Question

$\large{ \boxed{ \bold \red{ \left[ \begin{array} { l l l } { 3 } & { - 1 } & { 4 } \\ { 2 } & { 3 } & { 1 } \end{array} \right] \left[ \begin{array} { l l l } { 1 } & { 3 } & { 4 } \\ { 2 } & { 1 } & { 0 } \\ { - 3 } & { 2 } & { 3 } \end{array} \right]}}}$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given matrix is

$\rm :\longmapsto\:\left[ \begin{array} { l l l } { 3 } & { - 1 } & { 4 } \\ { 2 } & { 3 } & { 1 } \end{array} \right] \left[ \begin{array} { l l l } { 1 } & { 3 } & { 4 } \\ { 2 } & { 1 } & { 0 } \\ { - 3 } & { 2 } & { 3 } \end{array} \right]$

Now,

$\red{\rm :\longmapsto\:\left[ \begin{array} { l l l } { 3 } & { - 1 } & { 4 } \\ { 2 } & { 3 } & { 1 } \end{array} \right] _{2 \times 3} \left[ \begin{array} { l l l } { 1 } & { 3 } & { 4 } \\ { 2 } & { 1 } & { 0 } \\ { - 3 } & { 2 } & { 3 } \end{array} \right]_{3 \times 3}}$

Since, number of columns of pre multiplier is equal to number of rows of post multiplier.

So, it means matrix multiplication is defined.

So,

$\rm :\longmapsto\:\left[ \begin{array} { l l l } { 3 } & { - 1 } & { 4 } \\ { 2 } & { 3 } & { 1 } \end{array} \right] \left[ \begin{array} { l l l } { 1 } & { 3 } & { 4 } \\ { 2 } & { 1 } & { 0 } \\ { - 3 } & { 2 } & { 3 } \end{array} \right]$

$\rm \:  =  \: \begin{gathered}\sf\left[\begin{array}{ccc} 3 - 2 - 12&9 - 1 + 8& 12 + 0 + 12\\2 + 6 - 3& 6 + 3 + 2&8 + 0 + 3\end{array}\right]\end{gathered}$

$\rm \:  =  \: \begin{gathered}\sf\left[\begin{array}{ccc} - 11&16& 24\\5& 11&11\end{array}\right]\end{gathered}$

Hence,

$\boxed{\tt{ \left[ \begin{array} { l l l } { 3 } & { - 1 } & { 4 } \\ { 2 } & { 3 } & { 1 } \end{array} \right] \left[ \begin{array} { l l l } { 1 } & { 3 } & { 4 } \\ { 2 } & { 1 } & { 0 } \\ { - 3 } & { 2 } & { 3 } \end{array} \right] =\begin{gathered}\sf\left[\begin{array}{ccc} - 11&16& 24\\5& 11&11\end{array}\right]\end{gathered}}}$

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Additional Information :-

1. Matrix multiplication may or may not be Commutative

2. Matrix multiplication is Associative

3. Matrix multiplication is Distributive