Let M = [[0, 1, a], [1, 2, 3], [3, b, 1]] and adj M = [[-1, 1, -1], [8, -6, 2], [-5, 3, -1]] where a

Q: Wait, why do we equate the cofactor $C_{11}$ to the $(1,1)$ element of the adjoint? I thought the adjoint was just the transpose of the cofactor matrix?

You are absolutely right! The adjoint is indeed the transpose of the cofactor matrix. By definition, $(\text{adj } M)_{ij} = C_{ji}$. So, for the $(1,1)$ position, $i=1$ and $j=1$, which means $(\text{adj } M)_{11} = C_{11}$. It is the same element! You were already on the right track—trust your intuition.

Q: For Option B, why can't I just calculate $M^2$ and then find the adjoint?

You could, but that is the path of maximum resistance! In JEE Advanced, time is your most precious resource. Using the property $\text{det}(\text{adj } A) = (\text{det } A)^{n-1}$ allows you to bypass the tedious matrix multiplication and determinant calculation of $M^2$. Always look for the property that simplifies the problem before you start crunching numbers.

Q: Is there a trick to solving the system $MX = B$ in Option D without doing full matrix inversion?

Great question! While you could use Cramer's Rule, here we already have the adjoint matrix. Since $X = M^{-1}B$ and $M^{-1} = \frac{1}{\text{det } M} \text{adj } M$, we can simply multiply the adjoint matrix by the vector $B$ and then divide by the determinant. It is much faster than finding the inverse from scratch!

Question

Let $M = \begin{bmatrix} 0 & 1 & a \ 1 & 2 & 3 \ 3 & b & 1 \end{bmatrix}$ and $	ext{adj } M = \begin{bmatrix} -1 & 1 & -1 \ 8 & -6 & 2 \ -5 & 3 & -1 \end{bmatrix}$ where $a$ and $b$ are real numbers. Which of the following options is/are correct?

Options:

(A) $a + b = 3$

(B) $	ext{det}(	ext{adj } M^2) = 81$

(C) $(	ext{adj } M)^{-1} + 	ext{adj } M^{-1} = -M$

(D) If $M \begin{bmatrix} \alpha \ \beta \ \gamma \end{bmatrix} = \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix}$, then $\alpha - \beta + \gamma = 3$

Sigma Percentile · Accepted Answer

### The Symphony of Matrices: Unlocking the Hidden Structure

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a matrix problem; we are peeling back the layers of a mathematical structure to reveal the elegance hidden beneath.

Matrices are often seen as cold, rigid grids of numbers, but in reality, they are dynamic maps of linear transformations. When we are given a matrix $M$ and its adjoint, we are essentially being given a key and a lock. Our job is to find the missing pieces that make them fit perfectly.

### Phase 1: The Detective Work

We begin with our matrix:
$$M = \begin{bmatrix} 0 & 1 & a \ 1 & 2 & 3 \ 3 & b & 1 \end{bmatrix}$$

We have two unknowns, $a$ and $b$, lurking in the shadows. We are also given the adjoint matrix:
$$	ext{adj } M = \begin{bmatrix} -1 & 1 & -1 \ 8 & -6 & 2 \ -5 & 3 & -1 \end{bmatrix}$$

How do we find $a$ and $b$? We must remember the definition of the adjoint. The adjoint is the transpose of the cofactor matrix. This means the element at row $i$, column $j$ of the adjoint is the cofactor $C_{ji}$ of the original matrix.

Let us focus on the $(1,1)$ position. The cofactor $C_{11}$ is the determinant of the minor obtained by deleting the first row and first column:
$$\begin{vmatrix} 2 & 3 \ b & 1 \end{vmatrix} = 2(1) - 3(b) = 2 - 3b$$

Equating this to the $(1,1)$ element of our given adjoint, which is $-1$, we get $2 - 3b = -1$, leading us directly to $b = 1$.

