Question

Solve the following system of linear equations using Inverse Matrix Method.
x + 6y – z = 10
2x + 3y + 3z = 17
3x - 3y – 2z = -9


Please help me solve this problem, step by step as it is difficult for me to solve.

Answer 1

$\text{The given system of equations can be written as}$

$\left[\begin{array}{ccc}1&6&-1\\2&3&3\\3&-3&-2\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}10\\17\\-9\end{array}\right]$

$\text{This is of the form AX=B}$

$\text{Here}$

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

$|A|=1(-6+9)-6(-4-9)-1(-6-9)$

$|A|=3+78+15$

$|A|=96{\neq}0$

$\therefore\;A^{-1}\;\text{exists}$

$\text{Cofactor matrix of A is}$

$\left[\begin{array}{ccc}3&13&-15\\15&1&21\\21&-5&-9\end{array}\right]$

$\textbf{Adjoint of matrix A = Transpose of cofactor matrix of A}$

$\text{Then }$

$adjA=\left[\begin{array}{ccc}3&15&21\\13&1&-5\\-15&21&-9\end{array}\right]$

$\bf\,A^{-1}=\dfrac{1}{|A|}(adjA)$

$A^{-1}=\dfrac{1}{96}\left[\begin{array}{ccc}3&15&21\\13&1&-5\\-15&21&-9\end{array}\right]$

$\text{Now,}$

$X=A^{-1}B$

$X=\dfrac{1}{96}\left[\begin{array}{ccc}3&15&21\\13&1&-5\\-15&21&-9\end{array}\right]\left[\begin{array}{c}10\\17\\-9\end{array}\right]$

$X=\dfrac{1}{96}\left[\begin{array}{c}30+255-189\\130+17+45\\-150+357+81\end{array}\right]$

$X=\dfrac{1}{96}\left[\begin{array}{c}96\\192\\288\end{array}\right]$

$X=\left[\begin{array}{ccc}1\\2\\3\end{array}\right]$

$\therefore\textbf{The solution is}$

$\textbf{x= 1, y= 2, z= 3}$