જો $A=\left[\begin{array}{cc}8 & 0 \\ 4 & -2 \\ 3 & 6\end{array}\right]$ અને $B=\left[\begin{array}{cc}2 & -2 \\ 4 & 2 \\ -5 & 1\end{array}\right]$ તો $2 \mathrm{A}+3 \mathrm{X}=5 \mathrm{B}$ થાય એવો શ્રેણિક $X$ શોધો.
Medium
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We have $2A+3X=5 B$
or $2A+3X-2A=5B-2A$
or $2 A-2A+3X=5B-2A$ $($ Matrix addition is commutative $)$
or $O+3 X=5B-2 A$ $(-2A$ is the additive inverse of $2A)$
or $3 \mathrm{X}=5 \mathrm{B}-2 \mathrm{A}$ ( $O$ is the additive identity)
or $X=\frac{1}{3}(5 B-2 A)$
or ${\text{X}} = $ $\frac{1}{3}\left( {5\left[ {\begin{array}{*{20}{c}}
2&{ - 2} \\
4&2 \\
{ - 5}&1
\end{array}} \right] - 2\left[ {\begin{array}{*{20}{l}}
8&0 \\
4&{ - 2} \\
3&6
\end{array}} \right]} \right)$ $ = \frac{1}{3}\left( {\left[ {\begin{array}{*{20}{c}}
{10}&{ - 10} \\
{20}&{10} \\
{ - 25}&5
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
{ - 16}&0 \\
{ - 8}&4 \\
{ - 6}&{ - 12}
\end{array}} \right]} \right)$
$ = \frac{1}{3}\left[ {\begin{array}{*{20}{c}}
{10 - 16}&{ - 10 + 0} \\
{20 - 8}&{10 + 4} \\
{ - 25 - 6}&{5 - 12}
\end{array}} \right]$
$ = \frac{1}{3}\left[ {\begin{array}{*{20}{c}}
{ - 6}&{ - 10} \\
{12}&{14} \\
{ - 31}&{ - 7}
\end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}}
{ - 2}&{\frac{{ - 10}}{3}} \\
4&{\frac{{14}}{3}} \\
{\frac{{ - 31}}{3}}&{\frac{{ - 7}}{3}}
\end{array}} \right]$
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