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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 $A-3 I_{3}$ is
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The correct answer is:
non-invertible
Given, that $A-3 J_{3}$
$$
=\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]-3\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
$$
$=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]-\left[\begin{array}{ccc}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right]$
$=\left[\begin{array}{ccc}-2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2\end{array}\right]$
$=[-2(4-1)-1(-2-1)+1(1+2)]$
$=[-2(3)-1(-3)+1(3)]$
$=[-6+3+3]=0$
$\mathrm{Now}, \mathrm{det}=0$
So, matrix $A-3 I_{3}$ is non-invertible matrix.
$$
=\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]-3\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
$$
$=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]-\left[\begin{array}{ccc}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right]$
$=\left[\begin{array}{ccc}-2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2\end{array}\right]$
$=[-2(4-1)-1(-2-1)+1(1+2)]$
$=[-2(3)-1(-3)+1(3)]$
$=[-6+3+3]=0$
$\mathrm{Now}, \mathrm{det}=0$
So, matrix $A-3 I_{3}$ is non-invertible matrix.
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