 [ 4 5 2 1 ] \dfrac{1}{14}\begin{bmatrix} 4 & 5 \\ 2 & 1 \end{bmatrix} 14 1 [ 4 2 5 1 ] D − 1 6 [ 4 5 2 1 ] -\dfrac{1}{6}\begin{bmatrix} 4 & 5 \\ 2 & 1 \end{bmatrix} − 6 1 [ 4 2 5 1 ] #40 · Inverse of a Matrix — Adjoint Formula, Products and Verification
If A = [ 3 2 6 1 1 2 2 2 5 ] , B = [ 1 0 1 ] A = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}, B = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} A = 3 1 2 2 1 2 6 2 5 , B = 1 0 1 such that X A = B T XA = B^T X A = B T and A − 1 Y = B A^{-1}Y = B A − 1 Y = B , then X Y = XY = X Y = #41 · Inverse of a Matrix — Adjoint Formula, Products and Verification
The inverse of the matrix A = [ 2 − 1 4 4 − 3 1 1 2 1 ] A = \begin{bmatrix} 2 & -1 & 4 \\ 4 & -3 & 1 \\ 1 & 2 & 1 \end{bmatrix} A = 2 4 1 − 1 − 3 2 4 1 1 is B = 1 37 [ − 5 9 11 − 3 − 2 14 11 − 5 k ] B = \dfrac{1}{37}\begin{bmatrix} -5 & 9 & 11 \\ -3 & -2 & 14 \\ 11 & -5 & k \end{bmatrix} B = 37 1 − 5 − 3 11 9 − 2 − 5 11 14 k , then the value of k k k is... #42 · Inverse of a Matrix — Adjoint Formula, Products and Verification
The inverse of matrix [ 1 + p q p 0 q 1 + p q p 0 q 1 ] \begin{bmatrix} 1 + pq & p & 0 \\ q & 1 + pq & p \\ 0 & q & 1 \end{bmatrix} 1 + pq q 0 p 1 + pq q 0 p 1 is ... A [ 1 + p q p 0 q 1 + p q p 0 q 1 ] \begin{bmatrix} 1 + pq & p & 0 \\ q & 1 + pq & p \\ 0 & q & 1 \end{bmatrix} 1 + pq q 0 p 1 + pq q 0 p 1 B [ 1 p p 2 q 1 + p q p + p 2 q q 2 q + p q 2 1 + p q + p 2 q 2 ] \begin{bmatrix} 1 & p & p^2 \\ q & 1 + pq & p + p^2q \\ q^2 & q + pq^2 & 1 + pq + p^2q^2 \end{bmatrix} 1 q q 2 p 1 + pq q + p q 2 p 2 p + p 2 q 1 + pq + p 2 q 2 C [ 1 − p p 2 − q 1 + p q − ( p + p 2 q ) q 2 − ( q + p q 2 ) 1 + p q + p 2 q 2 ] \begin{bmatrix} 1 & -p & p^2 \\ -q & 1 + pq & -(p + p^2q) \\ q^2 & -(q + pq^2) & 1 + pq + p^2q^2 \end{bmatrix} 1 − q q 2 − p 1 + pq − ( q + p q 2 ) p 2 − ( p + p 2 q ) 1 + pq + p 2 q 2 D [ 1 − p p 2 − q 1 + p q p + p 2 q q 2 q + p q 2 1 + p q + p 2 q 2 ] \begin{bmatrix} 1 & -p & p^2 \\ -q & 1 + pq & p + p^2q \\ q^2 & q + pq^2 & 1 + pq + p^2q^2 \end{bmatrix} 1 − q q 2 − p 1 + pq q + p q 2 p 2 p + p 2 q 1 + pq + p 2 q 2 #43 · Inverse of a Matrix — Adjoint Formula, Products and Verification
If matrix A and its inverse A − 1 A^{-1} A − 1 are given by A = [ 0 1 2 1 2 3 3 x 1 ] A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1 \end{bmatrix} A = 0 1 3 1 2 x 2 3 1 and A − 1 = [ 1 2 − 1 2 1 2 − 4 3 y 5 2 − 3 2 1 2 ] A^{-1} = \begin{bmatrix} \dfrac{1}{2} & -\dfrac{1}{2} & \dfrac{1}{2} \\ -4 & 3 & y \\ \dfrac{5}{2} & -\dfrac{3}{2} & \dfrac{1}{2} \end{bmatrix} A − 1 = 2 1 − 4 2 5 − 2 1 3 − 2 3 2 1 y 2 1 , then the polar co-ordinates of the points whose Cartesian co-ordinates are ( x , y ) (x, y) ( x , y ) are … \ldots … A ( 2 , 7 π 4 ) \left(2, \dfrac{7\pi}{4}\right) ( 2 , 4 7 π ) B ( 2 , π 4 ) \left(\sqrt{2}, \dfrac{\pi}{4}\right) ( 2 , 4 π ) C ( 2 , 7 π 4 ) \left(\sqrt{2}, \dfrac{7\pi}{4}\right) ( 2 , 4 7 π ) D ( 2 , π 4 ) \left(2, \dfrac{\pi}{4}\right) ( 2 , 4 π ) MHT-CET · 2026 · 18 April Shift I · Q66 Hard Add #44 · Inverse of a Matrix — Adjoint Formula, Products and Verification
