JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Inverse of a Conjugated Matrix Power

Q: 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, and by the 2 × 2 matrix inverse formula 1/(ad-bc)[ d -b; -c a ], the inverse is directly [ 1 -k; 0 1 ].

Let A = [ 1/(√(10)) 3/(√(10)); (-3)/(√(10)) 1/(√(10)) ] and B = [ 1 -i; 0 1 ], where i = √(-1). If M = A^()T BA, then the inverse of the matrix AM^2023A^()T is

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A[ 1 0; -2023i 1 ]B[ 1 -2023i; 0 1 ]C[ 1 0; 2023i 1 ]D[ 1 2023i; 0 1 ]

Step-by-step solutionView

Correct answer

Since $A$ is orthogonal, $A M^{2023} A^{T} = B^{2023} = [\begin{array}{cc} 1 & - 2023 i \\ 0 & 1 \end{array}]$ , and its inverse is $[\begin{array}{cc} 1 & 2023 i \\ 0 & 1 \end{array}]$ .

Option analysis

Why each option works or fails

A · [ 1 0; -2023i 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 $[\begin{array}{cc} 1 & k \\ 0 & 1 \end{array}]$ is $[\begin{array}{cc} 1 & - k \\ 0 & 1 \end{array}]$ , not its transpose.

B · [ 1 -2023i; 0 1 ]

Computing the simplified matrix expression $A M^{2023} A^{T} = B^{2023}$ correctly but forgetting to take its inverse as required by the stem. Always check the final operation requested in the stem; here the question asks for the inverse of $A M^{2023} A^{T}$ , not the matrix itself.

C · [ 1 0; 2023i 1 ]

Confusing transposition with inversion, transposing the matrix and inverting the off-diagonal sign simultaneously. The matrix $B^{2023}$ has zero in the lower-left entry, so its inverse also maintains zeros on the subdiagonal; do not swap rows and columns.

D · [ 1 2023i; 0 1 ]

The student correctly recognized $A A^{T} = I$ , simplified the power to $B^{2023}$ , and inverted it by negating the off-diagonal entry. Correct approach.

Reviewed route

Solution

StepWorking

01Given

$A = [\begin{array}{cc} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ - \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{array}]$ , $B = [\begin{array}{cc} 1 & - i \\ 0 & 1 \end{array}]$ , and $M = A^{T} B A$ .

02Approach

Check if $A$ is an orthogonal matrix. Since $A A^{T} = I$ , simplify $M^{k} = (A^{T} B A)^{k} = A^{T} B^{k} A$ , which collapses $A M^{2023} A^{T}$ to $B^{2023}$ . Finally, find the inverse of $B^{2023}$ .

03Execute

Verify orthogonality of $A$ : $A A^{T} = [\begin{array}{cc} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ - \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{array}] [\begin{array}{cc} \frac{1}{\sqrt{10}} & - \frac{3}{\sqrt{10}} \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{array}] = [\begin{array}{cc} \frac{1}{10} + \frac{9}{10} & - \frac{3}{10} + \frac{3}{10} \\ - \frac{3}{10} + \frac{3}{10} & \frac{9}{10} + \frac{1}{10} \end{array}] = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}] = I$ . Thus, $A^{T} A = A A^{T} = I$ .

04Execute

Express $M^{2023}$ and evaluate the required matrix $D = A M^{2023} A^{T}$ : $M = A^{T} B A ⟹ M^{2} = (A^{T} B A) (A^{T} B A) = A^{T} B (A A^{T}) B A = A^{T} B^{2} A$ . By induction, $M^{2023} = A^{T} B^{2023} A$ . Therefore, $D = A M^{2023} A^{T} = A (A^{T} B^{2023} A) A^{T} = (A A^{T}) B^{2023} (A A^{T}) = B^{2023}$ .

05Execute

Compute $B^{2023}$ and its inverse: For a shear matrix $B = [\begin{array}{cc} 1 & - i \\ 0 & 1 \end{array}]$ , $B^{n} = [\begin{array}{cc} 1 & - n i \\ 0 & 1 \end{array}]$ . So $D = B^{2023} = [\begin{array}{cc} 1 & - 2023 i \\ 0 & 1 \end{array}]$ . The inverse is $D^{- 1} = (B^{2023})^{- 1} = B^{- 2023} = [\begin{array}{cc} 1 & 2023 i \\ 0 & 1 \end{array}]$ .

✓Verify

Multiply $D D^{- 1} = [\begin{array}{cc} 1 & - 2023 i \\ 0 & 1 \end{array}] [\begin{array}{cc} 1 & 2023 i \\ 0 & 1 \end{array}] = [\begin{array}{cc} 1 & 2023 i - 2023 i \\ 0 & 1 \end{array}] = I$ . Matches Option D.

