Question

The sum of three numbers is 6. Thrice the third number when added to the first number gives 7. On adding the sum of second and third number to three times the first number, we get 12. Find the three numbers using determinants.

Answer 1

Given let $a,b,c$ be the three numbers

Then according to question

$a+b+c=6----\left(1\right)$

$a+3c=7-----\left(2\right)$

$3a+b+c=12-------\left(3\right)$

Using determinant property for finding value of $a,b,c$ we get.

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

$\left[\begin{array}{ccc}1& 1& 1\\ 1& 0& 3\\ 3& 1& 1\end{array}\right]\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\left[\begin{array}{c}6\\ 7\\ 12\end{array}\right]$ where $A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 0& 3\\ 3& 1& 1\end{array}\right]$

Now $A^{-1}=\frac{adjA}{|A|}$ we know

$|A|=1(0-3)-1(1-9)+1(1-0)$

$|A|=-3+8+1$

$|A|=6$

And $adjA=\left[\begin{array}{ccc}-3& +8& 1\\ 0& -2& +2\\ 3& -2& -1\end{array}\right]$

$adjA=\left[\begin{array}{ccc}-3& 0& 3\\ 8& -2& -2\\ 1& 2& -1\end{array}\right]\Rightarrow A^{-1}=\frac{1}{6}\left[\begin{array}{ccc}-3& 0& 3\\ 8& -2& -2\\ 1& 2& -1\end{array}\right]$

Now

$\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\frac{1}{6}\left[\begin{array}{ccc}-3& 0& 3\\ 8& -2& -2\\ 1& 2& -1\end{array}\right]\left[\begin{array}{c}6\\ 7\\ 12\end{array}\right]=\frac{1}{6}\left[\begin{array}{ccc}-18& +3& 6\\ 48& -14& -24\\ 6& +14& -12\end{array}\right]$

$\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\frac{1}{6}\left[\begin{array}{c}18\\ 10\\ 8\end{array}\right]$

On comparing we get

$[a=3,b=\frac{5}{3}\&c=\frac{4}{3}]$