Let B = [[1, 3, α], [1, 2, 3], [α, α, 4]], α 2 be the adjoint of a matrix A and |A| = 2. Then [α -

Q: Wait, why can't we just find the inverse of $B$ to get $A$ and then calculate $|A|$?

That is a classic trap! Finding the inverse of a $3 \times 3$ matrix with an unknown variable $\alpha$ is algebraically exhausting and highly prone to calculation errors. In JEE Advanced, we always look for properties of determinants first. The property $|\text{adj}(A)| = |A|^{n-1}$ is your shortcut. It bypasses the need to find $A$ entirely, saving you precious time and mental energy.

Q: Why do we reject $\alpha = 2$?

In JEE Advanced, constraints are not just suggestions; they are part of the problem's definition. The problem explicitly states $\alpha > 2$. If you include $\alpha = 2$, you are solving a different problem than the one asked. Always check your final values against the initial constraints given in the question stem.

Q: Is there a faster way to do the final matrix multiplication? It looks tedious.

It might look tedious, but it is actually a test of your precision. Once you have $\alpha = 4$, the matrix $B$ becomes a matrix of integers. The multiplication is just a series of dot products. Stay calm, write down each step clearly, and don't rush the arithmetic. The beauty of these problems is that the numbers often simplify beautifully if you have the correct value of $\alpha$.

Question

Let $B = \begin{bmatrix} 1 & 3 & \alpha \ 1 & 2 & 3 \ \alpha & \alpha & 4 \end{bmatrix}$, $\alpha > 2$ be the adjoint of a matrix $A$ and $|A| = 2$. Then $[\alpha \quad -2\alpha \quad \alpha] B \begin{bmatrix} \alpha \ -2\alpha \ \alpha \end{bmatrix}$ is equal to

Options:

(A) 0

(B) 16

(C) -16

(D) 32

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are going to dissect a problem that, at first glance, looks like a standard matrix manipulation exercise. But beneath the surface, it is a test of your ability to connect abstract properties with concrete calculation.

We are given a matrix $B$, which is the adjoint of some unknown matrix $A$, and we are tasked with evaluating a specific quadratic form involving $B$. Let us embark on this journey together.

### The Determinant Bridge

The first instinct for many students is to try and find the matrix $A$. But stop! Take a breath. We do not know $A$; we only know its adjoint, $B$.

If you try to invert $B$ to find $A$, you will be trapped in a labyrinth of algebraic expressions involving $\alpha$. Instead, we must use the 'Bridge Property' of determinants. We know that for any square matrix $A$ of order $n$, the determinant of its adjoint is given by the elegant relation:

$$|	ext{adj}(A)| = |A|^{n-1}$$

In our case, $n = 3$. Therefore, $|B| = |A|^{3-1} = |A|^2$. We are given that $|A| = 2$. Substituting this, we find:

$$|B| = 2^2 = 4$$

Just like that, the mystery of the determinant of $B$ is solved. It is not an expression; it is a solid, unshakeable number: $4$.

### The Algebraic Hunt

Now that we know $|B| = 4$, we must look at the matrix $B$ itself. It contains the variable $\alpha$. To find $\alpha$, we must expand the determinant of $B$ and equate it to $4$.

Let us expand along the first row:

$$|B| = 1(2 \cdot 4 - 3 \cdot \alpha) - 3(1 \cdot 4 - 3 \cdot \alpha) + \alpha(1 \cdot \alpha - 2 \cdot \alpha)$$

Simplifying this, we get $1(8 - 3\alpha) - 3(4 - 3\alpha) + \alpha(\alpha - 2\alpha)$. This expands to $8 - 3\alpha - 12 + 9\alpha - \alpha^2$. Combining like terms, we arrive at the quadratic expression:

$$-\alpha^2 + 6\alpha - 4$$

Now, we equate this to our known value: $-\alpha^2 + 6\alpha - 4 = 4$. Rearranging this gives us the quadratic equation:

$$\alpha^2 - 6\alpha + 8 = 0$$

Factoring this, we get $(\alpha - 4)(\alpha - 2) = 0$. This gives us two candidates: $\alpha = 4$ and $\alpha = 2$.

But wait! Look at the problem statement again. The constraint is $\alpha > 2$. This is the gatekeeper of the problem. We must reject $\alpha = 2$. Thus, the only valid value is $\alpha = 4$.

### The Final Matrix Dance

With $\alpha = 4$, our matrix $B$ is fully defined:

$$B = \begin{bmatrix} 1 & 3 & 4 \ 1 & 2 & 3 \ 4 & 4 & 4 \end{bmatrix}$$

We are asked to evaluate the product:

$$[\alpha \quad -2\alpha \quad \alpha] B \begin{bmatrix} \alpha \ -2\alpha \ \alpha \end{bmatrix}$$

Substituting $\alpha = 4$, this becomes:

$$[4 \quad -8 \quad 4] B \begin{bmatrix} 4 \ -8 \ 4 \end{bmatrix}$$

Let us perform the multiplication in two steps. First, the row vector multiplied by $B$:

$$[4 \quad -8 \quad 4] \begin{bmatrix} 1 & 3 & 4 \ 1 & 2 & 3 \ 4 & 4 & 4 \end{bmatrix}$$

The first element is $4(1) - 8(1) + 4(4) = 12$. The second element is $4(3) - 8(2) + 4(4) = 12$. The third element is $4(4) - 8(3) + 4(4) = 8$. We now have the row vector $[12 \quad 12 \quad 8]$.

Finally, we multiply this by the column vector:

$$[12 \quad 12 \quad 8] \begin{bmatrix} 4 \ -8 \ 4 \end{bmatrix} = 12(4) + 12(-8) + 8(4) = 48 - 96 + 32$$

Calculating this sum, $48 + 32 = 80$, and $80 - 96 = -16$. The final answer is **$-16$**.

Animated Solution for Mathematics - Matrices and Determinants: Let $B = 11 α 32 α α 34$ , $α > 2$ be the adjoint of a matrix $A$ and $∣ A ∣ = 2$ . Then $[α - 2 α α] B α - 2 α α$ is equal to

Select Answer:

Visualized Solution

Introduction to Matrix $B$

Determinant of Adjoint Matrix

Calculating $∣ B ∣$

Expanding $∣ B ∣$

Simplifying the Expansion

Forming the Quadratic Equation

Solving for $α$

Applying the Constraint

Setting Up the Matrix Product

First Matrix Multiplication

Final Matrix Multiplication

Final Calculation

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Determinant Bridge

The Algebraic Hunt

The Final Matrix Dance

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