The trace of a square matrix is defined to be the sum of its diagonal entries. If A is a 2 × 2 matri

Q: Wait, why can't we just find the matrix $A$ itself? Can't we solve for the entries?

That is a brilliant instinct, but here is the trap: the system of equations you would get for the entries of $A$ is underdetermined. There are infinitely many matrices that satisfy these trace conditions. By focusing on the eigenvalues, we bypass the need to know the individual entries and go straight to the invariant properties of the matrix.

Q: What if the eigenvalues are complex numbers? Does the identity still work?

Absolutely! The beauty of algebraic identities like $\lambda_1^3 + \lambda_2^3 = (\lambda_1 + \lambda_2)^3 - 3\lambda_1\lambda_2(\lambda_1 + \lambda_2)$ is that they are universal. They hold true whether the roots are real, complex, or even repeated. Mathematics is elegant that way; it doesn't care about the nature of the numbers, only their relationships.

Q: Is there a way to solve this without using eigenvalues? Maybe using the characteristic equation?

Yes, you can use the Cayley-Hamilton theorem! For a $2 \times 2$ matrix, $A^2 - \text{tr}(A)A + \det(A)I = 0$. You could multiply by $A$ to get $A^3$ in terms of $A$ and $I$, then take the trace of both sides. It leads to the exact same result, but the eigenvalue method is often faster and less prone to algebraic errors.

Question

The trace of a square matrix is defined to be the sum of its diagonal entries. If $A$ is a $2 	imes 2$ matrix such that the trace of $A$ is 3 and the trace of $A^3$ is $-18$, then the value of the determinant of $A$ is ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the precipice of a complex matrix transformation. You have a $2 	imes 2$ matrix $A$, and you are given two pieces of information: the trace of $A$ is $3$, and the trace of $A^3$ is $-18$.

At first glance, this might seem like a puzzle about individual entries, but I want you to shift your perspective. Think of eigenvalues as the DNA of a matrix; they encode the fundamental behavior of the transformation, independent of the basis you choose.

### The Eigenvalue Insight

Let the eigenvalues of our matrix $A$ be $\lambda_1$ and $\lambda_2$. The trace of a matrix is not just the sum of its diagonal elements; it is, by definition, the sum of its eigenvalues.

So, we immediately have our first anchor:
$$\lambda_1 + \lambda_2 = 3$$

Now, consider the power of the matrix $A^3$. If $A$ has eigenvalues $\lambda_1$ and $\lambda_2$, then $A^3$ has eigenvalues $\lambda_1^3$ and $\lambda_2^3$.

Consequently, the trace of $A^3$ is simply the sum of these cubed eigenvalues:
$$\lambda_1^3 + \lambda_2^3 = -18$$

We are now looking at a classic symmetric polynomial problem. We know the sum of the roots and the sum of their cubes. Our goal is to find the determinant, which is the product of the eigenvalues: $\det(A) = \lambda_1 \lambda_2$.

### The Algebraic Bridge

This is where we pull a powerful tool from our mathematical arsenal: the algebraic identity for the sum of cubes. We know that:
$$\lambda_1^3 + \lambda_2^3 = (\lambda_1 + \lambda_2)^3 - 3\lambda_1\lambda_2(\lambda_1 + \lambda_2)$$

This identity is our bridge. It connects the sum of the cubes (which we know is $-18$) to the sum of the roots (which we know is $3$) and the product of the roots (our target, $\det(A)$).

By substituting our known values, we get:
$$-18 = (3)^3 - 3(\lambda_1\lambda_2)(3)$$

### The Final Calculation

Now, let's simplify this with precision. We have:
$$-18 = 27 - 9(\lambda_1\lambda_2)$$

To isolate our target, the product $\lambda_1\lambda_2$, we rearrange the terms. Moving the $9\lambda_1\lambda_2$ to the left and the $-18$ to the right, we get:
$$9\lambda_1\lambda_2 = 27 + 18$$
$$9\lambda_1\lambda_2 = 45$$

Dividing both sides by $9$, we find that $\lambda_1\lambda_2 = 5$.

Since the determinant of a $2 	imes 2$ matrix is the product of its eigenvalues, we have arrived at our destination:
$$\mathbf{\det(A) = 5}$$

It is a beautiful, clean result that emerges from the interplay of traces and powers. Remember, in JEE Advanced, the complexity of the problem is often just a veil; once you identify the underlying symmetry, the path to the solution becomes clear and inevitable.

Animated Solution for Mathematics - Matrices and Determinants: The trace of a square matrix is defined to be the sum of its diagonal entries. If $A$ is a $2 \times 2$ matrix such that the trace of $A$ is 3 and the trace of $A^{3}$ is $- 18$ , then the value of the determinant of $A$ is ____.

Enter Numerical Value:

Visualized Solution

$Eigenvalues of A$

$Trace of A$

$Trace of A^{3}$

$Determinant of A$

$Algebraic Identity$

$Substitution$

$Simplification$

$Isolating the Product$

$Conclusion$

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Eigenvalue Insight

The Algebraic Bridge

The Final Calculation

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Enter Numerical Value:

Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Trace of A3

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Isolating the Product

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