The options involve , and , but not and . Since , substituting eliminates and directly.

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Matrices and Determinants: Mathematics | JEE Main

Q: Can we always write ?

Yes, because , and since , is invertible.

Let

A = \begin{pmatrix} m & n \\ p & q \end{pmatrix}

d = |A| \neq 0

and

|A - d(\text{Adj} A)| = 0

. Then

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What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

1 + d^2 = m^2 + q^2

1 + d^2 = (m + q)^2

(1 + d)^2 = m^2 + q^2

(1 + d)^2 = (m + q)^2

Step-by-step solutionView

Correct answer

Using $A(\text{adj } A) = dI$ , we rewrite $|A - d(\text{adj } A)| = 0$ 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 $2d$ and $2mq$ , falsely simplifying to sums of squares. Retain the cross-terms when completing the square: $(mq - np) = d$ , so $2mq - 2d = 2np$ , 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 $(1 + d)^2 = 1 + 2d + d^2$ . Notice that expanding $|A - d(\text{adj } A)|$ yields $(m - qd)(q - md) - np(1 + d)^2 = 0$ , which factors symmetrically in $(1 + d)^2$ .

C ·

(1 + d)^2 = m^2 + q^2

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

D ·

(1 + d)^2 = (m + q)^2

None. This option correctly relates the trace squared $(m + q)^2$ to $(1 + d)^2$ . Directly evaluate $|A - d(\text{adj } A)| = (1 + d^2)d - d(m+q)^2 + 2d^2 = 0$ , divide by $d \neq 0$ , and complete the square to get $(1 + d)^2 = (m + q)^2$ .

Reviewed route

Solution

StepWorking

01given

$A = \begin{pmatrix} m & n \\ p & q \end{pmatrix}$ , $d = |A| = mq - np \neq 0$ , and $|A - d(\text{adj } A)| = 0$ .

02approach

Compute the explicit 2x2 matrix expression for $A - d(\text{adj } A)$ , set its determinant to zero, and eliminate $np$ using $np = mq - d$ to relate $d$ with $(m+q)$ .

03execute

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

04execute

Setting determinant to zero: $|A - d(\text{adj } A)| = (m - dq)(q - dm) - np(1 + d)^2 = 0$ . Expanding the first product gives $mq - d(m^2 + q^2) + d^2 mq - np(1 + d)^2 = 0$ . Since $np = mq - d$ , this becomes $mq(1 + d^2) - d(m^2 + q^2) - (mq - d)(1 + d)^2 = 0$ .

05execute

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

✓verify

Let $A = I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ . Here $m=1, q=1, n=0, p=0$ , so $d = 1$ . Then $\text{adj } A = I_2$ , giving $A - d(\text{adj } A) = I - I = O$ , so $|O| = 0$ . Checking option (3): $(1+1)^2 = 4$ and $(m+q)^2 = (1+1)^2 = 4$ , which holds identically.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

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.

Can we always write

\text{adj } A = d A^{-1}

Yes, because $A (\text{adj } A) = |A| I = d I$ , and since $d \neq 0$ , $A$ is invertible.

Answer

Using $A(\text{adj } A) = dI$ , we rewrite $|A - d(\text{adj } A)| = 0$ 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 $2d$ and $2mq$ , falsely simplifying to sums of squares. Retain the cross-terms when completing the square: $(mq - np) = d$ , so $2mq - 2d = 2np$ , 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 $(1 + d)^2 = 1 + 2d + d^2$ . Notice that expanding $|A - d(\text{adj } A)|$ yields $(m - qd)(q - md) - np(1 + d)^2 = 0$ , which factors symmetrically in $(1 + d)^2$ .
C: $(1 + d)^2 = m^2 + q^2$ - Failing to account for the $2mq$ cross-term when rewriting $(m+q)^2 - 2mq$ , incorrectly equating $(1 + d)^2$ to $m^2 + q^2$ . Express the determinant directly in terms of trace $\text{tr}(A) = m + q$ , which appears squared as $(m + q)^2$ .
D · correct: $(1 + d)^2 = (m + q)^2$ - None. This option correctly relates the trace squared $(m + q)^2$ to $(1 + d)^2$ . Directly evaluate $|A - d(\text{adj } A)| = (1 + d^2)d - d(m+q)^2 + 2d^2 = 0$ , divide by $d \neq 0$ , and complete the square to get $(1 + d)^2 = (m + q)^2$ .

