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(C) Given $A = \begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{bmatrix}$ and $I_{3} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \en

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non-invertible

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(C) Given $A = \begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{bmatrix}$ and $I_{3} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$.Then $A-3I_{3} = \begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{bmatrix} - \begin{bmatrix} 3 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 3 \end{bmatrix} = \begin{bmatrix} -2 & 1 & 1 \ 1 & -2 & 1 \ 1 & 1 & -2 \end{bmatrix}$.To check if the matrix is invertible, we calculate its determinant:$\det(A-3I_{3}) = -2((-2)(-2) - (1)(1)) - 1((1)(-2) - (1)(1)) + 1((1)(1) - (-2)(1))$$\det(A-3I_{3}) = -2(4-1) - 1(-2-1) + 1(1+2)$$\det(A-3I_{3}) = -2(3) - 1(-3) + 1(3) = -6 + 3 + 3 = 0$.Since the determinant of the matrix is $0$, the matrix $A-3I_{3}$ is non-invertible.

Let $A$ be a square matrix of order $3$ whose all entries are $1$ and let $I_{3}$ be the identity matrix of order $3$. Then, the matrix $A-3I_{3}$ is

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