JEE MainMathematics

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Determinant Condition on a Linear Combination of a $2 \times 2$ Matrix and its Adjugate

Q: Why replace np with mq - d?

The options involve m, q, and d, but not n and p. Since d = mq - np, substituting np = mq - d eliminates n and p directly.

Q: Can we always write adj A = d A^(-1)?

Yes, because A (adj A) = |A| I = d I, and since d ≠ 0, A is invertible.

Let A = [ m n; p q ], d = |A| ≠ 0 and |A - d(Adj A)| = 0. Then

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What feels right?

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A1 + d^2 = m^2 + q^2B1 + d^2 = (m + q)^2C(1 + d)^2 = m^2 + q^2D(1 + d)^2 = (m + q)^2

Step-by-step solutionView

Correct answer

Using $A (adj A) = d I$ , we rewrite $∣ A - d (adj A) ∣ = 0$ as $∣ A^{2} - d^{2} I ∣ = 0$ , which evaluates via the characteristic equation to $(1 + d)^{2} = (m + q)^{2}$ .

Option analysis

Why each option works or fails

A · 1 + d^2 = m^2 + q^2

Believing that expanding $(1 + d)^{2}$ and $(m + q)^{2}$ omits the cross-terms $2 d$ and $2 m q$ , falsely simplifying to sums of squares. Retain the cross-terms when completing the square: $(m q - n p) = d$ , so $2 m q - 2 d = 2 n p$ , giving $(m + q)^{2} = (1 + d)^{2}$ , not $m^{2} + q^{2} = 1 + d^{2}$ .

B · 1 + d^2 = (m + q)^2

Failing to distribute the square correctly over $(1 + d)$ , writing $1 + d^{2}$ instead of $(1 + d)^{2} = 1 + 2 d + d^{2}$ . Notice that expanding $∣ A - d (adj A) ∣$ yields $(m - q d) (q - m d) - n p (1 + d)^{2} = 0$ , which factors symmetrically in $(1 + d)^{2}$ .

C · (1 + d)^2 = m^2 + q^2

Failing to account for the $2 m q$ cross-term when rewriting $(m + q)^{2} - 2 m q$ , incorrectly equating $(1 + d)^{2}$ to $m^{2} + q^{2}$ . Express the determinant directly in terms of trace $tr (A) = m + q$ , which appears squared as $(m + q)^{2}$ .

D · (1 + d)^2 = (m + q)^2

This option correctly relates the trace squared $(m + q)^{2}$ to $(1 + d)^{2}$ . Directly evaluate $∣ A - d (adj A) ∣ = (1 + d^{2}) d - d (m + q)^{2} + 2 d^{2} = 0$ , divide by $d \neq 0$ , and complete the square to get $(1 + d)^{2} = (m + q)^{2}$ .

Reviewed route

Solution

StepWorking

01Given

$A = (\begin{array}{cc} m & n \\ p & q \end{array})$ , $d = ∣ A ∣ = m q - n p \neq 0$ , and $∣ A - d (adj A) ∣ = 0$ .

02Approach

Compute the explicit 2x2 matrix expression for $A - d (adj A)$ , set its determinant to zero, and eliminate $n p$ using $n p = m q - d$ to relate $d$ with $(m + q)$ .

03Execute

For $A = (\begin{array}{cc} m & n \\ p & q \end{array})$ , we have $adj A = (\begin{array}{cc} q & - n \\ - p & m \end{array})$ . Thus, $A - d (adj A) = (\begin{array}{cc} m - d q & n (1 + d) \\ p (1 + d) & q - d m \end{array})$ .

04Execute

Setting determinant to zero: $∣ A - d (adj A) ∣ = (m - d q) (q - d m) - n p (1 + d)^{2} = 0$ . Expanding the first product gives $m q - d (m^{2} + q^{2}) + d^{2} m q - n p (1 + d)^{2} = 0$ . Since $n p = m q - d$ , this becomes $m q (1 + d^{2}) - d (m^{2} + q^{2}) - (m q - d) (1 + d)^{2} = 0$ .

