Sample 2024-25 Class 12 Maths Q1 Solution: |A| + |adj A| when A(adj A) = 2025 I

The question

If for a square matrix $A$ ⁠,

\begin{aligned} &A.(\operatorname{adj} A) \\ &= \begin{bmatrix} 2025 & 0 & 0 \\ 0 & 2025 & 0 \\ 0 & 0 & 2025 \end{bmatrix}, \end{aligned}

then the value of $|A| + |\operatorname{adj} A|$ is equal to:

(A) $1$

(B) $2025 + 1$

(C) $(2025)^2 + 45$

(D) $2025 + (2025)^2$

Exam answer

For a square matrix $A$ of order $n$ ⁠, $A\,(\operatorname{adj} A) = |A|\,I_n$ ⁠.

Why?

This is the basic property of the adjoint. Each diagonal entry of $A\,(\operatorname{adj} A)$ is a row of $A$ times the cofactors of that same row, which is the expansion of $|A|$ ⁠. Each other entry uses the cofactors of a different row, and that sum is always 0.

The given matrix has 2025 on the diagonal and 0 elsewhere, so

A\,(\operatorname{adj} A) = 2025\,I_3

Why?

Taking 2025 out of every entry leaves 1 on the diagonal and 0 elsewhere, which is the identity matrix $I_3$ ⁠. So $A$ is of order $n = 3$ ⁠.

\begin{aligned} &\begin{bmatrix} 2025 & 0 & 0 \\ 0 & 2025 & 0 \\ 0 & 0 & 2025 \end{bmatrix} \\ &= 2025\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ &= 2025\,I_3 \end{aligned}

Now $A\,(\operatorname{adj} A) = |A|\,I_3$ and also $A\,(\operatorname{adj} A) = 2025\,I_3$ ⁠. The numbers in front of $I_3$ must match, so $|A| = 2025$ ⁠.

Comparing with $|A|\,I_3$ ⁠, $|A| = 2025$ ⁠.

Using $|\operatorname{adj} A| = |A|^{n - 1}$ with $n = 3$ ⁠,

|\operatorname{adj} A| = |A|^{3 - 1} = (2025)^2

Why?

Where the rule comes from: take determinants of both sides of $A\,(\operatorname{adj} A) = |A|\,I_3$ ⁠. Since $|kI_3| = k^3$ ⁠,

\begin{aligned} |A\,(\operatorname{adj} A)| &= \big|\,|A|\,I_3\big| \\ |A|\,|\operatorname{adj} A| &= |A|^3 \end{aligned}

Here $|A| = 2025 \ne 0$ ⁠, so we may divide by $|A|$ ⁠. That leaves $|\operatorname{adj} A| = |A|^2$ ⁠.

\begin{aligned} |\operatorname{adj} A| &= \frac{|A|^3}{|A|} = |A|^2 \\ &= (2025)^2 \end{aligned}

Watch out: the power is $n - 1 = 2$ ⁠, not $n = 3$ ⁠.

\begin{aligned} \therefore\ &|A| + |\operatorname{adj} A| \\ &= 2025 + (2025)^2 \end{aligned}

Answer: (D) $2025 + (2025)^2$

Where this comes from

Chapter 4, Determinants: Adjoint: |A| and |adj A| from A(adj A).

Question type: One-mark questions on $|kA|$ ⁠, $|\operatorname{adj}A|$ and $|A^{-1}|$

How to spot it: click to see

Type 10. One-mark questions on $|kA|$ ⁠, $|\operatorname{adj}A|$ and $|A^{-1}|$

Spot it: an MCQ about a matrix you are not given, only its order.

Here: $A(\operatorname{adj}A) = |A|I$ ⁠, and the given matrix is $2025I$ of order 3, so $|A| = 2025$ and $|\operatorname{adj}A| = |A|^{3 - 1} = 2025^2$ ⁠.

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Q1: ∣A∣+∣adj⁡A∣|A| + |\operatorname{adj} A|∣A∣+∣adjA∣ when A (adj⁡A)=2025 IA\,(\operatorname{adj} A) = 2025\,IA(adjA)=2025I

Q1: $|A| + |\operatorname{adj} A|$ when $A\,(\operatorname{adj} A) = 2025\,I$