Let A=[a_i j] be a 3 \times 3 matrix such that A[\beginarrayl0 1 0\endarray]=[\beginarrayl0 0 1\endarray], A[\beginarrayl4 1 3\endarray]=[\beginarrayl0 1 0\endarray] and A[\beginarrayl2 1 2\endarray]=[\beginarrayl1 0 0\endarray] , then a_23 equals :

Matrices and Determinants PYQ — JEE Main 2025 · 23 Jan · Shift 2 · Q38 with Solution

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Correct answer: B

A=[\mathbf{c}_1\ \mathbf{c}_2\ \mathbf{c}_3]

where $\mathbf{c}_1,\mathbf{c}_2,\mathbf{c}_3$ are the columns of $A$ .

For any vector $\begin{bmatrix}x\\y\\z\end{bmatrix}$ ,

A\begin{bmatrix}x\\y\\z\end{bmatrix}=x\mathbf{c}_1+y\mathbf{c}_2+z\mathbf{c}_3.

A\begin{bmatrix}0\\1\\0\end{bmatrix}=\begin{bmatrix}0\\0\\1\end{bmatrix},

we directly get the second column:

\mathbf{c}_2=\begin{bmatrix}0\\0\\1\end{bmatrix}.

A\begin{bmatrix}4\\1\\3\end{bmatrix}=\begin{bmatrix}0\\1\\0\end{bmatrix}.

So,

4\mathbf{c}_1+\mathbf{c}_2+3\mathbf{c}_3=\begin{bmatrix}0\\1\\0\end{bmatrix}.

Substitute $\mathbf{c}_2=\begin{bmatrix}0\\0\\1\end{bmatrix}$ :

4\mathbf{c}_1+3\mathbf{c}_3=\begin{bmatrix}0\\1\\-1\end{bmatrix}. \qquad (1)

A\begin{bmatrix}2\\1\\2\end{bmatrix}=\begin{bmatrix}1\\0\\0\end{bmatrix}.

Thus,

2\mathbf{c}_1+\mathbf{c}_2+2\mathbf{c}_3=\begin{bmatrix}1\\0\\0\end{bmatrix}.

Again substituting $\mathbf{c}_2$ :

2\mathbf{c}_1+2\mathbf{c}_3=\begin{bmatrix}1\\0\\-1\end{bmatrix}. \qquad (2)

Divide by $2$ :

\mathbf{c}_1+\mathbf{c}_3=\begin{bmatrix}\tfrac12\\0\\-\tfrac12\end{bmatrix}. \qquad (3)

4\mathbf{c}_1+4\mathbf{c}_3=\begin{bmatrix}2\\0\\-2\end{bmatrix}.

Subtract (1):

(4\mathbf{c}_1+4\mathbf{c}_3)-(4\mathbf{c}_1+3\mathbf{c}_3)=\begin{bmatrix}2\\0\\-2\end{bmatrix}-\begin{bmatrix}0\\1\\-1\end{bmatrix}.

Hence,

\mathbf{c}_3=\begin{bmatrix}2\\-1\\-1\end{bmatrix}.

The entry $a_{23}$ means row $2$ , column $3$ , i.e. the second component of $\mathbf{c}_3$ . Therefore,

a_{23}=-1.

So the correct option is B.