A steam plant burns two types of coal: anthracite (A) and bituminous (B). For each ton of A burned, the plant produces 27.6 million of Btu of heat, 3100 grams (g) of sulphur dioxide, and 250g of particulate matter (solid-particle pollutants). For each ton of B burned, the plant produces 30.2 million Btu, 6400g of sulphur dioxide, and 360g of particulate matter.

1. How much heat does the steam plant produce when it burns \({x_1}\) tons of \(A\) and \({x_2}\) tons of \(B\).
2. Suppose a vector that lists the amounts of heat, sulphur dioxide, and particulate matter describes the output of the steam plant. Express this output as a linear combination of two vectors, assuming that the plant burns \({x_1}\) tons of \(A\) and \({x_2}\) tons of \(B\).
3. [M] Over a certain time period, the steam plant produced 162 million Btu of heat, 23,610 g of sulphur dioxide, and 1623 g of particulate matter. Determine how many tons of each type of coal the steam plant must have burned. Include a vector equation as part of your solution.

a.

The amount of heat generated when the steam plant burns \({x_1}\) tons of anthracite and \({x_2}\) tons of bituminous coal is \(27.6{x_1} + 30.2{x_2}\) million Btu.

b.

A vector equation\({{\mathop{\rm x}\nolimits} _1}{a_1} + {x_2}{a_2} + ... + {x_n}{a_n} = b\)has the same solution set. The vector\({\mathop{\rm y}\nolimits} \)defined by\(y = {c_1}{v_1} + .... + {c_p}{v_p}\)is called alinear combination of\({{\mathop{\rm v}\nolimits} _1},{v_2},...,{v_p}\)with weights\({c_1},{c_2},...,{c_p}\).

Assume that the steam plant burns\({x_1}\)tons of anthracite and \({x_2}\) tons of bituminous coal.

The total output generated by\({x_1}\)tons of anthracite and \({x_2}\) tons of bituminous coal is represented by the linear combination as:

\({x_1}\left[ {\begin{array}{*{20}{c}}{27.6}\\{3100}\\{250}\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}{30.2}\\{6400}\\{360}\end{array}} \right]\)

c.

It is given that the steam plant produced 162 million Btu of heat, 23,610 g of sulphur dioxide, and 1623 g of particulate matter.

The linear combination of the heat of steam plant produced is:

\({x_1}\left[ {\begin{array}{*{20}{c}}{27.6}\\{3100}\\{250}\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}{30.2}\\{6400}\\{360}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{162}\\{23610}\\{1623}\end{array}} \right]\)

The augmented matrix for the obtained equation is represented as:

\(\left[ {\begin{array}{*{20}{c}}{27.6}&{30.2}&{162}\\{3100}&{6400}&{23610}\\{250}&{360}&{1623}\end{array}} \right]\)

Perform an elementary row operation to produce the first augmented matrix.

Multiply row 1 by \(\frac{1}{{27.6}}\).

\(\left[ {\begin{array}{*{20}{c}}1&{1.0942}&{5.869}\\{3100}&{6400}&{23610}\\{250}&{360}&{1623}\end{array}} \right]\)

Perform an elementary row operationto produce the second augmented matrix.

Apply the sum of \( - 3100\) times row 1 and row 2 at row 2, and the sum of \( - 250\) times row 1 and row 3 at row 3.

\(\left[ {\begin{array}{*{20}{c}}1&{1.0942}&{5.869}\\0&{3007.98}&{5416.1}\\0&{86.449}&{155.60}\end{array}} \right]\)

Multiply row 2 by \(\frac{1}{{3007.98}}\).

\(\left[ {\begin{array}{*{20}{c}}1&{1.0942}&{5.869}\\0&1&{1.8}\\0&{86.449}&{155.60}\end{array}} \right]\)

Perform an elementaryrow operation to produce the third augmented matrix.

Apply the sum of \( - 1.0942\) times row 2 and row 1 at row 1.

\(\left[ {\begin{array}{*{20}{c}}1&0&{3.9}\\0&1&{1.8}\\0&{86.449}&{155.60}\end{array}} \right]\)

Apply the sum of \(86.449\) times row 2 and row 3 at row 3.

\(\left[ {\begin{array}{*{20}{c}}1&0&{3.9}\\0&1&{1.8}\\0&0&0\end{array}} \right]\)

If vector\({{\mathop{\rm v}\nolimits} _1},{v_2},...,{v_p}\)in\({\mathbb{R}^n}\)are given with scalars\({c_1},{c_2},...,{c_p}\), then the vector\({\mathop{\rm y}\nolimits} \)defined by\(y = {c_1}{v_1} + .... + {c_p}{v_p}\)is called alinear combination of\({{\mathop{\rm v}\nolimits} _1},{v_2},...,{v_p}\)with weights\({c_1},{c_2},...,{c_p}\).

To obtain the solution of the vector equations, it is required to convert the augmented matrix into the system of equations.

Write the obtained matrix \(\left[ {\begin{array}{*{20}{c}}1&0&{3.9}\\0&1&{1.8}\\0&0&0\end{array}} \right]\)into the equation notation.

\(\begin{array}{l}{x_1} = 3.9\\{x_2} = 1.8\end{array}\)

Thus, the steam plant consumes 3.9 tonnes of anthracite coal and 1.8 tonnes of bituminous coal.