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CBSE Board Examination 2025 · Section A · 1 mark

Q1: Inverse of A=diag(−1,1,1)A = \text{diag}(-1, 1, 1)

Board 2025Ch 3 · Matrices
The question

If A=[−100010001]A = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}⁠, then A−1A^{-1} is

(A) [−1000−1000−1]\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}

(B) [1000−1000−1]\begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}

(C) [−1000−10001]\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

(D) [−100010001]\begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

Exam answer

∣A∣=(−1)(1)(1)=−1≠0|A| = (-1)(1)(1) = -1 \ne 0⁠, so A−1A^{-1} exists.

Why?

A is a diagonal matrix, so its determinant is the product of its diagonal entries. Expanding along the first row gives the same value: −1×(1−0)=−1-1 \times (1 - 0) = -1⁠.

A square matrix has an inverse exactly when its determinant is not zero.

A is diagonal, so A2A^2 has diagonal entries (−1)2,12,12(-1)^2, 1^2, 1^2⁠, that is, A2=AA=IA^2 = AA = I⁠.

AA=[100010001]=IAA = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I
Why?

The product of two diagonal matrices is diagonal. Each diagonal entry is the product of the matching diagonal entries:

(−1)(−1)=1(1)(1)=1\begin{aligned} (-1)(-1) &= 1 \\ (1)(1) &= 1 \end{aligned}

Every entry off the diagonal stays 0.

AA=IAA = I⁠, so A−1=AA^{-1} = A⁠.

Why?

By definition, B is the inverse of A when AB=BA=IAB = BA = I⁠. Here A itself does the job, so A is its own inverse.

Watch out: the inverse of a diagonal matrix replaces each diagonal entry d by 1d\frac{1}{d}⁠. Since 1−1=−1\frac{1}{-1} = -1 and 11=1\frac{1}{1} = 1⁠, no entry changes. Changing signs, as in (A), (B) and (C), is wrong.

Of the options, only (D) is equal to A.

Answer: (D) [−100010001]\begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

Where this comes from

Chapter 3, Matrices: Inverse of a diagonal matrix.

Question type: Inverse of a matrix

How to spot it: click to see

Type 6. Inverse of a matrix: ∣A∣|A|⁠, then the adjoint, then divide

Spot it: "find the inverse" or "find A−1A^{-1}".

Here: A−1A^{-1} is asked, so first ∣A∣=−1≠0|A| = -1 \ne 0⁠. This AA is diagonal with entries ±1\pm 1⁠, so AA=IAA = I and A−1=AA^{-1} = A⁠, with no cofactors needed.

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