Now, let us turn our gaze to the $(2,2)$ position. The cofactor $C_{22}$ is the determinant of the minor obtained by deleting the second row and second column:
$$\begin{vmatrix} 0 & a \ 3 & 1 \end{vmatrix} = 0(1) - 3(a) = -3a$$

Equating this to the $(2,2)$ element of the adjoint, which is $-6$, we find $-3a = -6$, so $a = 2$. Just like that, the fog clears, and our matrix $M$ is fully revealed:
$$M = \begin{bmatrix} 0 & 1 & 2 \ 1 & 2 & 3 \ 3 & 1 & 1 \end{bmatrix}$$

### Phase 2: The Heartbeat of the Matrix

Before we can evaluate the complex options, we need the determinant of $M$. Expanding along the first row, we calculate:
$$	ext{det } M = 0(2-3) - 1(1-9) + 2(1-6) = 0 - 1(-8) + 2(-5) = 8 - 10 = -2$$

This value, $-2$, is the heartbeat of our matrix. It will be the denominator in our inverse calculations and the scaling factor for our adjoint properties.

### Phase 3: Demystifying the Properties

Option B asks about $	ext{det}(	ext{adj } M^2)$. Many students panic here, thinking they must square the matrix. Do not fall into that trap!

We use the powerful property $	ext{det}(	ext{adj } A) = (	ext{det } A)^{n-1}$. With $A = M^2$ and $n=3$, this becomes:
$$(	ext{det } M^2)^{3-1} = (	ext{det } M^2)^2 = (	ext{det } M)^4$$

Substituting our heartbeat, $(-2)^4 = 16$. Since $16 
eq 81$, Option B is incorrect. See how the property saved us from a mountain of arithmetic?

Now, consider the expression $(	ext{adj } M)^{-1} + 	ext{adj } M^{-1} = -M$. We know $	ext{adj } M = (	ext{det } M) M^{-1}$.

Therefore, $(	ext{adj } M)^{-1} = ((	ext{det } M) M^{-1})^{-1} = \frac{1}{	ext{det } M} M$. For the second term, $	ext{adj } M^{-1} = (	ext{det } M^{-1}) (M^{-1})^{-1} = \frac{1}{	ext{det } M} M$.

Adding them together, we get:
$$\frac{2}{	ext{det } M} M = \frac{2}{-2} M = -M$$

It is a perfect, elegant cancellation! Option C is correct.

### Phase 4: The Final Victory

Finally, Option D asks us to solve $MX = B$ where $B = [1, 2, 3]^T$. We know $X = M^{-1}B = \frac{1}{	ext{det } M} (	ext{adj } M) B$.

Substituting our values:
$$X = \frac{1}{-2} \begin{bmatrix} -1 & 1 & -1 \ 8 & -6 & 2 \ -5 & 3 & -1 \end{bmatrix} \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix}$$

Performing the multiplication, we get:
$$X = -\frac{1}{2} \begin{bmatrix} -2 \ 2 \ -2 \end{bmatrix} = \begin{bmatrix} 1 \ -1 \ 1 \end{bmatrix}$$

Thus, $\alpha=1, \beta=-1, \gamma=1$. The expression $\alpha - \beta + \gamma = 1 - (-1) + 1 = 3$. Option D is correct.

We have navigated the logic, verified the properties, and arrived at the truth. Remember, in JEE Advanced, the math is not just about calculation; it is about seeing the connections. Keep practicing, keep questioning, and keep falling in love with the process.

Animated Solution for Mathematics - Matrices and Determinants: Let $M = 013 12 b a 31$ and $adj M = - 1 8 - 5 1 - 6 3 - 1 2 - 1$ where $a$ and $b$ are real numbers. Which of the following options is/are correct?

Select Answer:

Visualized Solution

Problem Setup

Adjoint Property for $b$

Calculating $b$

Adjoint Property for $a$

Calculating $a$

Checking Option A

Calculating $det M$

Checking Option B

Checking Option C: First Term

Checking Option C: Second Term

Option C: Final Verification

Checking Option D: Setup

Option D: Matrix Multiplication

Option D: Final Verification

The Sigma Insight: Adjoint and Inverse of a Matrix

The Symphony of Matrices

Phase 1

Phase 2

Phase 3

Phase 4

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