Skip to the question JEE Main › Mathematics › Matrices and Determinants › Answered the equation, not the question +4 marks−1 if incorrectSingle correct pyq
Matrices and Determinants: Mathematics | JEE Main Let
A = [ 1 10 3 10 − 3 10 1 10 ] \mathrm{A} = \begin{bmatrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{bmatrix} A = [ 10 1 10 − 3 10 3 10 1 ] and
B = [ 1 − i 0 1 ] \mathrm{B} = \begin{bmatrix} 1 & -i \\ 0 & 1 \end{bmatrix} B = [ 1 0 − i 1 ] , where
i = − 1 i = \sqrt{-1} i = − 1 .
If
M = A T B A \mathrm{M} = \mathrm{A}^\mathrm{T} \mathrm{BA} M = A T BA , then the inverse of the matrix
A M 2023 A T \mathrm{AM}^{2023}\mathrm{A}^\mathrm{T} AM 2023 A T is
Hint 1 of 4
What is the product A A T AA^T A A T A = [ 1 10 3 10 − 3 10 1 10 ] A = \begin{bmatrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{bmatrix} A = [ 10 1 10 − 3 10 3 10 1 ]
A A T = I AA^T = I A A T = I A A A A A T = − I AA^T = -I A A T = − I Step-by-step solution View Correct answer
Since A A A A M 2023 A T = B 2023 = [ 1 − 2023 i 0 1 ] AM^{2023}A^T = B^{2023} = \begin{bmatrix} 1 & -2023i \\ 0 & 1 \end{bmatrix} A M 2023 A T = B 2023 = [ 1 0 − 2023 i 1 ] [ 1 2023 i 0 1 ] \begin{bmatrix} 1 & 2023i \\ 0 & 1 \end{bmatrix} [ 1 0 2023 i 1 ] Option analysis
Why each option works or fails A · [ 1 0 − 2023 i 1 ] \begin{bmatrix} 1 & 0 \\ -2023i & 1 \end{bmatrix} [ 1 − 2023 i 0 1 ] Believing that matrix inversion transposes the matrix without changing the sign of off-diagonal elements. Recall that the inverse of an upper triangular shear matrix [ 1 k 0 1 ] \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix} [ 1 0 k 1 ] [ 1 − k 0 1 ] \begin{bmatrix} 1 & -k \\ 0 & 1 \end{bmatrix} [ 1 0 − k 1 ]
B · [ 1 − 2023 i 0 1 ] \begin{bmatrix} 1 & -2023i \\ 0 & 1 \end{bmatrix} [ 1 0 − 2023 i 1 ] Computing the simplified matrix expression A M 2023 A T = B 2023 AM^{2023}A^T = B^{2023} A M 2023 A T = B 2023 A M 2023 A T AM^{2023}A^T A M 2023 A T
C · [ 1 0 2023 i 1 ] \begin{bmatrix} 1 & 0 \\ 2023i & 1 \end{bmatrix} [ 1 2023 i 0 1 ] Confusing transposition with inversion, transposing the matrix and inverting the off-diagonal sign simultaneously. The matrix B 2023 B^{2023} B 2023
D · [ 1 2023 i 0 1 ] \begin{bmatrix} 1 & 2023i \\ 0 & 1 \end{bmatrix} [ 1 0 2023 i 1 ] None. The student correctly recognized A A T = I AA^T = I A A T = I B 2023 B^{2023} B 2023
Step Working
01 given A = [ 1 10 3 10 − 3 10 1 10 ] A = \begin{bmatrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ -\frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{bmatrix} A = [ 10 1 − 10 3 10 3 10 1 ] B = [ 1 − i 0 1 ] B = \begin{bmatrix} 1 & -i \\ 0 & 1 \end{bmatrix} B = [ 1 0 − i 1 ] M = A T B A M = A^T B A M = A T B A
