One serving of Post Shredded Wheat supplies 160 calories, 5 g of protein, 6 g of fiber, and 1 g of fat. One serving of Crispix supplies 110 calories, 2 g of protein, .1 g of fiber, and .4 g of f\(\left( {\begin{array}{*{20}{c}}{160}&{110}\\5&2\\6&{.1}\\1&{.4}\end{array}} \right)\)at.

1. Set up a matrix B and a vector u such that Bu gives the amounts of calories, protein, fiber, and fat contained in a mixture of three servings of Shredded Wheat and two servings of Crispix.
2. (M) Suppose that you want a cereal with more fiber than Crispix but fewer calories than Shredded Wheat. Is it possible for a mixture of the two cereals to supply 130 calories, 3.20 g of protein, 2.46 g of fiber, and .64 g of fat? If so, what is the mixture?

Let S and C be the matrices for one serving of Post Shredded Wheat, and one serving of Crispix, respectively. Then,

\(S = \left( {\begin{array}{*{20}{c}}{160}\\5\\6\\1\end{array}} \right)\) and \(C = \left( {\begin{array}{*{20}{c}}{110}\\2\\{.1}\\{.4}\end{array}} \right)\).

(a)

The matrix for 3S+2C is obtained as shown below.

\(\begin{array}{c}3S + 2C = 3\left( {\begin{array}{*{20}{c}}{160}\\5\\6\\1\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}{110}\\2\\{.1}\\{.4}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{160}&{110}\\5&2\\6&{.1}\\1&{.4}\end{array}} \right)\left( {\begin{array}{*{20}{c}}3\\2\end{array}} \right)\end{array}\)

Hence, the required matrix B is \(\left( {\begin{array}{*{20}{c}}{160}&{110}\\5&2\\6&{.1}\\1&{.4}\end{array}} \right)\), and vector u is \(\left( {\begin{array}{*{20}{c}}3\\2\end{array}} \right)\).

(b)

Let\({x_1}\)and\({x_2}\)be the number of servings of Post Shredded Wheat and Crispix, respectively. Then, the system is

\(\left( {\begin{array}{*{20}{c}}{160}&{110}\\5&2\\6&{.1}\\1&{.4}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{130}\\{3.20}\\{2.46}\\{.64}\end{array}} \right)\).

Its augmented matrix is \(\left( {\begin{array}{*{20}{c}}{160}&{110}&{130}\\5&2&{3.20}\\6&{.1}&{2.46}\\1&{.4}&{.64}\end{array}} \right)\).\(\)

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

\(\left( {\begin{array}{*{20}{c}}{160}&{110}&{130}\\5&2&{3.20}\\6&{.1}&{2.46}\\1&{.4}&{.64}\end{array}} \right) \sim \left( {\begin{array}{*{20}{c}}1&{.4}&{.64}\\5&2&{3.20}\\6&{.1}&{2.46}\\{160}&{110}&{130}\end{array}} \right)\)

At row two, multiply row one by 5 and subtract row two from it, i.e.,\({R_2} \to 5{R_1} - {R_2}\). At row three, multiply row one by 6 and subtract row three from it, i.e.,\({R_3} \to 6{R_1} - {R_3}\). And at row four, multiply row one by 160 and subtract row four from it, i.e.,\({R_4} \to 160{R_1} - {R_4}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&{.4}&{.64}\\0&0&0\\0&{2.3}&{1.38}\\0&{ - 46}&{101.1}\end{array}} \right)\)

Interchange rows two and four, i.e.,\({R_2} \leftrightarrow {R_4}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&{.4}&{.64}\\0&{ - 46}&{101.1}\\0&{2.3}&{1.38}\\0&0&0\end{array}} \right)\)

At row three, multiply row three by 20 and add it to row two, i.e.,\({R_3} \to 20{R_3} + {R_2}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&{.4}&{.64}\\0&{ - 46}&{101.1}\\0&0&{128.7}\\0&0&0\end{array}} \right)\)

Note that\(0 \ne 128.7\). Hence, the given system is inconsistent. This implies that the system has no solution.

Therefore, it is not possible for a mixture of the two cereals to supply 130 calories, 3.20 g of protein, 2.46 g of fiber, and .64 g of fat.