Question

$\left\begin{array}{ccc}4x-3y+2z=4\\3x-2y+3z=8\\4x+2y-2z=2\end{array}\right ,Use matrix method to solve the given system of equation.$

Answer 1

equations are :
4x - 3y + 2z = 4
3x - 2y + 3z = 8
4x + 2y - 2z = 2

so, $A=\left[\begin{array}{ccc}4&-3&2\\3&-2&3\\4&2&-2\end{array}\right]$

$X=\left[\begin{array}{c}x\\y\\z\end{array}\right]$

and $B=\left[\begin{array}{c}4\\8\\2\end{array}\right]$

here, equation is written as $AX=B$ (in matrix form)
taking A^-1 both sides, we get,
$X=A^{-1}B$

so, we have to find out inverse of A,
first of all, take transpose of A.
$A^T=\left[\begin{array}{ccc}4&3&4\\-3&-2&2\\2&3&-2\end{array}\right]$

now we have to get adjoint of A.
so, $adj(A)=\left[\begin{array}{ccc}-2&-2&-5\\18&-16&-6\\14&-20&1\end{array}\right]$

after that find determinant of A.
$det(A)=\left|\begin{array}{ccc}4&-3&2\\3&-2&3\\4&2&-2\end{array}\right|$

=4(4 - 6) + 3(-6 - 12) + 2(6 + 8)
= -8 - 54 + 28
= -62 + 28
= -34

now, $A^{-1}=\frac{adj(A)}{det(A)}$

$A^{-1}=\frac{\left[\begin{array}{ccc}-2&-2&-5\\18&-16&-6\\14&-20&1\end{array}\right]}{-34}$

=$\left[\begin{array}{ccc}1/17&1/17&5/34\\-9/17&8/17&3/17\\-7/17&10/17&-1/34\end{array}\right]$

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

$X=\left[\begin{array}{ccc}1/17&1/17&5/34\\-9/17&8/17&3/17\\-7/17&10/17&-1/34\end{array}\right]\left[\begin{array}{c}4\\8\\2\end{array}\right]$

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

hence, x = 1 , y = 2 and z = 3