MCQ
If $A = \left( {\begin{array}{*{20}{c}}1&2&3\\3&1&2\\2&3&1\end{array}} \right)$ and $B = \left( {\begin{array}{*{20}{c}}{ - 5}&7&1\\1&{ - 5}&7\\7&1&{ - 5}\end{array}} \right)$ then $AB$ is equal to
  • A
    ${I_3}$
  • B
    $2{I_3}$
  • C
    $4{I_3}$
  • $18{I_3}$

Answer

Correct option: D.
$18{I_3}$
d
(d) We have $A = \left[ {\begin{array}{*{20}{c}}1&2&3\\3&1&2\\2&3&1\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}{ - 5}&7&1\\1&{ - 5}&7\\7&1&{ - 5}\end{array}} \right]$

$\therefore $ $AB = \left[ {\begin{array}{*{20}{c}}1&2&3\\3&1&2\\2&3&1\end{array}} \right]\,\,\left[ {\begin{array}{*{20}{c}}{ - 5}&7&1\\1&{ - 5}&7\\7&1&{ - 5}\end{array}} \right]$

$AB = \left[ {\begin{array}{*{20}{c}}{18}&0&0\\0&{18}&0\\0&0&{18}\end{array}} \right] = 18\,\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$

$AB = 18\,{I_3}$.

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