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यदि $P = \left[ {\begin{array}{*{20}{c}}1&2&3\\2&3&4\\3&4&5\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 2}\\{ - 2}&0\\0&{ - 4}\end{array}} \right]$$\left[ {\begin{array}{*{20}{c}}{ - 4}&{ - 5}&{ - 6}\\0&0&1\end{array}} \right]$, तब ${P_{22}} = $
$40$
$-40$
$-20$
$20$
Solution
(a) $P = {\left[ {\begin{array}{*{20}{c}}1&2&3\\2&3&4\\3&4&5\end{array}} \right]_{3 \times 3}}{\left[ {\begin{array}{*{20}{c}}{ – 1}&{ – 2}\\{ – 2}&0\\0&{ – 4}\end{array}} \right]_{3 \times 2}}{\left[ {\begin{array}{*{20}{c}}{ – 4}&{ – 5}&{ – 6}\\0&0&1\end{array}} \right]_{2 \times 3}}$
$P = {\left[ {\begin{array}{*{20}{c}}{ – 3}&{ – 14}\\{ – 8}&{ – 20}\\{ – 11}&{ – 26}\end{array}} \right]_{3 \times 2}}{\left[ {\begin{array}{*{20}{c}}{ – 4}&{ – 5}&{ – 6}\\0&0&1\end{array}} \right]_{2 \times 3}}$
$P = {\left[ {\begin{array}{*{20}{c}}{12}&{15}&4\\{32}&{40}&{28}\\{44}&{55}&{40}\end{array}} \right]_{3 \times 3}}$$ \Rightarrow $ ${P_{22}} = 40$.