Question

If $A=\left[\begin{array}{ccc}3& 2& 1\\ 4& -1& 2\\ 7& 3& -3\end{array}\right],$ then find $A^{-1}$ and hence solve the following system of equations: 3x + 4y + 7z = 14, 2x - y + 3z = 4, x + 2y - 3z = 0.

Answer 1

Here $A=\left[\begin{array}{ccc}3& 2& 1\\ 4& -1& 2\\ 7& 3& -3\end{array}\right]\Rightarrow |A|=\left|\begin{array}{ccc}3& 2& 1\\ 4& -1& 2\\ 7& 3& -3\end{array}\right|=3(3-6)-\left(2\right)(-12-14)+1(12+7)=62\ne 0$
Therefore $A^{-1}$ exists.
Consider $C_{ij}$ be the cofactor of $a_{ij}$ for matrix A.
$C_{11}=-3,C_{12}=26,C_{13}=19;C_{21}=9,C_{22}=-16,C_{23}=5;C_{31}=5,C_{32}=-2,C_{33}=-11.$
So, adj. $A=\left[\begin{array}{ccc}-3& 9& 5\\ 26& -16& -2\\ 19& 5& -11\end{array}\right]\therefore A^{-1}=\frac{1}{62}\left[\begin{array}{ccc}-3& 9& 5\\ 26& -16& -2\\ 19& 5& -11\end{array}\right]...\left(i\right)$
Now 3x + 4y + 7z = 14, 2x - y + 3z = 4, x + 2y - 3z = 0
Let $P=\left[\begin{array}{ccc}3& 4& 7\\ 2& -1& 3\\ 1& 2& -3\end{array}\right]=A^T,X=\left(\begin{array}{c}x\\ y\\ z\end{array}\right),B=\left(\begin{array}{c}14\\ 4\\ 0\end{array}\right).$
As $PX=Bi.e.,X=(A^T{)}^{-1}B=(A^{-1}{)}^{T}B$
So by (i), $X=\frac{1}{62}\left[\begin{array}{ccc}-3& 26& 19\\ 9& -16& 5\\ 5& -2& -11\end{array}\right]\left(\begin{array}{c}14\\ 4\\ 0\end{array}\right)=\frac{1}{62}\left(\begin{array}{c}62\\ 62\\ 62\end{array}\right)\Rightarrow \left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)\therefore x=1,y=1,z=1.$