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In a third order matrix $A$, $a_{0}$ denotes the element in the ith row and $j$ th column. If $a_{i j}=0$ for $i=j,1$ for $i>j,-1$ for $i < j$ Then the matrix is
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The correct answers are:
skew symmetric, not invertible
$$
A=\left[\begin{array}{ccc}
0 & -1 & -1 \\
1 & 0 & -1 \\
1 & 1 & 0
\end{array}\right]
$$
$\Rightarrow$
$A^{\prime}=\left[\begin{array}{ccc}0 & 1 & 1 \\ -1 & 0 & 1 \\ -1 & -1 & 0\end{array}\right]=-A$
$\Rightarrow A$ is a skew-symmetric matrix. $\begin{aligned}|A| &=0+1(0+1)-1(-0) \\ &=0+1-1=0=\text { singular } \end{aligned}$
$\Rightarrow A$ is not invertible.
A=\left[\begin{array}{ccc}
0 & -1 & -1 \\
1 & 0 & -1 \\
1 & 1 & 0
\end{array}\right]
$$
$\Rightarrow$
$A^{\prime}=\left[\begin{array}{ccc}0 & 1 & 1 \\ -1 & 0 & 1 \\ -1 & -1 & 0\end{array}\right]=-A$
$\Rightarrow A$ is a skew-symmetric matrix. $\begin{aligned}|A| &=0+1(0+1)-1(-0) \\ &=0+1-1=0=\text { singular } \end{aligned}$
$\Rightarrow A$ is not invertible.
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