Let A = [ 3 − 4 1 − 1 ] A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} A = [ 3 1 − 4 − 1 ] and B = [ 6 − 13 5 − 10 ] B = \begin{bmatrix} 6 & -13 \\ 5 & -10 \end{bmatrix} B = [ 6 5 − 13 − 10 ] be two matrices. If the variables x x x and y y y satisfy the matrix equation ( ( A − 1 ) 2 + B ) [ x y ] = [ 0 0 ] ((A^{-1})^2 + B)\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} (( A − 1 ) 2 + B ) [ x y ] = [ 0 0 ] , then the ordered pair ( x , y ) = (x, y) = ( x , y ) = #45 · Inverse of a Matrix — Adjoint Formula, Products and Verification
If A = [ 1 − tan θ 2 tan θ 2 1 ] A = \begin{bmatrix} 1 & -\tan\dfrac{\theta}{2} \\ \tan\dfrac{\theta}{2} & 1 \end{bmatrix} A = 1 tan 2 θ − tan 2 θ 1 and B = [ 1 tan θ 2 − tan θ 2 1 ] B = \begin{bmatrix} 1 & \tan\dfrac{\theta}{2} \\ -\tan\dfrac{\theta}{2} & 1 \end{bmatrix} B = 1 − tan 2 θ tan 2 θ 1 then A − 1 B A^{-1}B A − 1 B is equal to A [ cos θ sin θ − sin θ cos θ ] \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} [ cos θ − sin θ sin θ cos θ ] B [ cos θ − sin θ sin θ cos θ ] \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} [ cos θ sin θ − sin θ cos θ ] C [ sin θ − cos θ cos θ sin θ ] \begin{bmatrix} \sin\theta & -\cos\theta \\ \cos\theta & \sin\theta \end{bmatrix} [ sin θ cos θ − cos θ sin θ ] D [ cos θ sin θ sin θ cos θ ] \begin{bmatrix} \cos\theta & \sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} [ cos θ sin θ sin θ cos θ ] #46 · Inverse of a Matrix — Adjoint Formula, Products and Verification
If A = [ 3 0 0 0 − 5 0 0 0 7 ] ; B = [ − 1 0 0 0 2 0 0 0 4 ] A = \begin{bmatrix} 3 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 7 \end{bmatrix}; B = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{bmatrix} A = 3 0 0 0 − 5 0 0 0 7 ; B = − 1 0 0 0 2 0 0 0 4 then, ( 2 A + 3 B ) − 1 = (2A + 3B)^{-1} = ( 2 A + 3 B ) − 1 = ______ A [ 1 3 0 0 0 − 1 4 0 0 0 1 26 ] \begin{bmatrix} \dfrac{1}{3} & 0 & 0 \\ 0 & -\dfrac{1}{4} & 0 \\ 0 & 0 & \dfrac{1}{26} \end{bmatrix} 3 1 0 0 0 − 4 1 0 0 0 26 1 B [ 1 3 0 0 0 1 4 0 0 0 1 26 ] \begin{bmatrix} \dfrac{1}{3} & 0 & 0 \\ 0 & \dfrac{1}{4} & 0 \\ 0 & 0 & \dfrac{1}{26} \end{bmatrix} 3 1 0 0 0 4 1 0 0 0 26 1 C [ 1 3 0 0 0 1 4 0 0 0 − 1 26 ] \begin{bmatrix} \dfrac{1}{3} & 0 & 0 \\ 0 & \dfrac{1}{4} & 0 \\ 0 & 0 & -\dfrac{1}{26} \end{bmatrix} 3 1 0 0 0 4 1 0 0 0 − 26 1 D [ − 1 3 0 0 0 1 4 0 0 0 − 1 26 ] \begin{bmatrix} -\dfrac{1}{3} & 0 & 0 \\ 0 & \dfrac{1}{4} & 0 \\ 0 & 0 & -\dfrac{1}{26} \end{bmatrix} − 3 1 0 0 0 4 1 0 0 0 − 26 1 #47 · Inverse of a Matrix — Adjoint Formula, Products and Verification
If A = [ 2 3 5 − 2 ] , B − 1 = [ 1 5 2 5 2 5 − 1 5 ] A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}, B^{-1} = \begin{bmatrix} \frac{1}{5} & \frac{2}{5} \\ \frac{2}{5} & -\frac{1}{5} \end{bmatrix} A = [ 2 5 3 − 2 ] , B − 1 = [ 5 1 5 2 5 2 − 5 1 ] , then ( A B ) − 1 = (AB)^{-1} = ( A B ) − 1 = A 1 95 [ 8 − 1 − 1 12 ] \dfrac{1}{95}\begin{bmatrix} 8 & -1 \\ -1 & 12 \end{bmatrix} 95 1 [ 8 − 1 − 1 12 ] B 1 95 [ 12 − 1 − 1 8 ] \dfrac{1}{95}\begin{bmatrix} 12 & -1 \\ -1 & 8 \end{bmatrix} 95 1 [ 12 − 1 − 1 8 ] C 1 95 [ − 12 1 1 − 8 ] \dfrac{1}{95}\begin{bmatrix} -12 & 1 \\ 1 & -8 \end{bmatrix} 95 1 [ − 12 1 1 − 8 ] D 1 95 [ − 8 1 1 − 12 ] \dfrac{1}{95}\begin{bmatrix} -8 & 1 \\ 1 & -12 \end{bmatrix} 95 1 [ − 8 1 1 − 12 ] #48 · Inverse of a Matrix — Adjoint Formula, Products and Verification
If A = [ a i j ] 3 × 3 A = [a_{ij}]_{3 \times 3} A = [ a ij ] 3 × 3 is a matrix such that a i j = ∣ 2 i − 5 j ∣ a_{ij} = |2i - 5j| a ij = ∣2 i − 5 j ∣ , where ∣ ⋅ ∣ |\cdot| ∣ ⋅ ∣ denotes the modulus function, then the element in the 2 n d 2^{nd} 2 n d row and 3 r d 3^{rd} 3 r d column of A − 1 A^{-1} A − 1 is ... #49 · Inverse of a Matrix — Adjoint Formula, Products and Verification
Inverse of [ 3 − 2 1 4 ] \begin{bmatrix} 3 & -2 \\ 1 & 4 \end{bmatrix} [ 3 1 − 2 4 ] is A [ 2 7 1 7 − 1 14 3 14 ] \begin{bmatrix} \dfrac{2}{7} & \dfrac{1}{7} \\ -\dfrac{1}{14} & \dfrac{3}{14} \end{bmatrix} 7 2 − 14 1 7 1 14 3 B [ 3 14 − 1 7 1 14 2 7 ] \begin{bmatrix} \dfrac{3}{14} & -\dfrac{1}{7} \\ \dfrac{1}{14} & \dfrac{2}{7} \end{bmatrix} 14 3 14 1 − 7 1 7 2 C [ 2 7 − 1 7 1 14 3 14 ] \begin{bmatrix} \dfrac{2}{7} & -\dfrac{1}{7} \\ \dfrac{1}{14} & \dfrac{3}{14} \end{bmatrix} 7 2 14 1 − 7 1 14 3 D [ − 3 14 1 7 1 14 2 7 ] \begin{bmatrix} -\dfrac{3}{14} & \dfrac{1}{7} \\ \dfrac{1}{14} & \dfrac{2}{7} \end{bmatrix} − 14 3 14 1 7 1 7 2 #50 · Inverse of a Matrix — Adjoint Formula, Products and Verification
If A = [ 3 1 − 1 2 ] , C = [ 7 3 0 6 ] A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}, C = \begin{bmatrix} 7 & 3 \\ 0 & 6 \end{bmatrix} A = [ 3 − 1 1 2 ] , C = [ 7 0 3 6 ] and A B = C AB = C A B = C , then the inverse of matrix B is A 1 42 [ 3 0 − 1 2 ] \dfrac{1}{42}\begin{bmatrix} 3 & 0 \\ -1 & 2 \end{bmatrix} 42 1 [ 3 − 1 0 2 ] B 1 6 [ 3 0 − 1 2 ] \dfrac{1}{6}\begin{bmatrix} 3 & 0 \\ -1 & 2 \end{bmatrix} 6 1 [ 3 − 1 0 2 ] C 1 42 [ 6 3 − 1 2 ] \dfrac{1}{42}\begin{bmatrix} 6 & 3 \\ -1 & 2 \end{bmatrix} 42 1 [ 6 − 1 3 2 ] D 1 6 [ 7 3 − 1 3 ] \dfrac{1}{6}\begin{bmatrix} 7 & 3 \\ -1 & 3 \end{bmatrix} 6 1 [ 7 − 1 3 3 ] Showing the first 50 of 102.
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