Which of the following is(are) NOT the square of a 3 x 3 matrix with real entries ? (A) [[1, 0, 0],

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

Actually, we aren't rejecting a root; we are rejecting the existence of the matrix itself! If $|M| < 0$, there is no real number $x$ such that $x^2 = |M|$. Thus, no real matrix $A$ can exist.

Q: Is the determinant condition sufficient? What if the determinant is positive?

Great intuition! The determinant condition is necessary, but not always sufficient. However, for diagonal matrices, it is a very strong indicator. Always check if you can construct the matrix explicitly if the determinant is positive.

Q: How did you know to use a rotation matrix for Option D?

When you see $-1$ on the diagonal, think of transformations. A $180^\circ$ rotation is just two $90^\circ$ rotations. It's about visualizing the geometry behind the algebra.

Question

Which of the following is(are) NOT the square of a $3 	imes 3$ matrix with real entries ?

Options:

(A) $\begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1 \end{bmatrix}$

(B) $\begin{bmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -1 \end{bmatrix}$

(C) $\begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$

(D) $\begin{bmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -1 \end{bmatrix}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we aren't just solving a matrix problem; we are detectives investigating the 'DNA' of matrices. We are looking for the square roots of matrices—a concept that feels abstract but is rooted in the solid ground of linear algebra.

We are given four matrices, and our mission is to identify which of them cannot be expressed as the square of a $3 	imes 3$ real matrix. Let's dive in.

### The Determinant Gatekeeper

Imagine we have a matrix $M$ that is the square of some real matrix $A$. Mathematically, we write this as $M = A^2$.

Now, how do we test if such an $A$ exists? We need a filter, a 'gatekeeper' property that separates the possible from the impossible. The most powerful tool in our arsenal is the determinant.

If we take the determinant of both sides, we get $|M| = |A^2|$. Using the fundamental property of determinants, $|A^n| = |A|^n$, we arrive at the elegant relation:

$$|M| = |A|^2$$

This is the key to the kingdom. Since $A$ is a real matrix, its determinant $|A|$ must be a real number.

And what is the most sacred rule of real numbers? The square of any real number is always non-negative. Therefore, for $M$ to be a square, it is absolutely mandatory that $|M| \ge 0$.

If $|M| < 0$, the matrix $M$ simply cannot be a square of a real matrix. This is our golden rule.

### The Filter in Action

Let's apply this rule to our candidates. Consider Option A:

$$M_A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1 \end{bmatrix}$$

Since this is a diagonal matrix, its determinant is simply the product of its diagonal elements: $(1)(1)(-1) = -1$. Because $-1 < 0$, Option A fails our test immediately. It cannot be a square.

Now, look at Option B:

$$M_B = \begin{bmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -1 \end{bmatrix}$$

Again, the determinant is $(-1)(-1)(-1) = -1$. This also fails the test. We have successfully identified two matrices that are not squares.

### The Subtle Trap

Now, we must be careful. We have Options C and D left. Option C is the identity matrix $I$. We know that $I^2 = I$, so it is clearly a square.

But what about Option D?

$$M_D = \begin{bmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & -1 \end{bmatrix}$$

Its determinant is $(1)(-1)(-1) = 1$. Since $1 \ge 0$, it passes our determinant test. But does that guarantee it is a square? Not necessarily.

The determinant condition is necessary, but not always sufficient. We must look deeper. Notice the structure: the top-left is $1$, and the bottom-right is a $2 	imes 2$ block of $-1$s.

This looks like a rotation. Specifically, it represents a $180^{\circ}$ rotation in the $y-z$ plane. Can we achieve this in two steps? Yes! A $180^{\circ}$ rotation is just two successive $90^{\circ}$ rotations.

If we define $A$ as:

$$A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & -1 & 0 \end{bmatrix}$$

Then $A^2$ will indeed yield $M_D$. Thus, Option D is a valid square.

### Conclusion

By using the determinant as our filter and geometric intuition to verify the remaining candidates, we have solved the mystery.

**Options A and B are the ones that cannot be the square of a real matrix.**

Keep this logic in your toolkit: when in doubt, check the determinant, but always keep an eye out for the geometric transformations hidden within the numbers.

Animated Solution for Mathematics - Matrices and Determinants: Which of the following is(are) NOT the square of a $3 \times 3$ matrix with real entries ?

Select Answer:

Visualized Solution

The Matrix Square Equation

The Determinant Condition

The Non-Negative Constraint

Checking Option A

Determinant of Option A

Checking Option B

Conclusion for Option B

Checking Option C

Conclusion for Option C

Checking Option D

The Trap in Option D

Constructing Matrix for Option D

Verifying the Rotation Matrix

Final Conclusion

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Determinant Gatekeeper

The Filter in Action

The Subtle Trap

Conclusion

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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; } Which of the following is(are) NOT the square of a 3×3 matrix with real entries ?