05Execute

Expanding $(m q - d) (1 + d)^{2} = m q (1 + 2 d + d^{2}) - d (1 + d)^{2}$ . Substituting back: $m q (1 + d^{2}) - d (m^{2} + q^{2}) - [m q (1 + d^{2}) + 2 d m q - d (1 + d)^{2}] = 0 ⟹ - d (m^{2} + q^{2} + 2 m q) + d (1 + d)^{2} = 0$ . Since $d \neq 0$ , dividing by $d$ yields $(1 + d)^{2} = (m + q)^{2}$ .

✓Verify

Let $A = I_{2} = (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array})$ . Here $m = 1, q = 1, n = 0, p = 0$ , so $d = 1$ . Then $adj A = I_{2}$ , giving $A - d (adj A) = I - I = O$ , so $∣ O ∣ = 0$ . Checking option D: $(1 + 1)^{2} = 4$ and $(m + q)^{2} = (1 + 1)^{2} = 4$ , which holds identically.

✓ 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 replace

n p

with

m q - d

The options involve $m, q$ , and $d$ , but not $n$ and $p$ . Since $d = m q - n p$ , substituting $n p = m q - d$ eliminates $n$ and $p$ directly.

Can we always write

adj A = d A^{- 1}

Yes, because $A (adj A) = ∣ A ∣ I = d I$ , and since $d \neq 0$ , $A$ is invertible.

Answer

Using $A (adj A) = d I$ , we rewrite $∣ A - d (adj A) ∣ = 0$ as $∣ A^{2} - d^{2} I ∣ = 0$ , which evaluates via the characteristic equation to $(1 + d)^{2} = (m + q)^{2}$ .

Why each option works or fails

A: 1 + d^2 = m^2 + q^2 - Believing that expanding $(1 + d)^{2}$ and $(m + q)^{2}$ omits the cross-terms $2 d$ and $2 m q$ , falsely simplifying to sums of squares. Retain the cross-terms when completing the square: $(m q - n p) = d$ , so $2 m q - 2 d = 2 n p$ , giving $(m + q)^{2} = (1 + d)^{2}$ , not $m^{2} + q^{2} = 1 + d^{2}$ .
B: 1 + d^2 = (m + q)^2 - Failing to distribute the square correctly over $(1 + d)$ , writing $1 + d^{2}$ instead of $(1 + d)^{2} = 1 + 2 d + d^{2}$ . Notice that expanding $∣ A - d (adj A) ∣$ yields $(m - q d) (q - m d) - n p (1 + d)^{2} = 0$ , which factors symmetrically in $(1 + d)^{2}$ .
C: (1 + d)^2 = m^2 + q^2 - Failing to account for the $2 m q$ cross-term when rewriting $(m + q)^{2} - 2 m q$ , incorrectly equating $(1 + d)^{2}$ to $m^{2} + q^{2}$ . Express the determinant directly in terms of trace $tr (A) = m + q$ , which appears squared as $(m + q)^{2}$ .
D · correct: (1 + d)^2 = (m + q)^2 - This option correctly relates the trace squared $(m + q)^{2}$ to $(1 + d)^{2}$ . Directly evaluate $∣ A - d (adj A) ∣ = (1 + d^{2}) d - d (m + q)^{2} + 2 d^{2} = 0$ , divide by $d \neq 0$ , and complete the square to get $(1 + d)^{2} = (m + q)^{2}$ .

Step-by-step solution

Given: $A = (\begin{array}{cc} m & n \\ p & q \end{array})$ , $d = ∣ A ∣ = m q - n p \neq 0$ , and $∣ A - d (adj A) ∣ = 0$ .
Approach: Compute the explicit 2x2 matrix expression for $A - d (adj A)$ , set its determinant to zero, and eliminate $n p$ using $n p = m q - d$ to relate $d$ with $(m + q)$ .
Execute: For $A = (\begin{array}{cc} m & n \\ p & q \end{array})$ , we have $adj A = (\begin{array}{cc} q & - n \\ - p & m \end{array})$ . Thus, $A - d (adj A) = (\begin{array}{cc} m - d q & n (1 + d) \\ p (1 + d) & q - d m \end{array})$ .
Execute: Setting determinant to zero: $∣ A - d (adj A) ∣ = (m - d q) (q - d m) - n p (1 + d)^{2} = 0$ . Expanding the first product gives $m q - d (m^{2} + q^{2}) + d^{2} m q - n p (1 + d)^{2} = 0$ . Since $n p = m q - d$ , this becomes $m q (1 + d^{2}) - d (m^{2} + q^{2}) - (m q - d) (1 + d)^{2} = 0$ .
Execute: Expanding $(m q - d) (1 + d)^{2} = m q (1 + 2 d + d^{2}) - d (1 + d)^{2}$ . Substituting back: $m q (1 + d^{2}) - d (m^{2} + q^{2}) - [m q (1 + d^{2}) + 2 d m q - d (1 + d)^{2}] = 0 ⟹ - d (m^{2} + q^{2} + 2 m q) + d (1 + d)^{2} = 0$ . Since $d \neq 0$ , dividing by $d$ yields $(1 + d)^{2} = (m + q)^{2}$ .
Verify: Let $A = I_{2} = (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array})$ . Here $m = 1, q = 1, n = 0, p = 0$ , so $d = 1$ . Then $adj A = I_{2}$ , giving $A - d (adj A) = I - I = O$ , so $∣ O ∣ = 0$ . Checking option D: $(1 + 1)^{2} = 4$ and $(m + q)^{2} = (1 + 1)^{2} = 4$ , which holds identically.

Shortcut:

When to use it: Fastest conceptual route using $A \cdot adj A = d I$

Given: $A$ is $2 \times 2$ , $\det (A) = d \neq 0$ , and $\det (A - d (adj A)) = 0$ .

Approach: For an invertible $2 \times 2$ matrix, $adj A = d A^{- 1}$ . Multiply inside the determinant by $A$ to convert the expression into the characteristic polynomial of $A$ .

Execute: $A - d (adj A) = A - d (d A^{- 1}) = A - d^{2} A^{- 1}$ . Therefore, $\det (A - d^{2} A^{- 1}) = 0 ⟹ \det (A^{2} - d^{2} I) / \det (A) = 0 ⟹ \det (A^{2} - d^{2} I) = 0$ .

Execute: This implies that $d^{2}$ is an eigenvalue of $A^{2}$ . If the eigenvalues of $A$ are $λ_{1}, λ_{2}$ , then $λ_{1} λ_{2} = d$ . The condition $\det (A^{2} - d^{2} I) = 0$ means either $λ_{1}^{2} = d^{2}$ or $λ_{2}^{2} = d^{2}$ , so $λ \in {d, - d}$ . Since $λ_{1} λ_{2} = d$ , one eigenvalue must be $1$ and the other $d$ (or $- 1$ and $- d$ ). In either case, the trace $tr (A) = m + q = λ_{1} + λ_{2} = \pm (1 + d)$ . Squaring gives $(m + q)^{2} = (1 + d)^{2}$ .

Verify: Both signs $λ_{1} + λ_{2} = \pm (1 + d)$ square to $(1 + d)^{2}$ , matching Option D.

Determinant Condition on a Linear Combination of a $2 \times 2$ Matrix and its Adjugate

What feels right?

Using $A (adj A) = d I$ , we rewrite $∣ A - d (adj A) ∣ = 0$ as $∣ A^{2} - d^{2} I ∣ = 0$ , which evaluates via the characteristic equation to $(1 + d)^{2} = (m + q)^{2}$ .

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

05Execute

✓Verify

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