Question

question 39 points 3

what is the type of the matrix \\(a = \

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

\\)?

• zero
• non-singular

Explanation:

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<reasoning>

Define the matrix and its properties

Using the Matrix Inverse Existence knowledge point
We are given the square matrix \(A\):
\[
A =

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

\]
To classify the type of matrix, we evaluate its determinant. If the determinant is non-zero, the matrix is non-singular (invertible). If it is zero, the matrix is singular.

Calculate the determinant of the matrix

Using the Matrix Determinant knowledge point
We compute the determinant \(\det(A)\) using cofactor expansion along the third row:
\[
\det(A) = 1 \cdot

$$\begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix}$$

• 0 \cdot

$$\begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix}$$

+ 0 \cdot

$$\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}$$

\]
\[
\det(A) = 1 \cdot (0 \cdot 1 - 1 \cdot 1) = -1
\]

Determine the matrix type

Since \(\det(A) = -1
eq 0\), the matrix is invertible, which means it is a non-singular matrix. It is clearly not a zero matrix because it contains non-zero elements.
</reasoning>

<answer>
<mcq-option>(A) Zero</mcq-option>
<mcq-correct>(B) Non-singular</mcq-correct>
</answer>

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Answer:

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<reasoning>

Define the matrix and its properties

Using the Matrix Inverse Existence knowledge point
We are given the square matrix \(A\):
\[
A =

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

\]
To classify the type of matrix, we evaluate its determinant. If the determinant is non-zero, the matrix is non-singular (invertible). If it is zero, the matrix is singular.

Calculate the determinant of the matrix

Using the Matrix Determinant knowledge point
We compute the determinant \(\det(A)\) using cofactor expansion along the third row:
\[
\det(A) = 1 \cdot

$$\begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix}$$

• 0 \cdot

$$\begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix}$$

+ 0 \cdot

$$\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}$$

\]
\[
\det(A) = 1 \cdot (0 \cdot 1 - 1 \cdot 1) = -1
\]

Determine the matrix type

Since \(\det(A) = -1
eq 0\), the matrix is invertible, which means it is a non-singular matrix. It is clearly not a zero matrix because it contains non-zero elements.
</reasoning>

<answer>
<mcq-option>(A) Zero</mcq-option>
<mcq-correct>(B) Non-singular</mcq-correct>
</answer>

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