Matrices — Class 12 Maths Solution

As we know,, in a square matrix, if ${b_{ij}} = 0$, when $i \ne j$,

then it is said to be a diagonal matrix. Here,

${b_{12}},{b_{13}}, \ldots \ne 0$, so the given matrix is not a diagonal matrix.

Now, $B = \left[ {\begin{array}{cccccccccccccccccccc}0&{ - 5}&8\\5&0&{12}\\{ - 8}&{ - 12}&0\end{array}} \right]$

$\therefore$ ${B^\prime } = \left[ {\begin{array}{cccccccccccccccccccc}0&5&{ - 8}\\{ - 5}&0&{ - 12}\\8&{12}&0\end{array}} \right]$

$= - \left[ {\begin{array}{cccccccccccccccccccc}0&{ - 5}&8\\5&0&{12}\\{ - 8}&{ - 12}&0\end{array}} \right] = - B$

Therefore, the given matrix is a skew-symmetric matrix,

since As we know, in a square matrix B,

if ${B^\prime } = - B$, then it is called skew-symmetric matrix.