Chapter 8 - Section 8.1 - Matrix Solutions to Linear Systems - Exercise Set - Page 894: 48

$15$ desks of first type, $10$ desks of second type and $20$ desks of third type should be produced each week.

Let $x$ be the number of desks of the first type, $y$ be the number of desks of the second type and $z$ be the number of desks of the third type. It is provided that the maximum hours for cutting three types of desks should be 100. This implies, $2x+3y+2z=100$ It is provided that the maximum hours for constructing the three types of desks should be 100. This implies, $2x+y+3z=100$ It is provided that the maximum hours for finishing the three types of desks should be 65. This implies, $x+y+2z=65$ Therefore, the system of equations is: $\begin{align} & 2x+3y+2z=100 \\ & 2x+y+3z=100 \\ & x+y+2z=65 \end{align}$ First write the augmented matrix for the given system of equations: The augmented matrix obtained from equations is, $\left[ \begin{matrix} 2 & 3 & 2 & 100 \\ 2 & 1 & 3 & 100 \\ 1 & 1 & 2 & 65 \\ \end{matrix} \right]$ Now, use row operation to reduce the matrix to row echelon form. ${{R}_{1}}\to \frac{1}{2}{{R}_{1}}$, we get $\left[ \begin{matrix} 1 & \frac{3}{2} & 1 & 50 \\ 2 & 1 & 3 & 100 \\ 1 & 1 & 2 & 65 \\ \end{matrix} \right]$ ${{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}$, we get $\left[ \begin{matrix} 1 & \frac{3}{2} & 1 & 50 \\ 0 & -2 & 1 & 0 \\ 1 & 1 & 2 & 65 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}-{{R}_{1}}$, we get $\left[ \begin{matrix} 1 & \frac{3}{2} & 1 & 50 \\ 0 & -2 & 1 & 0 \\ 0 & \frac{-1}{2} & 1 & 15 \\ \end{matrix} \right]$ ${{R}_{2}}\to \frac{-1}{2}{{R}_{2}}$, we get $\left[ \begin{matrix} 1 & \frac{3}{2} & 1 & 50 \\ 0 & 1 & \frac{-1}{2} & 0 \\ 0 & \frac{-1}{2} & 1 & 15 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}+\frac{1}{2}{{R}_{2}}$, we get $\left[ \begin{matrix} 1 & \frac{3}{2} & 1 & 50 \\ 0 & 1 & \frac{-1}{2} & 0 \\ 0 & 0 & \frac{3}{4} & 15 \\ \end{matrix} \right]$ ${{R}_{3}}\to \frac{4}{3}{{R}_{3}}$, we get $\left[ \begin{matrix} 1 & \frac{3}{2} & 1 & 50 \\ 0 & 1 & \frac{-1}{2} & 0 \\ 0 & 0 & 1 & 20 \\ \end{matrix} \right]$ ${{R}_{2}}\to {{R}_{2}}+\frac{1}{2}{{R}_{3}}$, we get $\left[ \begin{matrix} 1 & \frac{3}{2} & 1 & 50 \\ 0 & 1 & 0 & 10 \\ 0 & 0 & 1 & 20 \\ \end{matrix} \right]$ ${{R}_{1}}\to {{R}_{1}}-{{R}_{3}}$, we get $\left[ \begin{matrix} 1 & \frac{3}{2} & 0 & 50 \\ 0 & 1 & 0 & 10 \\ 0 & 0 & 1 & 20 \\ \end{matrix} \right]$ ${{R}_{1}}\to {{R}_{1}}-\frac{3}{92}{{R}_{1}}$, we get $\left[ \begin{matrix} 1 & 0 & 0 & 15 \\ 0 & 1 & 0 & 10 \\ 0 & 0 & 1 & 20 \\ \end{matrix} \right]$ Thus, $x=15,y=10,z=20$ Hence, $15$ desks of first type, $10$ desks of second type and $20$ desks of third type should be produced each week.