Question

26%20%7B%203%20%7D%20%5C%5C%20%7B%205%20%7D%20%26%20%7B%204%20%7D%20%5Cend%7Barray%7D%20%5Cend%7Bbmatrix%7D%20%5Cbegin%7Bbmatrix%7D%20%5Cbegin%7Barray%7D%20%7B%20l%20l%20l%20%7D%20%7B%202%20%7D%20%26%20%7B%200%20%7D%20%26%20%7B%203%20%7D%20%5C%5C%20%7B%20-%201%20%7D%20%26%20%7B%201%20%7D%20%26%20%7B%205%20%7D%20%5Cend%7Barray%7D%20%5Cend%7Bbmatrix%7D" id="TexFormula1" title=" \large\begin{bmatrix} \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \end{bmatrix} \begin{bmatrix} \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{array} \end{bmatrix}" alt=" \large\begin{bmatrix} \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \end{bmatrix} \begin{bmatrix} \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{array} \end{bmatrix}" align="absmiddle" class="latex-formula">
Solve using matrix multiplication rule. Please help! No spam.​

Answer 1

$\huge \boxed{\mathbb{QUESTION} \downarrow}$

$\begin{bmatrix} \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \end{bmatrix} \begin{bmatrix} \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{array} \end{bmatrix}$

$\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}$

$\begin{bmatrix} \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \end{bmatrix} \begin{bmatrix} \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{array} \end{bmatrix}$

In matrix multiplication, the number of columns in the 1st matrix is equal to the number of rows in the 2nd matrix.

$\left(\begin{matrix}2&3\\5&4\end{matrix}\right)\left(\begin{matrix}2&0&3\\-1&1&5\end{matrix}\right)$

Multiply each element of the 1st row of the 1st matrix by the corresponding element of the 1st column of the 2nd matrix. Then add these products to obtain the element in the 1st row, 1st column of the product matrix.

$\left(\begin{matrix}2\times 2+3\left(-1\right)&&\\&&\end{matrix}\right)$

The remaining elements of the product matrix are found in the same way.

$\left(\begin{matrix}2\times 2+3\left(-1\right)&3&2\times 3+3\times 5\\5\times 2+4\left(-1\right)&4&5\times 3+4\times 5\end{matrix}\right)$

Simplify each element by multiplying the individual terms.

$\left(\begin{matrix}4-3&3&6+15\\10-4&4&15+20\end{matrix}\right)$

Now, sum each element of the matrix.

$\large\boxed{\boxed{\left(\begin{matrix}1&3&21\\6&4&35\end{matrix}\right) }}$