Question

If three numbers are added, their sum is ${}^{\prime }{2}^{\prime }$. If $2$ times the second number is subtracted from the sum of first and third number we get '$8$' and if three times the first number is added to the sum of second and third number we get ${}^{\prime }{4}^{\prime }$. Find the numbers using matrices.

Answer 1

Let three numbers be $x,y,z$.

From given conditions, we have

$x+y+z=2$ $.............\left(1\right)$

$x+z-2y=8$

$x-2y+z=8$ $...........\left(2\right)$

And

$3x+y+z=4$ $.........\left(3\right)$

Given all equation can be written in matrix form as,

$\left[\begin{array}{ccc}1& 1& 1\\ 1& -2& 1\\ 3& 1& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}2\\ 8\\ 4\end{array}\right]$

Consider, $AX=B$

On multiplying $A^{-1}$ both sides, we get

$A^{-1}.A.X=A^{-1}.B$

$X=A^{-1}.B$ $.............\left(4\right)$

Now,

$A^{-1}=\frac{1}{|A|}Adj\left(A\right)$

$|A|=\left|\begin{array}{ccc}1& 1& 1\\ 1& -2& 1\\ 3& 1& 1\end{array}\right|$

$|A|=6$

Now, for $Adj\left(A\right)$, we need minors and co-factors.

$M_{11}=-3,M_{12}=-2,M_{13}=7$

$M_{21}=0,M_{22}=-2,M_{23}=-2$

$M_{31}=3,M_{32}=0,M_{33}=-3$

Therefore,

$A^{-1}=\frac{1}{6}\left[\begin{array}{ccc}-3& 0& 3\\ 2& -2& 0\\ 7& 2& -3\end{array}\right]$

From equation (4),

$X=\frac{1}{6}\left[\begin{array}{ccc}-3& 0& 3\\ 2& -2& 0\\ 7& 2& -3\end{array}\right]\left[\begin{array}{c}2\\ 8\\ 4\end{array}\right]$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ -2\\ 3\end{array}\right]$

Hence, $x=1,y=-2,z=3$.