Hints that build this answer step by step

What is the product $A A^{T}$ for the given matrix $A = [\begin{array}{cc} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{- 3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{array}]$ ?
$A A^{T} = I$ , so $A$ is an orthogonal matrix
Using $M = A^{T} B A$ and $A A^{T} = I$ , what does the expression $A M^{2023} A^{T}$ simplify to?
$B^{2023}$
Given $B = [\begin{array}{cc} 1 & - i \\ 0 & 1 \end{array}]$ , what is $B^{2023}$ ?
$[\begin{array}{cc} 1 & - 2023 i \\ 0 & 1 \end{array}]$
What is the inverse of the matrix $A M^{2023} A^{T} = [\begin{array}{cc} 1 & - 2023 i \\ 0 & 1 \end{array}]$ ?
$[\begin{array}{cc} 1 & 2023 i \\ 0 & 1 \end{array}]$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
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 $\det ([\begin{array}{cc} 1 & k \\ 0 & 1 \end{array}]) = 1$ , and by the $2 \times 2$ matrix inverse formula $\frac{1}{a d - b c} [\begin{array}{cc} d & - b \\ - c & a \end{array}]$ , the inverse is directly $[\begin{array}{cc} 1 & - k \\ 0 & 1 \end{array}]$ .

Answer

Since $A$ is orthogonal, $A M^{2023} A^{T} = B^{2023} = [\begin{array}{cc} 1 & - 2023 i \\ 0 & 1 \end{array}]$ , and its inverse is $[\begin{array}{cc} 1 & 2023 i \\ 0 & 1 \end{array}]$ .

Why each option works or fails

A: [ 1 0; -2023i 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 $[\begin{array}{cc} 1 & k \\ 0 & 1 \end{array}]$ is $[\begin{array}{cc} 1 & - k \\ 0 & 1 \end{array}]$ , not its transpose.
B: [ 1 -2023i; 0 1 ] - Computing the simplified matrix expression $A M^{2023} A^{T} = B^{2023}$ correctly but forgetting to take its inverse as required by the stem. Always check the final operation requested in the stem; here the question asks for the inverse of $A M^{2023} A^{T}$ , not the matrix itself.
C: [ 1 0; 2023i 1 ] - Confusing transposition with inversion, transposing the matrix and inverting the off-diagonal sign simultaneously. The matrix $B^{2023}$ has zero in the lower-left entry, so its inverse also maintains zeros on the subdiagonal; do not swap rows and columns.
D · correct: [ 1 2023i; 0 1 ] - The student correctly recognized $A A^{T} = I$ , simplified the power to $B^{2023}$ , and inverted it by negating the off-diagonal entry. Correct approach.

Step-by-step solution

Given: $A = [\begin{array}{cc} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ - \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{array}]$ , $B = [\begin{array}{cc} 1 & - i \\ 0 & 1 \end{array}]$ , and $M = A^{T} B A$ .
Approach: Check if $A$ is an orthogonal matrix. Since $A A^{T} = I$ , simplify $M^{k} = (A^{T} B A)^{k} = A^{T} B^{k} A$ , which collapses $A M^{2023} A^{T}$ to $B^{2023}$ . Finally, find the inverse of $B^{2023}$ .
Execute: Verify orthogonality of $A$ : $A A^{T} = [\begin{array}{cc} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ - \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{array}] [\begin{array}{cc} \frac{1}{\sqrt{10}} & - \frac{3}{\sqrt{10}} \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{array}] = [\begin{array}{cc} \frac{1}{10} + \frac{9}{10} & - \frac{3}{10} + \frac{3}{10} \\ - \frac{3}{10} + \frac{3}{10} & \frac{9}{10} + \frac{1}{10} \end{array}] = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}] = I$ . Thus, $A^{T} A = A A^{T} = I$ .
Execute: Express $M^{2023}$ and evaluate the required matrix $D = A M^{2023} A^{T}$ : $M = A^{T} B A ⟹ M^{2} = (A^{T} B A) (A^{T} B A) = A^{T} B (A A^{T}) B A = A^{T} B^{2} A$ . By induction, $M^{2023} = A^{T} B^{2023} A$ . Therefore, $D = A M^{2023} A^{T} = A (A^{T} B^{2023} A) A^{T} = (A A^{T}) B^{2023} (A A^{T}) = B^{2023}$ .
Execute: Compute $B^{2023}$ and its inverse: For a shear matrix $B = [\begin{array}{cc} 1 & - i \\ 0 & 1 \end{array}]$ , $B^{n} = [\begin{array}{cc} 1 & - n i \\ 0 & 1 \end{array}]$ . So $D = B^{2023} = [\begin{array}{cc} 1 & - 2023 i \\ 0 & 1 \end{array}]$ . The inverse is $D^{- 1} = (B^{2023})^{- 1} = B^{- 2023} = [\begin{array}{cc} 1 & 2023 i \\ 0 & 1 \end{array}]$ .
Verify: Multiply $D D^{- 1} = [\begin{array}{cc} 1 & - 2023 i \\ 0 & 1 \end{array}] [\begin{array}{cc} 1 & 2023 i \\ 0 & 1 \end{array}] = [\begin{array}{cc} 1 & 2023 i - 2023 i \\ 0 & 1 \end{array}] = I$ . Matches Option D.

Inverse of a Conjugated Matrix Power

What feels right?

Since $A$ is orthogonal, $A M^{2023} A^{T} = B^{2023} = [\begin{array}{cc} 1 & - 2023 i \\ 0 & 1 \end{array}]$ , and its inverse is $[\begin{array}{cc} 1 & 2023 i \\ 0 & 1 \end{array}]$ .

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

05Execute

✓Verify

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