Let αβ ≠ 0 and A = [[β, α, 3], [α, α, β], [-β, α, 2α]]. If B = [[3α, -9, 3α], [-α, 7, -2α], [-2α, 5,

Q: Wait, why can't we just multiply $A$ and $B$ to find the determinant?

Multiplying two $3 \times 3$ matrices is a recipe for algebraic disaster and wasted time. In JEE Advanced, we look for properties. The property $\det(AB) = \det(A) \det(B)$ is your best friend. By using $\det(B) = (\det(A))^2$, we reduce a massive multiplication problem into a simple cube of a single number.

Q: How do I know which elements of $B$ to use to find $\alpha$ and $\beta$?

Great question! You don't need to use all of them. Look for the elements that involve the fewest variables or the simplest expressions. $B_{21}$ and $B_{11}$ were chosen because they directly linked the variables $\alpha$ and $\beta$ to the given values in $B$. Always scan the matrix for the 'path of least resistance'.

Q: Is it always true that $\det(B) = (\det(A))^{n-1}$?

Yes, this is a fundamental property of the adjoint matrix. Since the cofactor matrix $B$ is the transpose of the adjoint matrix, and the determinant of the transpose is the same as the determinant of the original, $\det(B) = \det(\text{adj}(A))$. And we know $\det(\text{adj}(A)) = (\det(A))^{n-1}$. It's a beautiful, universal truth in linear algebra.

Question

Let $\alpha\beta 
eq 0$ and $A = \begin{bmatrix} \beta & \alpha & 3 \ \alpha & \alpha & \beta \ -\beta & \alpha & 2\alpha \end{bmatrix}$. If $B = \begin{bmatrix} 3\alpha & -9 & 3\alpha \ -\alpha & 7 & -2\alpha \ -2\alpha & 5 & -2\beta \end{bmatrix}$ is the matrix of cofactors of the elements of $A$, then $\det(AB)$ is equal to:

Options:

(A) 64

(B) 216

(C) 343

(D) 125

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex $3 	imes 3$ matrix $A$, filled with variables $\alpha$ and $\beta$. You are asked to find the determinant of the product $AB$, where $B$ is the matrix of cofactors.

Your first instinct might be to calculate the cofactor matrix $B$ element by element, then multiply it by $A$, and finally find the determinant. Stop! That is the trap.

In the world of JEE Advanced, brute force is rarely the intended path. Let us embark on a more elegant journey.

### The Beauty of Properties

The key to this problem lies in understanding the relationship between a matrix and its cofactor matrix. We know that for any $n 	imes n$ matrix $A$, the determinant of its cofactor matrix $B$ is given by:

$$\det(B) = (\det(A))^{n-1}$$

Since our matrix is $3 	imes 3$, we have $n=3$, which simplifies this to $\det(B) = (\det(A))^2$.

Now, look at our target: $\det(AB)$. By the multiplicative property of determinants, this is simply $\det(A) \cdot \det(B)$.

Substituting our property, we get:

$$\det(AB) = \det(A) \cdot (\det(A))^2 = (\det(A))^3$$

Suddenly, the problem transforms. We no longer need to perform matrix multiplication; we only need to find the value of $\det(A)$ and cube it.

### The Detective Work: Unlocking $\alpha$ and $\beta$

Now, how do we find $\alpha$ and $\beta$? We look at the given matrix $B$. The element $B_{21}$ is the cofactor of $A_{21}$.

By definition, $B_{21} = (-1)^{2+1} M_{21}$. Calculating the minor $M_{21}$ by hiding the second row and first column of $A$, we get:

$$\det \begin{bmatrix} \alpha & 3 \ \alpha & 2\alpha \end{bmatrix} = 2\alpha^2 - 3\alpha$$

Thus, $B_{21} = -(2\alpha^2 - 3\alpha)$. Comparing this to the given $B_{21} = -\alpha$, we get $-\alpha = -(2\alpha^2 - 3\alpha)$, which simplifies to $2\alpha^2 - 4\alpha = 0$.

Since $\alpha 
eq 0$, we find $\alpha = 2$.

With $\alpha$ in hand, we turn to $B_{11} = 3\alpha$. The cofactor $B_{11}$ is $(-1)^{1+1} M_{11} = \det \begin{bmatrix} \alpha & \beta \ \alpha & 2\alpha \end{bmatrix} = 2\alpha^2 - \alpha\beta$.

Setting $2\alpha^2 - \alpha\beta = 3\alpha$ and substituting $\alpha = 2$, we get $2(4) - 2\beta = 6$, which leads to $8 - 2\beta = 6$, so $\beta = 1$.

### The Grand Finale

With $\alpha = 2$ and $\beta = 1$, our matrix $A$ becomes:

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

Now, we calculate $\det(A)$ by expanding along the first row:

$$\det(A) = 1(8 - 2) - 2(8 + 1) + 3(4 + 2) = 6 - 18 + 18 = 6$$

Finally, we return to our initial simplification:

$$\det(AB) = (\det(A))^3 = 6^3 = 216$$

We have navigated the complexity, avoided the trap, and arrived at the solution with precision. The final answer is **216**.

Animated Solution for Mathematics - Matrices and Determinants: Let $α β \neq = 0$ and $A = β α - β α α α 3 β 2 α$ . If $B = 3 α - α - 2 α - 9 75 3 α - 2 α - 2 β$ is the matrix of cofactors of the elements of $A$ , then $det (A B)$ is equal to:

Select Answer:

Visualized Solution

Property of Cofactor Matrix

Determinant of Product $∣ A B ∣$

Defining Cofactor $B_{21}$

Solving for $α$

Defining Cofactor $B_{11}$

Solving for $β$

Constructing Matrix $A$

Calculating $det (A)$

Final Answer $∣ A B ∣$

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Beauty of Properties

The Detective Work

The Grand Finale

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Visualized Solution

Property of Cofactor Matrix

Determinant of Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣AB∣

Defining Cofactor .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B21​

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Defining Cofactor .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B11​

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Constructing Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } det(A)

Final Answer .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣AB∣