Let A=[a_ij] be 3x3 matrix such that A [0 1 0]^T = [0 0 1]^T, A [4 1 3]^T = [0 1 0]^T and A [2 1 2]^

Q: Why can't we just find the inverse of $A$ to solve for the matrix?

You absolutely could, but that is like using a sledgehammer to crack a nut! Finding the inverse of a $3 \times 3$ matrix is computationally expensive and prone to arithmetic errors. In JEE Advanced, time is your most precious currency. We only need one specific element, $a_{23}$, so we use targeted linear equations to isolate it directly.

Q: How did you know $a_{22} = 0$ so quickly?

Great observation! Remember that multiplying a matrix by the standard basis vector $\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$ acts as a selector. It extracts the entire second column of the matrix. Since the problem states that $A \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$, the second column of $A$ must be $\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$. Therefore, the middle element, $a_{22}$, is clearly $0$.

Q: What if I didn't notice the column extraction property?

If you missed that, you would have written out a system of nine linear equations. While that would eventually lead you to the correct answer, it would take significantly longer. Always look for the 'geometric' or 'transformational' meaning of matrix multiplication before diving into brute-force algebra. It is the hallmark of a top-tier JEE aspirant.

Question

Let $A=[a_{ij}]$ be $3	imes3$ matrix such that $A\begin{bmatrix}0\ 1\ 0\end{bmatrix}=\begin{bmatrix}0\ 0\ 1\end{bmatrix}$, $A\begin{bmatrix}4\ 1\ 3\end{bmatrix}=\begin{bmatrix}0\ 1\ 0\end{bmatrix}$ and $A\begin{bmatrix}2\ 1\ 2\end{bmatrix}=\begin{bmatrix}1\ 0\ 0\end{bmatrix}$, then $a_{23}$ equals:

Options:

(A) -1

(B) 2

(C) 1

(D) 0

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are going to dismantle a problem that, at first glance, looks like a tedious exercise in solving for nine unknowns. We have a $3 	imes 3$ matrix $A$, and we are given three specific transformations.

Many students would immediately write out $A$ as a grid of variables from $a_{11}$ to $a_{33}$ and start grinding through a massive system of equations. But you are not 'many students.' You are a strategist. Let us look at the elegance hidden within this problem.

### The Power of the Filter

Our first clue is the transformation:

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

In the language of linear algebra, multiplying a matrix by the vector $\begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix}$ is not just arithmetic; it is a filter. It effectively 'picks out' the second column of the matrix.

Imagine the matrix $A$ as a collection of three column vectors. When you multiply by this specific vector, the first and third columns are multiplied by zero and vanish, leaving only the second column.

Thus, we instantly know that the second column of $A$ is $\begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix}$. This gives us $a_{12} = 0$, $a_{22} = 0$, and $a_{32} = 1$. We have just unlocked a massive piece of the puzzle without breaking a sweat.

### The Art of Targeted Strikes

Now, we need to find $a_{23}$. Notice that we do not need the entire matrix; we only need the second row. This is where the 'Targeted Strike' method comes in.

We look at the second condition:

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

We only care about the second row of the resulting vector, which is $1$. The second row of the result is formed by the dot product of the second row of $A$ (which is $[a_{21}, a_{22}, a_{23}]$) and the input vector $\begin{bmatrix} 4 \ 1 \ 3 \end{bmatrix}$.

This gives us the equation:

$$4a_{21} + 1a_{22} + 3a_{23} = 1$$

Since we already discovered that $a_{22} = 0$, our equation simplifies beautifully to $4a_{21} + 3a_{23} = 1$. This is our first anchor equation.

### The Final Convergence

We repeat this logic for the third condition:

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

Again, we focus only on the second row. The dot product of $[a_{21}, a_{22}, a_{23}]$ and $\begin{bmatrix} 2 \ 1 \ 2 \end{bmatrix}$ must equal the second element of the result, which is $0$.

So, $2a_{21} + 1a_{22} + 2a_{23} = 0$. Substituting $a_{22} = 0$ again, we get $2a_{21} + 2a_{23} = 0$, which simplifies to $a_{21} = -a_{23}$.

Now, we have a simple system of two equations. Substituting $a_{21} = -a_{23}$ into our first equation, we get:

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

This leads us to $-a_{23} = 1$, or **$a_{23} = -1$**. We have arrived at the solution.

By refusing to solve for the entire matrix and instead focusing on the specific row interactions, we turned a daunting problem into a quick, elegant victory. Keep this mindset—always look for the shortcut, always look for the structure, and the math will always reward you.

Animated Solution for Mathematics - Matrices and Determinants: Let $A = [a_{ij}]$ be $3 \times 3$ matrix such that $A 010 = 001$ , $A 413 = 010$ and $A 212 = 100$ , then $a_{23}$ equals:

Select Answer:

Visualized Solution

Define Matrix $A$

Analyze First Condition

Extracting Elements

Analyze Second Condition

Formulating Equation 1

Substituting $a_{22} = 0$

Analyze Third Condition

Formulating Equation 2

Simplifying Equation 2

Solving for $a_{23}$

Final Answer

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Power of the Filter

The Art of Targeted Strikes

The Final Convergence

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