JEE Main 2025 Apr 2 Shift 1, Mathematics Q19: Algebra Of Determinants

JEE Main2025Apr 2, Shift 1Mathematics

Let $a \in R$ and $A$ be a matrix of order $3 \times 3$ such that $det (A) = - 4$ and $A + I = 1 a 1 210 a 12$ , where $I$ is the identity matrix of order $3 \times 3$ .

If $det ((a + 1) adj ((a - 1) A)) = 2^{m} 3^{n}$ , $m, n \in {0, 1, 2, \dots, 20}$ , then $m + n$ is equal to :

A
14
B
17
C
15
D
16

Solution

$A=\begin{bmatrix}1 & a & 1\\[2pt] 2 & 1 & 0\\[2pt] a & 1 & 2\end{bmatrix},\qquad A-I=\begin{bmatrix}0 & a & 1\\[2pt] 2 & 0 & 0\\[2pt] a & 1 & 1\end{bmatrix}$

$∣ A ∣ = - 4 \Rightarrow 2 - 2 a = - 4 \Rightarrow a = 3$

$\left|(a+1)\,\operatorname{adj}\!\left((a-1)A\right)\right| =\left|4\,\operatorname{adj}(3A)\right|$

$= 4^{3} ∣ adj (3 A) ∣$

$= 4^{3} ∣3 A ∣^{3 - 1} = 64 ∣3 A ∣^{2}$

$= 64 (3^{3})^{2} ∣ A ∣^{2}$

$= 2^{6} \times 3^{6} \times 16$

$= 2^{10} \times 3^{6}$

$∴ m = 10, n = 6$

$m + n = 16$

Concept behind this question

Algebra Of DeterminantsNotes, formulas and examples →

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Apr 2 Shift 1, Mathematics Q19: Algebra Of Determinants

Concept behind this question

More previous year questions on Algebra Of Determinants

Practice more Mathematics