Let A = [0 0 -1; 0 -1 0; -1 0 0]. The only correct statement about the matrix A is (a) A² = I (b) A

Q: Wait, why isn't $A = -I$? The elements are all -1 or 0, just like $-I$.

That is a common trap! The identity matrix $I$ has ones *only* on the main diagonal (from top-left to bottom-right). Your matrix $A$ has non-zero elements on the *anti-diagonal*. They are fundamentally different structures. Always check the position of the elements carefully!

Q: How do I know if $A^{-1}$ exists without calculating the inverse matrix?

You don't need the inverse itself! Just calculate the determinant $|A|$. If $|A| \neq 0$, the inverse exists. It's the gatekeeper property of matrices. In this case, $|A| = 1$, so the inverse definitely exists.

Q: Is there a faster way to check $A^2 = I$ than full multiplication?

For a $3 \times 3$ matrix, full multiplication is the most reliable method. However, you can look for patterns. Since $A$ is a permutation-like matrix, its square often results in the identity matrix. But always verify with the full product to be absolutely sure!

Question

Let $A = \begin{bmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{bmatrix}$. The only correct statement about the matrix $A$ is

Options:

(A) $A^2 = I$

(B) $A = (-1)I$ where $I$ is a unit matrix

(C) $A^{-1}$ does not exist

(D) $A$ is a zero matrix

Sigma Percentile · Accepted Answer

### The Dance of the Matrix: Unveiling the Hidden Symmetry

Welcome, future engineers! Today, we are going to dive into the elegant world of linear algebra. Often, when we see a matrix like

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

our first instinct might be to panic or start calculating blindly. But wait! Let's take a breath and look at the structure.

This is not just a random collection of numbers; it is a structured, symmetric entity. In this article, we will peel back the layers of this matrix to understand its true nature.

### Phase 1: The Elimination Game

Before we rush into heavy calculations, let's use the power of logic to eliminate the incorrect options. We are given four statements, and we need to find the one that is true.

First, is $A$ a zero matrix? A zero matrix, denoted as $O$, must have every single element equal to zero. Looking at our matrix $A$, we clearly see elements like $-1$.

Since $-1 
eq 0$, $A$ is definitely not a zero matrix. Option 4 is out!

Next, let's test if $A = -I$. The identity matrix $I$ is defined as

$$I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$$

Therefore, $-I$ would be

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

Comparing this with our matrix $A$, we see that the first element $a_{11}$ is $0$ in $A$, but it is $-1$ in $-I$. They don't match! So, option 2 is incorrect.

### Phase 2: The Gatekeeper of Invertibility

Now, let's tackle the existence of the inverse. A matrix is singular (meaning its inverse does not exist) if and only if its determinant is zero. Let's calculate the determinant of $A$ by expanding along the first row:

$$|A| = 0 - 0 + (-1) \begin{vmatrix} 0 & -1 \ -1 & 0 \end{vmatrix}$$

Calculating the $2 	imes 2$ determinant: $(0)(0) - (-1)(-1) = 0 - 1 = -1$. So, $|A| = -1 	imes (-1) = 1$.

Since $|A| = 1$, which is not zero, the inverse $A^{-1}$ definitely exists. Thus, option 3 is also incorrect.

### Phase 3: The Moment of Truth

By the process of elimination, option 1 must be the correct one. But as scientists and engineers, we don't just accept things; we prove them! Let's calculate $A^2 = A \cdot A$.

Don't be intimidated by the multiplication; it is just a systematic dance of rows and columns. We set up the multiplication:

$$A^2 = \begin{bmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{bmatrix}$$

Let's compute the first row of the result:
- $c_{11} = (0)(0) + (0)(0) + (-1)(-1) = 1$
- $c_{12} = (0)(0) + (0)(-1) + (-1)(0) = 0$
- $c_{13} = (0)(-1) + (0)(0) + (-1)(0) = 0$

Following this pattern for the second and third rows, we find that the resulting matrix is indeed the identity matrix:

$$I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$$

### Conclusion: The Involutory Elegance

We have successfully proven that **$A^2 = I$**. This is a beautiful result!

In linear algebra, a matrix that squares to the identity matrix is called an **Involutory Matrix**. It acts as its own inverse, which is a powerful property in transformations.

Keep this concept in your toolkit—it often appears in competitive exams to test your understanding of matrix properties. You've done great work today!

Animated Solution for Mathematics - Matrices and Determinants: Let $A = 00 - 1 0 - 1 0 - 1 00$ . The only correct statement about the matrix $A$ is

Select Answer:

Visualized Solution

Introduction to Matrix $A$

Checking Option: Zero Matrix

Checking Option: $A = - I$

Checking Option: Existence of $A^{- 1}$

Setup for $A^{2}$

Computing $A^{2}$ : First Row

Computing $A^{2}$ : Second Row

Computing $A^{2}$ : Third Row

Final Result: $A^{2} = I$

The Sigma Insight: Types of Matrices

The Dance of the Matrix

Phase 1

Phase 2

Phase 3

Conclusion

Similar Questions

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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; } Let A=​00−1​0−10​−100​​. The only correct statement about the matrix A is

Select Answer:

Visualized Solution

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

Checking Option: Zero Matrix

Checking Option: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A=−I

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

Setup for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2: First Row

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2: Second Row

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2: Third Row

Final Result: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2=I