02 approach Check if A A A A A T = I A A^T = I A A T = I M k = ( A T B A ) k = A T B k A M^k = (A^T B A)^k = A^T B^k A M k = ( A T B A ) k = A T B k A A M 2023 A T A M^{2023} A^T A M 2023 A T B 2023 B^{2023} B 2023 B 2023 B^{2023} B 2023
03 execute Verify orthogonality of A A A A A T = [ 1 10 3 10 − 3 10 1 10 ] [ 1 10 − 3 10 3 10 1 10 ] = [ 1 10 + 9 10 − 3 10 + 3 10 − 3 10 + 3 10 9 10 + 1 10 ] = [ 1 0 0 1 ] = I A A^T = \begin{bmatrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ -\frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{10}} & -\frac{3}{\sqrt{10}} \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{bmatrix} = \begin{bmatrix} \frac{1}{10}+\frac{9}{10} & -\frac{3}{10}+\frac{3}{10} \\ -\frac{3}{10}+\frac{3}{10} & \frac{9}{10}+\frac{1}{10} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I A A T = [ 10 1 − 10 3 10 3 10 1 ] [ 10 1 10 3 − 10 3 10 1 ] = [ 10 1 + 10 9 − 10 3 + 10 3 − 10 3 + 10 3 10 9 + 10 1 ] = [ 1 0 0 1 ] = I A T A = A A T = I A^T A = A A^T = I A T A = A A T = I
04 execute Express M 2023 M^{2023} M 2023 D = A M 2023 A T D = A M^{2023} A^T D = A M 2023 A T M = A T B A ⟹ M 2 = ( A T B A ) ( A T B A ) = A T B ( A A T ) B A = A T B 2 A M = A^T B A \implies M^2 = (A^T B A)(A^T B A) = A^T B (A A^T) B A = A^T B^2 A M = A T B A ⟹ M 2 = ( A T B A ) ( A T B A ) = A T B ( A A T ) B A = A T B 2 A M 2023 = A T B 2023 A M^{2023} = A^T B^{2023} A M 2023 = A T B 2023 A D = A M 2023 A T = A ( A T B 2023 A ) A T = ( A A T ) B 2023 ( A A T ) = B 2023 D = A M^{2023} A^T = A (A^T B^{2023} A) A^T = (A A^T) B^{2023} (A A^T) = B^{2023} D = A M 2023 A T = A ( A T B 2023 A ) A T = ( A A T ) B 2023 ( A A T ) = B 2023
05 execute Compute B 2023 B^{2023} B 2023 B = [ 1 − i 0 1 ] B = \begin{bmatrix} 1 & -i \\ 0 & 1 \end{bmatrix} B = [ 1 0 − i 1 ] B n = [ 1 − n i 0 1 ] B^n = \begin{bmatrix} 1 & -n i \\ 0 & 1 \end{bmatrix} B n = [ 1 0 − ni 1 ] D = B 2023 = [ 1 − 2023 i 0 1 ] D = B^{2023} = \begin{bmatrix} 1 & -2023i \\ 0 & 1 \end{bmatrix} D = B 2023 = [ 1 0 − 2023 i 1 ] D − 1 = ( B 2023 ) − 1 = B − 2023 = [ 1 2023 i 0 1 ] D^{-1} = (B^{2023})^{-1} = B^{-2023} = \begin{bmatrix} 1 & 2023i \\ 0 & 1 \end{bmatrix} D − 1 = ( B 2023 ) − 1 = B − 2023 = [ 1 0 2023 i 1 ]
✓ verify Multiply D D − 1 = [ 1 − 2023 i 0 1 ] [ 1 2023 i 0 1 ] = [ 1 2023 i − 2023 i 0 1 ] = I D D^{-1} = \begin{bmatrix} 1 & -2023i \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2023i \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2023i - 2023i \\ 0 & 1 \end{bmatrix} = I D D − 1 = [ 1 0 − 2023 i 1 ] [ 1 0 2023 i 1 ] = [ 1 0 2023 i − 2023 i 1 ] = I
Hints that build this answer step by step What is the product A A T AA^T A A T A = [ 1 10 3 10 − 3 10 1 10 ] A = \begin{bmatrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{bmatrix} A = [ 10 1 10 − 3 10 3 10 1 ]
A A T = I AA^T = I A A T = I A A A Using M = A T B A M = A^T B A M = A T B A A A T = I A A^T = I A A T = I A M 2023 A T A M^{2023} A^T A M 2023 A T
