Question 14 Marks
In an engineering workshop there are 10 machines for drilling, 8 machines for turning and 7 machines for grinding. Three types of brackets are made. Type I brackets require 0 minutes for drilling, 5 minutes for turning and 4 minutes for grinding. The corresponding times for type II and III brackets are 3, 3, 2 and 3, 2, 2, minutes respectively. How many brackets of each type should be produced per hour so that all the machines remain fully occupied during an hour? Solve by using matrix method.
Answer
View full question & answer→The information provided can be summarized in the tabular form as follows:
Let x, y and z denote the number of brackets produced of each type. Then, the time taken by drilling machine to produce x, y and z brackets of type I, II and III respectively, is 0x + 3y + 3z minutes. But the time for which the drilling machine is available is 600 minutes.
$
\therefore 0 x+3 y+3 z=600
$
Similarly, for turning and grinding machines, we obtain
$
5 x+3 y+2 z=480 \text { and } 4 x+2 y+2 z=420
$
Thus, we obtain the following system of linear equations
$
\begin{array}{l}
0 x+3 y+3 z=600 \\
5 x+3 y+2 z=480 \\
4 x+2 y+2 z+420
\end{array}
$
In matrix form, the above system can be written as
$
\left[\begin{array}{lll}
0 & 3 & 3 \\
5 & 3 & 2 \\
4 & 2 & 2
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
600 \\
480 \\
420
\end{array}\right]
$
or, $AX = B$ where $A =\left[\begin{array}{lll}0 & 3 & 3 \\ 5 & 3 & 2 \\ 4 & 2 & 2\end{array}\right], X =\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B =\left[\begin{array}{l}600 \\ 480 \\ 420\end{array}\right]$
Now, $| A |=\left|\begin{array}{lll}0 & 3 & 3 \\ 5 & 3 & 2 \\ 4 & 2 & 2\end{array}\right|=0(6-4)-3(10-8)+3(10-12)=-12 \neq 0$
So, A is invertible.
Let $C _{ ij }$ be cofactor of $a _{ ij }$ in $A =\left[ a _{ ij }\right]$. Then,
$
C_{11}=2, C_{12}=-2, C_{13}=-2, C_{21}=0, C_{22}=-12, C_{23}=12, C_{31}=-3, C_{32}=15, C_{33}=-15
$
$
\therefore \operatorname{adj} A=\left[\begin{array}{ccc}
2 & -2 & -2 \\
0 & -12 & 12 \\
-3 & 15 & -15
\end{array}\right]^T=\left[\begin{array}{ccc}
2 & 0 & -3 \\
-2 & -12 & 15 \\
-2 & 12 & -15
\end{array}\right]
$
Thus, $A ^{-1}=\frac{1}{|A|}$ adj $A =-\frac{1}{12}\left[\begin{array}{ccc}2 & 0 & -3 \\ -2 & -12 & 15 \\ -2 & 12 & -15\end{array}\right]$
Now, $AX = B$
$
\begin{array}{l}
\Rightarrow X=A^{-1} B \\
\Rightarrow\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{12}\left[\begin{array}{rrr}
2 & 0 & -3 \\
-2 & -12 & 15 \\
-2 & 12 & -15
\end{array}\right]\left[\begin{array}{l}
600 \\
480 \\
420
\end{array}\right]=-\frac{1}{12}\left[\begin{array}{rrr}
1200 & +0 & -1260 \\
-1200 & -5760 & +6300 \\
-1200 & +5760 & -6300
\end{array}\right]=-\frac{1}{12}\left[\begin{array}{c}
-60 \\
-660 \\
-1740
\end{array}\right]=\left[\begin{array}{c}
5 \\
55 \\
145
\end{array}\right] \\
\Rightarrow x=5, y=55 \text { and } z=145
\end{array}
$
Hence, 5 brackets of Type-I, 55 brackets of Type-II and 145 brackets of Type-III should be produced to keep all machines fully occupied.
| Brackets | Drilling | Turning | Grinding |
| Type I | 0 | 5 | 4 |
| Type II | 3 | 3 | 2 |
| Type III | 3 | 2 | 2 |
| Time available (in minutes) | 10 $\times$ 60 = 600 | 8 $\times$ 60 = 480 | 7 $\times$ 60 = 420 |
$
\therefore 0 x+3 y+3 z=600
$
Similarly, for turning and grinding machines, we obtain
$
5 x+3 y+2 z=480 \text { and } 4 x+2 y+2 z=420
$
Thus, we obtain the following system of linear equations
$
\begin{array}{l}
0 x+3 y+3 z=600 \\
5 x+3 y+2 z=480 \\
4 x+2 y+2 z+420
\end{array}
$
In matrix form, the above system can be written as
$
\left[\begin{array}{lll}
0 & 3 & 3 \\
5 & 3 & 2 \\
4 & 2 & 2
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
600 \\
480 \\
420
\end{array}\right]
$
or, $AX = B$ where $A =\left[\begin{array}{lll}0 & 3 & 3 \\ 5 & 3 & 2 \\ 4 & 2 & 2\end{array}\right], X =\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B =\left[\begin{array}{l}600 \\ 480 \\ 420\end{array}\right]$
Now, $| A |=\left|\begin{array}{lll}0 & 3 & 3 \\ 5 & 3 & 2 \\ 4 & 2 & 2\end{array}\right|=0(6-4)-3(10-8)+3(10-12)=-12 \neq 0$
So, A is invertible.
Let $C _{ ij }$ be cofactor of $a _{ ij }$ in $A =\left[ a _{ ij }\right]$. Then,
$
C_{11}=2, C_{12}=-2, C_{13}=-2, C_{21}=0, C_{22}=-12, C_{23}=12, C_{31}=-3, C_{32}=15, C_{33}=-15
$
$
\therefore \operatorname{adj} A=\left[\begin{array}{ccc}
2 & -2 & -2 \\
0 & -12 & 12 \\
-3 & 15 & -15
\end{array}\right]^T=\left[\begin{array}{ccc}
2 & 0 & -3 \\
-2 & -12 & 15 \\
-2 & 12 & -15
\end{array}\right]
$
Thus, $A ^{-1}=\frac{1}{|A|}$ adj $A =-\frac{1}{12}\left[\begin{array}{ccc}2 & 0 & -3 \\ -2 & -12 & 15 \\ -2 & 12 & -15\end{array}\right]$
Now, $AX = B$
$
\begin{array}{l}
\Rightarrow X=A^{-1} B \\
\Rightarrow\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{12}\left[\begin{array}{rrr}
2 & 0 & -3 \\
-2 & -12 & 15 \\
-2 & 12 & -15
\end{array}\right]\left[\begin{array}{l}
600 \\
480 \\
420
\end{array}\right]=-\frac{1}{12}\left[\begin{array}{rrr}
1200 & +0 & -1260 \\
-1200 & -5760 & +6300 \\
-1200 & +5760 & -6300
\end{array}\right]=-\frac{1}{12}\left[\begin{array}{c}
-60 \\
-660 \\
-1740
\end{array}\right]=\left[\begin{array}{c}
5 \\
55 \\
145
\end{array}\right] \\
\Rightarrow x=5, y=55 \text { and } z=145
\end{array}
$
Hence, 5 brackets of Type-I, 55 brackets of Type-II and 145 brackets of Type-III should be produced to keep all machines fully occupied.
