Question

Matrix inversion method ​

Answer 1

Answer:

The solution is

x=-1

y=1

z=2

Step-by-step explanation:

The given system of equations can be written as

$\left(\begin{array}{ccc}1&3&2\\3 &-2&5\\2&-3&6\end{array}\right)\left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}6\\5\\7\end{array}\right)$

This is of the form AX=B

$|A|=\left|\begin{array}{ccc}1&3&2\\3 &-2&5\\2&-3&6\end{array}\right|$

Expanding along first row

$|A|=1(-12+15)-3(18-10)+2(-9+4)$

$\implies\,|A|=1(3)-3(8)+2(-5)$

$\implies\,|A|=3-24-10$

$\implies\,|A|=-31\neq\,0$

$\therefore\,\bf\,A^{-1}\:exists$

Cofactor matrix of A

$[a_{ij}]=\left(\begin{array}{ccc}3&-8&-5\\-24&2&9\\19&1&-11\end{array}\right)$

$adj(A)=[a_{ij}]^T$

$\implies\,adj(A)=\left(\begin{array}{ccc}3&-24&19\\-8&2&1\\-5&9&-11\end{array}\right)$

$A^{-1}=\frac{1}{|A|}(adjA)$

$\implies\bf\,A^{-1}=\frac{-1}{31}\left(\begin{array}{ccc}3&-24&19\\-8&2&1\\-5&9&-11\end{array}\right)$

Now,

$X=A^{-1}\,B$

$X=\frac{-1}{31}\left(\begin{array}{ccc}3&-24&19\\-8&2&1\\-5&9&-11\end{array}\right)\left(\begin{array}{c}6\\5\\7\end{array}\right)$

$X=\frac{-1}{31}\left(\begin{array}{c}18-120+133\\-48+10+7\\-30+45-77\end{array}\right)\left$

$X=\frac{-1}{31}\left(\begin{array}{c}31\\-31\\-62\end{array}\right)\left$

$X=\left(\begin{array}{c}-1\\1\\2\end{array}\right)\left$

$\implies\,\left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}-1\\1\\2\end{array}\right)\left$

$\therefore$The solution is

x=-1, y=1 and z=2