Question

The sum of three numbers is $6$ when second number is subtracted from thrice the sum of first and third number, we get numbers $10$. Four times the third number is subtracted from five times the sum of first and second number, the result is $3$. Using above information, find these three numbers by matrix method.

Answer 1

Let the numbers be $x,y$ and $z$

According to question, $x+y+z=6$ ...... $\left(i\right)$

$3(x+z)-y=10$

or $3x-y+3z=10$ ........ $\left(ii\right)$

Also, $5(x+y)-4z=3$

or $5x+5y-4z=3$ ........ $\left(iii\right)$

Thus, $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\left[\begin{array}{ccc}1& 1& 1\\ 3& -1& 3\\ 5& 5& -4\end{array}\right]=\left[\begin{array}{c}6\\ 10\\ 3\end{array}\right]$

Applying $R_3\to R_3-5R_1$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\left[\begin{array}{ccc}1& 1& 1\\ 3& -1& 3\\ 0& 0& -9\end{array}\right]=\left[\begin{array}{c}6\\ 10\\ -27\end{array}\right]$

Thus, the equations reduce to

$x+y+z=6$ ..... $\left(iv\right)$

$3x-y+3z=10$ ..... $\left(v\right)$

$-9z=-27\Rightarrow z=3$ .... $\left(vi\right)$

Adding $\left(iv\right)$ and $\left(v\right)$, we get

$4x+4z=16\Rightarrow x+z=4$ ..... $\left(vii\right)$

Substitute value of $z$ from $\left(vi\right)$ in $\left(vii\right)$,

$x+3=4\Rightarrow x=1$

Substitute the values of $x$ and $z$ in $\left(iv\right)$, we get

$1+y+3=6\Rightarrow y=2$

Hence, the numbers are $1,2$ and $3$.