Question

Two factories decided to award their employees for three values of (a) adaptable to new techniques, (b) careful and alert in difficult situations and (c) keeping calm in tense situations, at the rate Rs. $x$, Rs. $y$ and Rs. $z$ per person respectively. The first factory decided to honor respectively $2,4$ and $3$ employees with a tota prize money of Rs. $29000$. The second factory decided to honor respectively $5,2$ and$3$ employees with the prize money of Rs. $30500$. If the three per person together cost Rs. $9500$, then
(i) represent the above situation by a matrix equation and form linear equations using matrix multiplication
(ii) Solve these equations using matrices?

Answer 1

Let the numbers are $x,y,z$ be the prize amount per person for adaptability, carefulness and calmness respectively

As per the given data we get,

$2x+4y+3z=29000$

$5x+2y+3z=30500$

$x+y+z=9500$

These three equations can be written as

$\left[\begin{array}{ccc}2& 4& 3\\ 5& 2& 3\\ 1& 1& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}29000\\ 30500\\ 9500\end{array}\right]$

$AX=B$

$\left|A\right|=2(2–3)–4(5–3)+3(5–2)=2(–1)–4\left(2\right)+3\left(3\right)=-2-8+9=-1$

Hence, the unique solution given by $x=A^{–1}B$

$C_{11}={(-1)}^{1+1}\left|\begin{array}{cc}2& 3\\ 1& 1\end{array}\right|=2-3=-1$

$C_{12}={(-1)}^{1+2}\left|\begin{array}{cc}5& 3\\ 1& 1\end{array}\right|=-(5-3)=-2$

$C_{13}={(-1)}^{1+3}\left|\begin{array}{cc}5& 2\\ 1& 1\end{array}\right|=5-1=4$

$C_{21}={(-1)}^{2+1}\left|\begin{array}{cc}4& 3\\ 1& 1\end{array}\right|=-(4-3)=-1$

$C_{22}={(-1)}^{2+2}\left|\begin{array}{cc}2& 3\\ 1& 1\end{array}\right|=2-3=-1$

$C_{23}={(-1)}^{2+3}\left|\begin{array}{cc}2& 4\\ 1& 1\end{array}\right|=-(2-4)=2$

$C_{31}={(-1)}^{3+1}\left|\begin{array}{cc}4& 3\\ 2& 3\end{array}\right|=12-6=6$

$C_{32}={(-1)}^{3+2}\left|\begin{array}{cc}2& 3\\ 5& 3\end{array}\right|=-(6-15)=9$

$C_{33}={(-1)}^{3+3}\left|\begin{array}{cc}2& 4\\ 5& 2\end{array}\right|=4-20=-16$

$AdjA={\left[\begin{array}{ccc}-1& -2& 3\\ -1& -1& 2\\ 6& 9& -16\end{array}\right]}^T=\left[\begin{array}{ccc}-1& -1& 6\\ -2& -1& 9\\ 3& 2& -16\end{array}\right]$

$X=A^{-1}B=\frac{1}{\left|A\right|}\left(AdjA\right)B$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{-1}\left[\begin{array}{ccc}-1& -1& 6\\ -2& -1& 9\\ 3& 2& -16\end{array}\right]\left[\begin{array}{c}29000\\ 30500\\ 9500\end{array}\right]$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{ccc}1& 1& -6\\ 2& 1& -9\\ -3& -2& 16\end{array}\right]\left[\begin{array}{c}29000\\ 30500\\ 9500\end{array}\right]$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}29000+30500-57000\\ 58000+30500-85500\\ -87000-61000+152000\end{array}\right]$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}2500\\ 3000\\ 4000\end{array}\right]$

Hence $x=2500,y=3000,$ and $z=4000$