If P = [1 α 3; 1 3 3; 2 4 4] is the adjoint of a 3 x 3 matrix A and |A| = 4, then α is equal to: (a)

Q: Wait, why can't we just find the inverse of $P$ to get $A$? Isn't $A = \text{adj}(P) / |P|$?

That is a mathematically sound approach, but it is a trap! Finding the adjoint of a $3 \times 3$ matrix involves calculating nine cofactors. That is a massive amount of work for a JEE exam. Always look for properties first. The property $|\text{adj}(A)| = |A|^{n-1}$ is designed to bypass the need to find $A$ entirely.

Q: Why is the determinant of the adjoint $|A|^{n-1}$? Where does this formula come from?

Think about the definition of the inverse: $A^{-1} = \frac{1}{|A|} \text{adj}(A)$. If you take the determinant of both sides, you get $|A^{-1}| = |\frac{1}{|A|} \text{adj}(A)|$. Since $|kA| = k^n |A|$ for an $n \times n$ matrix, this becomes $\frac{1}{|A|} = \frac{1}{|A|^n} |\text{adj}(A)|$. Rearranging this gives $|\text{adj}(A)| = |A|^{n-1}$. It is a beautiful derivation that saves you minutes of calculation!

Q: I keep getting a different value for $\alpha$. Is there a sign convention I am missing?

Yes, the most common mistake is forgetting the alternating sign in the cofactor expansion. When expanding along the first row, the signs are $(+), (-), (+)$. So, for the second element $\alpha$, you must subtract the product of its minor. If you missed that negative sign, your equation will be wrong. Always write out the expansion carefully!

Question

If $P = \begin{bmatrix} 1 & \alpha & 3 \ 1 & 3 & 3 \ 2 & 4 & 4 \end{bmatrix}$ is the adjoint of a $3 	imes 3$ matrix $A$ and $|A| = 4$, then $\alpha$ is equal to:

Options:

(A) $4$

(B) $11$

(C) $5$

(D) $0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to demystify a classic JEE Advanced problem. It is a beautiful example of how linear algebra rewards those who know their properties rather than those who brute-force their way through calculations.

We are given a matrix $P = 	ext{adj}(A)$ and told that the determinant of the original matrix $A$ is $4$. Our mission is to find the value of $\alpha$ hidden inside $P$.

Many students see a matrix and immediately want to perform row operations or find the inverse. But stop! In the world of competitive exams, the path of least resistance is usually the path of properties.

### The Secret Key: The Adjoint Property

The most powerful tool in your arsenal for this problem is the relationship between the determinant of a matrix and the determinant of its adjoint. For any $n 	imes n$ matrix $A$, the property is $|	ext{adj}(A)| = |A|^{n-1}$.

Why is this true? It stems from the definition of the inverse matrix, $A^{-1} = \frac{1}{|A|} 	ext{adj}(A)$. By taking the determinant of both sides and applying the scalar multiplication rule for determinants, we arrive at this elegant identity.

In our case, the matrix $A$ is $3 	imes 3$, so $n=3$. We are given $|A|=4$. Substituting these values, we get:

$$|P| = |	ext{adj}(A)| = 4^{3-1} = 4^2 = 16$$

Just like that, we have reduced a matrix problem to a simple numerical value. The determinant of $P$ is $16$.

### The Calculus of Arithmetic: Determinant Expansion

Now that we know $|P| = 16$, we can set up the determinant expansion of $P$ using the elements provided:

$$|P| = \begin{vmatrix} 1 & \alpha & 3 \ 1 & 3 & 3 \ 2 & 4 & 4 \end{vmatrix} = 16$$

To solve for $\alpha$, we expand along the first row. Remember the alternating sign convention: $(+), (-), (+)$.

Let us take it step by step. For the first element, $1$, we hide its row and column to get the minor: $1(3 	imes 4 - 3 	imes 4) = 1(12 - 12) = 0$.

For the second element, $\alpha$, we must remember the negative sign: $-\alpha(1 	imes 4 - 3 	imes 2) = -\alpha(4 - 6) = -\alpha(-2) = 2\alpha$.

Finally, for the third element, $3$, we have: $+3(1 	imes 4 - 3 	imes 2) = 3(4 - 6) = 3(-2) = -6$.

### The Victory Lap: Solving for $\alpha$

We now combine these terms to form our equation: $0 + 2\alpha - 6 = 16$. This is the moment where the complexity vanishes.

We add $6$ to both sides to get $2\alpha = 22$. Dividing by $2$, we find that **$\alpha = 11$**.

It is that simple! We did not need to find the inverse of $P$, nor did we need to perform complex row reductions.

By leveraging the adjoint determinant property, we turned a potentially daunting matrix problem into a straightforward algebraic equation. Remember, in JEE Advanced, the goal is not just to solve, but to solve elegantly. Keep this property in your toolkit, and you will be ready for any matrix challenge that comes your way!

Animated Solution for Mathematics - Matrices and Determinants: If $P = 112 α 34 334$ is the adjoint of a $3 \times 3$ matrix $A$ and $∣ A ∣ = 4$ , then $α$ is equal to:

Select Answer:

Visualized Solution

Identify the Given Information

Recall the Adjoint Property

Calculate the Determinant of $P$

Set up Determinant Expansion of $P$

Expand the First Term

Expand the Second Term

Expand the Third Term

Simplify the Determinant Terms

Solve for $α$

Final Conclusion

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Secret Key

The Calculus of Arithmetic

The Victory Lap

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Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If P=​112​α34​334​​ is the adjoint of a 3×3 matrix A and ∣A∣=4, then α is equal to: