Question

The sum of three numbers is $9$. If we multiply third number by $3$ and add to the second number, we get $16$. By adding the first and the third number and then subtracting twice the second number from this sum, we get $6$. Use this information and find the system of linear equations. Hence, find the three numbers using matrices

Answer 1

Let the three numbers be $x,y$ and $z$. Then, according to the question,
$x+y+z=9$ ... (i)
$y+3z=16$ .... (ii)
$x-2y+z=6$ .... (iii)

Then, $AX=B$

Now, $|A|=\left|\begin{array}{ccc}1& 1& 1\\ 0& 1& 3\\ 1& -2& 1\end{array}\right|$
$=1(1+6)+1(3-0)+1(0-1)$
$=7+3-1=9\ne 0$

$\therefore A^{-1}$ exists.

$\therefore$ The system $AX=B$ has a unique solution.

For A, $A_{11}=\left|\begin{array}{cc}1& 3\\ -2& 1\end{array}\right|=(1+6)=7$

$A_{12}=-\left|\begin{array}{cc}0& 3\\ 1& 1\end{array}\right|=-(0-3)=3$

$A_{13}=\left|\begin{array}{cc}0& 1\\ 1& -2\end{array}\right|=(0-1)=-1$

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

$A_{22}=\left|\begin{array}{cc}1& 1\\ 1& 1\end{array}\right|=(1-1)=0$

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

$A_{31}=\left|\begin{array}{cc}1& 1\\ 1& 3\end{array}\right|=(3-1)=2$

$A_{32}=-\left|\begin{array}{cc}1& 1\\ 0& 3\end{array}\right|=-(3-0)=-3$

$A_{33}=\left|\begin{array}{cc}1& 1\\ 0& 1\end{array}\right|=(1-0)=1$

$\therefore adjA=\left[\begin{array}{ccc}7& 3& -1\\ -3& 0& 3\\ 2& -3& 1\end{array}\right]$

$=\left[\begin{array}{ccc}7& -3& 2\\ 3& 0& -3\\ -1& 3& 1\end{array}\right]$
$\therefore A^{-1}=\frac{adjA}{|A|}$

$=\frac{1}{9}\left[\begin{array}{ccc}7& -3& 2\\ 3& 0& -3\\ -1& 3& 1\end{array}\right]$

$=\left[\begin{array}{ccc}\frac{7}{9}& -\frac{1}{3}& \frac{2}{9}\\ \frac{1}{3}& 0& -\frac{1}{3}\\ -\frac{1}{9}& \frac{1}{3}& \frac{1}{9}\end{array}\right]$

$\therefore$ The unique solution is given by $X=A^{-1}B$

$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{ccc}\frac{7}{9}& -\frac{1}{3}& \frac{2}{9}\\ \frac{1}{3}& 0& -\frac{1}{3}\\ -\frac{1}{9}& \frac{1}{3}& \frac{1}{9}\end{array}\right]\left[\begin{array}{c}9\\ 16\\ 6\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{ccc}7& -\frac{16}{3}& +\frac{4}{3}\\ 3& +0& -2\\ -1& +\frac{16}{3}& +\frac{2}{3}\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}3\\ 1\\ 5\end{array}\right]$

$\Rightarrow x=3,y=1,z=5$

Hence, the required three numbers are $3,1$ and $5$.