If the adjoint of a 3 x 3 matrix P is [1 4 4; 2 1 7; 1 1 3], then the possible value(s) of the deter

Q: Wait, why do we reject the negative root here? Or do we?

Actually, you don't! That is exactly where the trap lies. When you solve $|P|^2 = 4$, the solutions are $|P| = 2$ and $|P| = -2$. Both are valid. Always remember that the square root of a positive number yields both a positive and a negative result.

Q: Why can't we just find the inverse of the adjoint matrix to get $P$?

You could, but it is computationally expensive and prone to arithmetic errors. In JEE Advanced, time is your most precious resource. Using the property $|\text{adj}(P)| = |P|^{n-1}$ is the elegant, intended path that saves you minutes of tedious calculation.

Q: Does the order of the matrix matter for this property?

Absolutely! The exponent $n-1$ depends entirely on the dimension of the matrix. If it were a $2 \times 2$ matrix, the exponent would be $1$. Since it is a $3 \times 3$ matrix, the exponent is $2$. Always check the dimensions before applying the formula.

Question

If the adjoint of a $3 	imes 3$ matrix $P$ is $\begin{bmatrix} 1 & 4 & 4 \ 2 & 1 & 7 \ 1 & 1 & 3 \end{bmatrix}$, then the possible value(s) of the determinant of $P$ is (are)

Options:

(A) $-2$

(B) $-1$

(C) $1$

(D) $2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer! Today, we are going to peel back the layers of a classic linear algebra problem. Many students look at a matrix and immediately want to invert it or find its elements.

But in the world of JEE Advanced, the secret to success is often not brute force, but recognizing the hidden symmetries. We are given the adjoint of a $3 	imes 3$ matrix $P$ and asked for its determinant. Let us embark on this journey.

### The Master Identity

The relationship between a matrix $P$ and its adjoint $	ext{adj}(P)$ is one of the most beautiful results in linear algebra. It is defined by the identity:

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

Here, $n$ represents the order of the matrix. Think of this as a bridge; we are standing on the side of the adjoint, and we need to cross over to the side of the original matrix $P$.

Since our matrix is $3 	imes 3$, we have $n=3$. Substituting this into our identity, we get:

$$|	ext{adj}(P)| = |P|^{3-1} = |P|^2$$

This is our master equation. It tells us that the determinant of the adjoint is simply the square of the determinant of the original matrix.

### The Calculation

Now, we must calculate the determinant of the given adjoint matrix:

$$\begin{vmatrix} 1 & 4 & 4 \ 2 & 1 & 7 \ 1 & 1 & 3 \end{vmatrix}$$

Let us expand along the first row. We take the first element, $1$, and multiply it by the determinant of the minor:

$$1 	imes \begin{vmatrix} 1 & 7 \ 1 & 3 \end{vmatrix} = 1 	imes (3 - 7) = -4$$

Next, we take the second element, $4$, and apply the sign convention (which is negative for the second position). We multiply by the minor:

$$-4 	imes \begin{vmatrix} 2 & 7 \ 1 & 3 \end{vmatrix} = -4 	imes (6 - 7) = -4 	imes (-1) = 4$$

Finally, we take the third element, $4$, and multiply by the minor:

$$4 	imes \begin{vmatrix} 2 & 1 \ 1 & 1 \end{vmatrix} = 4 	imes (2 - 1) = 4$$

Summing these up, we get:

$$|	ext{adj}(P)| = -4 + 4 + 4 = 4$$

### The Final Twist

We are almost there. We have found that $|	ext{adj}(P)| = 4$. Plugging this into our master equation, we get:

$$|P|^2 = 4$$

Now, take a deep breath. This is the moment where many brilliant students lose marks. When you take the square root of both sides, you must account for both the positive and negative roots.

Thus, $|P| = \pm \sqrt{4}$, which means $|P| = 2$ or $|P| = -2$. Both are perfectly valid values for the determinant.

**The possible values for the determinant of $P$ are $\pm 2$.** Mathematics is not just about finding one answer; it is about finding all possible truths. You have successfully navigated the trap and arrived at the solution.

Animated Solution for Mathematics - Matrices and Determinants: If the adjoint of a $3 \times 3$ matrix $P$ is $121411473$ , then the possible value(s) of the determinant of $P$ is (are)

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Visualized Solution

The Adjoint Matrix

Determinant of Adjoint

Substituting $n = 3$

Expansion along Row 1

Evaluating the First Minor

The Second Term

Evaluating the Second Minor

The Third Term

Evaluating the Third Minor

Total Determinant Value

Finding the Determinant of $P$

The Possible Values

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Master Identity

The Calculation

The Final Twist

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