B 2023 B^{2023} B 2023 Given B = [ 1 − i 0 1 ] B = \begin{bmatrix} 1 & -i \\ 0 & 1 \end{bmatrix} B = [ 1 0 − i 1 ] B 2023 B^{2023} B 2023
[ 1 − 2023 i 0 1 ] \begin{bmatrix} 1 & -2023i \\ 0 & 1 \end{bmatrix} [ 1 0 − 2023 i 1 ] What is the inverse of the matrix A M 2023 A T = [ 1 − 2023 i 0 1 ] AM^{2023}A^T = \begin{bmatrix} 1 & -2023i \\ 0 & 1 \end{bmatrix} A M 2023 A T = [ 1 0 − 2023 i 1 ]
[ 1 2023 i 0 1 ] \begin{bmatrix} 1 & 2023i \\ 0 & 1 \end{bmatrix} [ 1 0 2023 i 1 ] Your next move We think you should solve this next ✓ Source and academic review↓
Question type Single correct
Exam relevance JEE Main · Mathematics
Concepts assessed Mathematics
Academic status Reviewed by official_key
Source pyq
Editorial review 9 September 2026 Quick checks
Students also ask Why does inverting a matrix of the form [1, k; 0, 1] simply change the sign of k? Because det ( [ 1 k 0 1 ] ) = 1 \det(\begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}) = 1 det ( [ 1 0 k 1 ] ) = 1 2 × 2 2 \times 2 2 × 2 1 a d − b c [ d − b − c a ] \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} a d − b c 1 [ d − c − b a ] [ 1 − k 0 1 ] \begin{bmatrix} 1 & -k \\ 0 & 1 \end{bmatrix} [ 1 0 − k 1 ]
Answer Since A A A A M 2023 A T = B 2023 = [ 1 − 2023 i 0 1 ] AM^{2023}A^T = B^{2023} = \begin{bmatrix} 1 & -2023i \\ 0 & 1 \end{bmatrix} A M 2023 A T = B 2023 = [ 1 0 − 2023 i 1 ] [ 1 2023 i 0 1 ] \begin{bmatrix} 1 & 2023i \\ 0 & 1 \end{bmatrix} [ 1 0 2023 i 1 ]
Why each option works or fails A: [ 1 0 − 2023 i 1 ] \begin{bmatrix} 1 & 0 \\ -2023i & 1 \end{bmatrix} [ 1 − 2023 i 0 1 ] - Believing that matrix inversion transposes the matrix without changing the sign of off-diagonal elements. Recall that the inverse of an upper triangular shear matrix [ 1 k 0 1 ] \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix} [ 1 0 k 1 ] [ 1 − k 0 1 ] \begin{bmatrix} 1 & -k \\ 0 & 1 \end{bmatrix} [ 1 0 − k 1 ] B: [ 1 − 2023 i 0 1 ] \begin{bmatrix} 1 & -2023i \\ 0 & 1 \end{bmatrix} [ 1 0 − 2023 i 1 ] - Computing the simplified matrix expression A M 2023 A T = B 2023 AM^{2023}A^T = B^{2023} A M 2023 A T = B 2023 A M 2023 A T AM^{2023}A^T A M 2023 A T C: [ 1 0 2023 i 1 ] \begin{bmatrix} 1 & 0 \\ 2023i & 1 \end{bmatrix} [ 1 2023 i 0 1 ] - Confusing transposition with inversion, transposing the matrix and inverting the off-diagonal sign simultaneously. The matrix B 2023 B^{2023} B 2023 D · correct: [ 1 2023 i 0 1 ] \begin{bmatrix} 1 & 2023i \\ 0 & 1 \end{bmatrix} [ 1 0 2023 i 1 ] - None. The student correctly recognized A A T = I AA^T = I A A T = I B 2023 B^{2023} B 2023 Step-by-step solution given: A = [ 1 10 3 10 − 3 10 1 10 ] A = \begin{bmatrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ -\frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{bmatrix} A = [ 10 1 − 10 3 10 3 10 1 ] B = [ 1 − i 0 1 ] B = \begin{bmatrix} 1 & -i \\ 0 & 1 \end{bmatrix} B = [ 1 0 − i 1 ] M = A T B A M = A^T B A M = A T B A approach: Check if A A A A A T = I A A^T = I A A T = I M k = ( A T B A ) k = A T B k A M^k = (A^T B A)^k = A^T B^k A M k = ( A T B A ) k = A T B k A A M 2023 A T A M^{2023} A^T A M 2023 A T B 2023 B^{2023} B 2023 B 2023 B^{2023} B 2023 execute: Verify orthogonality of A A A A A T = [ 1 10 3 10 − 3 10 1 10 ] [ 1 10 − 3 10 3 10 1 10 ] = [ 1 10 + 9 10 − 3 10 + 3 10 − 3 10 + 3 10 9 10 + 1 10 ] = [ 1 0 0 1 ] = I A A^T = \begin{bmatrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ -\frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{10}} & -\frac{3}{\sqrt{10}} \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{bmatrix} = \begin{bmatrix} \frac{1}{10}+\frac{9}{10} & -\frac{3}{10}+\frac{3}{10} \\ -\frac{3}{10}+\frac{3}{10} & \frac{9}{10}+\frac{1}{10} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I A A T = [ 10 1 − 10 3 10 3 10 1 ] [ 10 1 10 3 − 10 3 10 1 ] = [ 10 1 + 10 9 − 10 3 + 10 3 − 10 3 + 10 3 10 9 + 10 1 ] = [ 1 0 0 1 ] = I A T A = A A T = I A^T A = A A^T = I A T A = A A T = I execute: Express M 2023 M^{2023} M 2023 D = A M 2023 A T D = A M^{2023} A^T D = A M 2023 A T M = A T B A ⟹ M 2 = ( A T B A ) ( A T B A ) = A T B ( A A T ) B A = A T B 2 A M = A^T B A \implies M^2 = (A^T B A)(A^T B A) = A^T B (A A^T) B A = A^T B^2 A M = A T B A ⟹ M 2 = ( A T B A ) ( A T B A ) = A T B ( A A T ) B A = A T B 2 A M 2023 = A T B 2023 A M^{2023} = A^T B^{2023} A M 2023 = A T B 2023 A D = A M 2023 A T = A ( A T B 2023 A ) A T = ( A A T ) B 2023 ( A A T ) = B 2023 D = A M^{2023} A^T = A (A^T B^{2023} A) A^T = (A A^T) B^{2023} (A A^T) = B^{2023} D = A M 2023 A T = A ( A T B 2023 A ) A T = ( A A T ) B 2023 ( A A T ) = B 2023 execute: Compute B 2023 B^{2023} B 2023 B = [ 1 − i 0 1 ] B = \begin{bmatrix} 1 & -i \\ 0 & 1 \end{bmatrix} B = [ 1 0 − i 1 ] B n = [ 1 − n i 0 1 ] B^n = \begin{bmatrix} 1 & -n i \\ 0 & 1 \end{bmatrix} B n = [ 1 0 − ni 1 ] D = B 2023 = [ 1 − 2023 i 0 1 ] D = B^{2023} = \begin{bmatrix} 1 & -2023i \\ 0 & 1 \end{bmatrix} D = B 2023 = [ 1 0 − 2023 i 1 ] D − 1 = ( B 2023 ) − 1 = B − 2023 = [ 1 2023 i 0 1 ] D^{-1} = (B^{2023})^{-1} = B^{-2023} = \begin{bmatrix} 1 & 2023i \\ 0 & 1 \end{bmatrix} D − 1 = ( B 2023 ) − 1 = B − 2023 = [ 1 0 2023 i 1 ] verify: Multiply D D − 1 = [ 1 − 2023 i 0 1 ] [ 1 2023 i 0 1 ] = [ 1 2023 i − 2023 i 0 1 ] = I D D^{-1} = \begin{bmatrix} 1 & -2023i \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2023i \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2023i - 2023i \\ 0 & 1 \end{bmatrix} = I D D − 1 = [ 1 0 − 2023 i 1 ] [ 1 0 2023 i 1 ] = [ 1 0 2023 i − 2023 i 1 ] = I