Let A = [[x, y, z], [y, z, x], [z, x, y]], where x, y and z are real numbers such that x + y + z 0

Q: Wait, why do we reject the positive root for the determinant?

Excellent question! We found $|A|^2 = 1$, so $|A|$ could be $1$ or $-1$. However, we expressed $|A|$ as $-(x+y+z) \cdot \frac{1}{2}[(x-y)^2 + (y-z)^2 + (z-x)^2]$. Since $x+y+z > 0$ and the sum of squares is non-negative, the entire expression is $\leq 0$. Thus, $|A|$ must be $-1$.

Q: Can't we just assume $x=y=z$ to solve this?

That is a tempting shortcut! If $x=y=z$, then $x^3=2$, so $x = \sqrt[3]{2}$. But does this satisfy $A^2=I_3$? If $x=y=z$, the matrix becomes a matrix of all equal elements. Calculating $A^2$ would show it's not the identity matrix. Always check if your assumption satisfies the original matrix equation!

Q: Why is the determinant of a cyclic matrix $3xyz - (x^3+y^3+z^3)$?

This is a standard result for a $3 \times 3$ cyclic matrix. When you expand the determinant $\begin{vmatrix} x & y & z \\ y & z & x \\ z & x & y \end{vmatrix}$, you get $x(zy-x^2) - y(y^2-xz) + z(xy-z^2)$. Simplifying this yields $xyz - x^3 - y^3 + xyz + xyz - z^3$, which is $3xyz - (x^3+y^3+z^3)$. It is a very useful identity to memorize for JEE!

Question

Let $A = \begin{bmatrix} x & y & z \ y & z & x \ z & x & y \end{bmatrix}$, where $x, y$ and $z$ are real numbers such that $x + y + z > 0$ and $xyz = 2$. If $A^2 = I_3$, then the value of $x^3 + y^3 + z^3$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to unravel a problem that looks like a standard matrix puzzle but hides a beautiful, deep algebraic secret.

We are given a $3 	imes 3$ matrix:
$$A = \begin{bmatrix} x & y & z \ y & z & x \ z & x & y \end{bmatrix}$$

Notice the symmetry? This is a **cyclic matrix**. The variables $x, y,$ and $z$ dance around in a perfect, repeating pattern.

We are also given two vital clues: $x + y + z > 0$ and $xyz = 2$. Our mission is to find the value of $x^3 + y^3 + z^3$, given the constraint $A^2 = I_3$.

### The Determinant Expansion

To connect $A^2 = I_3$ to our target expression, we need to look at the determinant. Let us expand $|A|$ along the first row.

The expansion is:
$$|A| = x(yz - x^2) - y(y^2 - xz) + z(xy - z^2)$$

When we distribute these terms, we get:
$$|A| = xyz - x^3 - y^3 + xyz + xyz - z^3$$

Grouping these, we find the elegant identity:
$$|A| = 3xyz - (x^3 + y^3 + z^3)$$

This is our bridge. If we know $|A|$, we are one step away from finding the sum of the cubes.

### The Matrix Constraint

Now, let us use the given equation $A^2 = I_3$. If we take the determinant of both sides, we get $|A^2| = |I_3|$.

Using the property $|A^n| = |A|^n$, this becomes $|A|^2 = 1$. This is the moment where most students pause.

We have two possibilities: $|A| = 1$ or $|A| = -1$. How do we choose? This is where the JEE examiner tests your attention to detail. We must look at the constraints $x+y+z > 0$ and $xyz=2$.

### The Sign Trap

Let us revisit our determinant expression: $|A| = -(x^3 + y^3 + z^3 - 3xyz)$. We know the famous factorization:
$$x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx)$$

The second factor can be rewritten as:
$$\frac{1}{2}[(x-y)^2 + (y-z)^2 + (z-x)^2]$$

So, the determinant becomes:
$$|A| = -(x+y+z) \cdot \frac{1}{2}[(x-y)^2 + (y-z)^2 + (z-x)^2]$$

Since $x+y+z > 0$ and the sum of squares is always non-negative, the entire expression is $\leq 0$. Therefore, $|A|$ cannot be $1$. It must be $-1$.

### The Final Calculation

Now, we are in the home stretch. We have:
$$3xyz - (x^3 + y^3 + z^3) = |A| = -1$$

We know $xyz = 2$. Substituting this in, we get:
$$3(2) - (x^3 + y^3 + z^3) = -1$$

This simplifies to:
$$6 - (x^3 + y^3 + z^3) = -1$$

Rearranging the terms, we get:
$$x^3 + y^3 + z^3 = 6 + 1 = 7$$

And there it is! The answer is **7**. A beautiful, clean result from a problem that seemed daunting at first. Remember, in JEE, always look for the structure, respect the constraints, and trust the algebra.

Animated Solution for Mathematics - Matrices and Determinants: Let $A = x y z y z x z x y$ , where $x, y$ and $z$ are real numbers such that $x + y + z > 0$ and $x y z = 2$ . If $A^{2} = I_{3}$ , then the value of $x^{3} + y^{3} + z^{3}$ is

Enter Numerical Value:

Visualized Solution

Introduction to the Problem

Expanding the Determinant $∣ A ∣$

Simplifying the Determinant Expression

Using the Property $A^{2} = I_{3}$

Possible Values of $∣ A ∣$

Relating to Algebraic Identity

Analyzing the Sign of $∣ A ∣$

Determining the Final Sign

Setting up the Final Equation

Substituting the Value of $x y z$

Final Calculation

Summary and Key Takeaways

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Determinant Expansion

The Matrix Constraint

The Sign Trap

The Final Calculation

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Enter Numerical Value:

Visualized Solution

Introduction to the Problem

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

Simplifying the Determinant Expression

Using the Property .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2=I3​

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