Question 1

24 32 1 230 20) (² 3 9) - (60 10 0a12

Question 2

Solution

$a_(12)=element\text{ }in\text{ }the\text{ }first\text{ }row\text{ }and\text{ }second\text{ }column\text{ }after\text{ }multiplication$
$\begin{bmatrix}{24} & {32} \\ {60} & {10}\end{bmatrix}\begin{bmatrix}{1} & {2} & {0} \\ {0} & {3} & {1} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{} & {} & {} \\ {24*1+32*0} & {24*2+32*3} & {24*0+32*1} \\ {\placeholder{⬚}} & {\placeholder{⬚}} & {\placeholder{⬚}}\end{bmatrix}$

Along the first row , we have the elements

(24 144 32)

The element on the first row ; second column is 144

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