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(a)

Assume the number of servings of Cheerios as \({x_1}\), and the number of servings of \(100\% \) Natural Cereal as \({x_2}\). Then,

\(\left( {\begin{array}{*{20}{c}}{110}\\4\\{20}\\2\end{array}} \right){x_1} + \left( {\begin{array}{*{20}{c}}{130}\\3\\{18}\\5\end{array}} \right){x_2} = \left( {\begin{array}{*{20}{c}}{295}\\9\\{48}\\8\end{array}} \right)\).

This is the required vector equation.

(b)

The equivalent matrix equation for part (a) can be written as shown below.

\(\left( {\begin{array}{*{20}{c}}{110}&{130}\\4&3\\{20}&{18}\\2&5\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{295}\\9\\{48}\\8\end{array}} \right)\)

Its augmented matrix is \(\left( {\begin{array}{*{20}{c}}{110}&{130}&{295}\\4&3&9\\{20}&{18}&{48}\\2&5&8\end{array}} \right)\).

Reduce the augmented matrix into the row echelon form.

Interchange rows one and four, i.e., \({R_1} \leftrightarrow {R_4}\).

\(\left( {\begin{array}{*{20}{c}}{110}&{130}&{295}\\4&3&9\\{20}&{18}&{48}\\2&5&8\end{array}} \right) \sim \left( {\begin{array}{*{20}{c}}2&5&8\\4&3&9\\{20}&{18}&{48}\\{110}&{130}&{295}\end{array}} \right)\)

Divide row one by row two.

\( \sim \left( {\begin{array}{*{20}{c}}1&{2.5}&4\\4&3&9\\{20}&{18}&{48}\\{110}&{130}&{295}\end{array}} \right)\)

At row two, multiply row one by 4 and subtract row two from it, i.e., \({R_2} \to 4{R_1} - {R_2}\). At row three, multiply row one by 20 and subtract row three from it, i.e., \({R_3} \to 20{R_1} - {R_3}\).

And at row four, multiply row one by 110 and subtract row four from it, i.e., \({R_4} \to 110{R_1} - {R_4}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&{2.5}&4\\0&7&7\\0&{32}&{32}\\0&{145}&{145}\end{array}} \right)\)

Divide row two by 7.

\( \sim \left( {\begin{array}{*{20}{c}}1&{2.5}&4\\0&1&1\\0&{32}&{32}\\0&{145}&{145}\end{array}} \right)\)

At row three, multiply row two by 32 and subtract row three from it, i.e., \({R_3} \to 32{R_2} - {R_3}\). And at row four, multiply row two by 145 and subtract row four from it, i.e., \({R_4} \to 145{R_2} - {R_4}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&{2.5}&4\\0&1&1\\0&0&0\\0&0&0\end{array}} \right)\)

At row one, multiply row two by 2.5 and subtract it from row one, i.e., \({R_1} \to {R_1} - 2.5{R_2}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&0&{1.5}\\0&1&1\\0&0&0\\0&0&0\end{array}} \right)\)

This implies that the desired nutrients are given as 1.5 servings of Cheerios with one serving of 100% Natural Cereal.