For a 3x3 matrix M, let trace(M) denote the sum of all the diagonal elements of M. Let A be a 3x3 ma

Q: Wait, why does the scalar k come out as k^3 for the determinant but just k for the trace?

This is a fundamental distinction in linear algebra! Think of the determinant as a volume scaling factor in n-dimensional space. If you scale each of the 3 dimensions of a 3D object by k, the volume scales by k * k * k = k^3. The trace, however, is just the sum of the diagonal elements. If you multiply the matrix by k, every element on the diagonal is multiplied by k, so the sum simply becomes k times the original sum. It is linear, not volumetric!

Q: Can I just assume A is a diagonal matrix with entries like a, b, c to solve this?

That is a brilliant intuition! Yes, for a general problem like this, if the result must hold for any matrix A satisfying the conditions, it must hold for a diagonal matrix. You could set A = diag(a, b, c) such that a+b+c = 3 and abc = 1/2. While this works, mastering the property-based approach is faster and more robust for JEE Advanced, as it saves you from solving for specific values.

Q: Why can't we just use energy conservation or some other physics principle here?

I love the enthusiasm! While physics problems often use energy conservation, this is a pure linear algebra problem. In JEE, we must switch gears. Think of matrix properties as the 'laws of motion' for linear algebra. We don't need energy here; we need the 'laws' of determinants and traces to navigate the structure of the matrix.

Question

For a $3	imes3$ matrix M, let trace(M) denote the sum of all the diagonal elements of M. Let A be a $3	imes3$ matrix such that $|A|=\frac{1}{2}$ and trace$(A)=3$. If $B=adj(adj(2A))$, then the value of $|B|+$ trace$(B)$ equals :

Options:

(A) 56

(B) 132

(C) 174

(D) 280

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, my dear student, to a journey through the heart of linear algebra. Today, we are not just solving a matrix problem; we are uncovering the hidden symmetries that govern the behavior of matrices.

When you first look at the expression $B = 	ext{adj}(	ext{adj}(2A))$, it is natural to feel a sense of trepidation. It looks like a labyrinth of operations, but in the world of JEE Advanced, complexity is often just a mask for underlying simplicity.

### The Double Adjoint Mystery

The first step in our quest is to simplify the double adjoint. We have a powerful identity in our toolkit: for any $n 	imes n$ matrix $X$, the adjoint of the adjoint is given by:

$$ 	ext{adj}(	ext{adj}(X)) = |X|^{n-2} \cdot X $$

In our case, the matrix is $3 	imes 3$, so $n = 3$. This simplifies our identity beautifully to:

$$ 	ext{adj}(	ext{adj}(X)) = |X|^{3-2} \cdot X = |X| \cdot X $$

By defining $X = 2A$, we transform our daunting expression into $B = |X| \cdot X$. The complexity has already begun to dissolve.

### The Scaling Trap

Now, we must find $|X|$. Here is where many students stumble. We know $X = 2A$, and we need the determinant $|X| = |2A|$.

The scaling property of determinants is a classic JEE trap. For an $n 	imes n$ matrix, the property is $|kA| = k^n \cdot |A|$. Since our matrix is $3 	imes 3$, the scalar $2$ must be raised to the power of $3$:

$$ |2A| = 2^3 \cdot |A| = 8 \cdot |A| $$

Given that $|A| = 1/2$, we find that $|X| = 8 \cdot (1/2) = 4$. Substituting this back, we obtain:

$$ B = 4 \cdot (2A) = 8A $$

### The Synthesis

We are now in the home stretch. We need to find $|B| + 	ext{trace}(B)$. We have established that $B = 8A$.

For the determinant, we apply the scaling property again:

$$ |B| = |8A| = 8^3 \cdot |A| = 512 \cdot \left( \frac{1}{2} ight) = 256 $$

For the trace, we use the linear property $	ext{trace}(kA) = k \cdot 	ext{trace}(A)$:

$$ 	ext{trace}(B) = 	ext{trace}(8A) = 8 \cdot 	ext{trace}(A) $$

Given $	ext{trace}(A) = 3$, we calculate $	ext{trace}(B) = 8 \cdot 3 = 24$. Finally, we sum these values:

$$ |B| + 	ext{trace}(B) = 256 + 24 = \mathbf{280} $$

We navigated the double adjoint, handled the scaling properties with precision, and arrived at the solution. This is the power of understanding the 'why' behind the formulas.

Select Answer:

Visualized Solution

Analyze the Given Information

The Double Adjoint Property

Defining the Intermediate Matrix $X$

Scaling Property of Determinants

Calculating the Determinant of $X$

Expressing $B$ in terms of $A$

Finding the Determinant of $B$

Scaling Property of Trace

Calculating the Trace of $B$

Final Summation and Conclusion

The Sigma Insight: Adjoint and Inverse of a Matrix

Analyzing the Setup

The Double Adjoint Mystery

The Scaling Trap

The Synthesis

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Select Answer:

Visualized Solution

Analyze the Given Information

The Double Adjoint Property

Defining the Intermediate Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X

Scaling Property of Determinants

Calculating the Determinant of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Finding the Determinant of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Scaling Property of Trace

Calculating the Trace of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B