Step-by-step solution

given: $A = \begin{pmatrix} m & n \\ p & q \end{pmatrix}$ , $d = |A| = mq - np \neq 0$ , and $|A - d(\text{adj } A)| = 0$ .
approach: Compute the explicit 2x2 matrix expression for $A - d(\text{adj } A)$ , set its determinant to zero, and eliminate $np$ using $np = mq - d$ to relate $d$ with $(m+q)$ .
execute: For $A = \begin{pmatrix} m & n \\ p & q \end{pmatrix}$ , we have $\text{adj } A = \begin{pmatrix} q & -n \\ -p & m \end{pmatrix}$ . Thus, $A - d(\text{adj } A) = \begin{pmatrix} m - dq & n(1 + d) \\ p(1 + d) & q - dm \end{pmatrix}$ .
execute: Setting determinant to zero: $|A - d(\text{adj } A)| = (m - dq)(q - dm) - np(1 + d)^2 = 0$ . Expanding the first product gives $mq - d(m^2 + q^2) + d^2 mq - np(1 + d)^2 = 0$ . Since $np = mq - d$ , this becomes $mq(1 + d^2) - d(m^2 + q^2) - (mq - d)(1 + d)^2 = 0$ .
execute: Expanding $(mq - d)(1 + d)^2 = mq(1 + 2d + d^2) - d(1 + d)^2$ . Substituting back: $mq(1 + d^2) - d(m^2 + q^2) - [mq(1 + d^2) + 2dmq - d(1 + d)^2] = 0 \implies -d(m^2 + q^2 + 2mq) + d(1 + d)^2 = 0$ . Since $d \neq 0$ , dividing by $d$ yields $(1 + d)^2 = (m + q)^2$ .
verify: Let $A = I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ . Here $m=1, q=1, n=0, p=0$ , so $d = 1$ . Then $\text{adj } A = I_2$ , giving $A - d(\text{adj } A) = I - I = O$ , so $|O| = 0$ . Checking option (3): $(1+1)^2 = 4$ and $(m+q)^2 = (1+1)^2 = 4$ , which holds identically.

Shortcut:

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

given: $A$ is $2 \times 2$ , $\det(A) = d \neq 0$ , and $\det(A - d(\text{adj } A)) = 0$ .

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

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

execute: This implies that $d^2$ is an eigenvalue of $A^2$ . If the eigenvalues of $A$ are $\lambda_1, \lambda_2$ , then $\lambda_1 \lambda_2 = d$ . The condition $\det(A^2 - d^2 I) = 0$ means either $\lambda_1^2 = d^2$ or $\lambda_2^2 = d^2$ , so $\lambda \in \{d, -d\}$ . Since $\lambda_1 \lambda_2 = d$ , one eigenvalue must be $1$ and the other $d$ (or $-1$ and $-d$ ). In either case, the trace $\text{tr}(A) = m + q = \lambda_1 + \lambda_2 = \pm(1 + d)$ . Squaring gives $(m + q)^2 = (1 + d)^2$ .

verify: Both signs $\lambda_1 + \lambda_2 = \pm(1+d)$ square to $(1+d)^2$ , matching Option (3).

Matrices and Determinants: Mathematics | JEE Main

What feels right?

Using $A(\text{adj } A) = dI$ , we rewrite $|A - d(\text{adj } A)| = 0$ 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

We think you should solve this next

Let…

Consider the system of linear equation…

If…

Students also ask

Matrices and Determinants: Mathematics | JEE Main

Using A(adj A)=dIA(\text{adj } A) = dIA(adj A)=dI, we rewrite ∣A−d(adj A)∣=0|A - d(\text{adj } A)| = 0∣A−d(adj A)∣=0 as ∣A2−d2I∣=0|A^2 - d^2 I| = 0∣A2−d2I∣=0, which evaluates via the characteristic equation to (1+d)2=(m+q)2(1 + d)^2 = (m + q)^2(1+d)2=(m+q)2.

01given

02approach

03execute

04execute

05execute

✓verify

Let…

Consider the system of linear equation…

If…

Students also ask

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