Let $A=[a_{ij}]$ be a $3 \times 3$ matrix such that $A\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right], A\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right]=\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]$ and $A\left[\begin{array}{l}2 \\ 1 \\ 2\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$, then $a_{23}$ equals:

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Matrices

Let

A = [a_{ij}]

be a

3 \times 3

matrix such that

A 010 = 001, A 413 = 010

and

A 212 = 100

, then

a_{23}

equals:

Answer: (A)

Let

A = [a_{ij}]

. From

A 010 = 001

, the second column of

A

(0, 0, 1)^{T}

, so

a_{12} = 0, a_{22} = 0, a_{32} = 1

.

From

A 413 = 010

, the second row gives

4 a_{21} + a_{22} + 3 a_{23} = 1 \Rightarrow 4 a_{21} + 3 a_{23} = 1 (i)

From

A 212 = 100

, the second row gives

2 a_{21} + a_{22} + 2 a_{23} = 0 \Rightarrow a_{21} + a_{23} = 0 \Rightarrow a_{21} = - a_{23} (ii)

Substituting (ii) into (i):

4 (- a_{23}) + 3 a_{23} = 1 \Rightarrow - a_{23} = 1 \Rightarrow a_{23} = - 1

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option