Question

Use the inverse matrix method to solve the system of equations
2y + 2z = 3
x + 3z = 1
4x - 3y + 8z = 0​

Answer 1

Answer:

x=5/2

y=2

z=-1/2

Step-by-step explanation:

We are given that

$2y+2x=3$

$x+3z=1$

$4x-3y+8z=0$

We have to solve the system of equation by using inverse matrix method.

We know that

AX=B

$X=A^{-1} B$

Where

A

=$\left[\begin{array}{ccc}0&2&2\\1&0&3\\4&-3&8\end{array}\right]$

$B=\left[\begin{array}{ccc}3\\1\\0\end{array}\right]$

$|A|=0(0+9)-2(8-12)+2(-3-0)=2$

Now,Minor

$M_{11}=9$,$M_{21}=22$,$M_{31}=6$

$M_{12}=-4$,$M_{22}=-8$,$M_{32}=-2$

$M_{13}=-3$,$M_{23}=-8$,$M_{33}=-2$

Co-factor

$A_{ij}=(-1)^{i+j}M_{ij}$

$A_{11}=9,A_{12}=4,A_{13}=-3$

$A_{21}=-22,A_{22}=-8,A_{23}=8$

$A_{31}=6,A_{32}=2,A_{33}=-2$

Now, adjoint

$adj A=\left[\begin{array}{ccc}9&4&-3\\-22&-8&8\\6&2&-2\end{array}\right]^{T$

$adj A=\left[\begin{array}{ccc}9&-22&6\\4&-8&2\\-3&8&-2\end{array}\right]$

Now,

$A^{-1}=\frac{1}{|A|}adj A$

$A^{-1}=\frac{1}{2}\left[\begin{array}{ccc}9&-22&6\\4&-8&2\\-3&8&-2\end{array}\right]$

Now,

$X=\frac{1}{2}\left[\begin{array}{ccc}9&-22&6\\4&-8&2\\-3&8&-2\end{array}\right]\left[\begin{array}{ccc}3\\1\\0\end{array}\right]$

$X=\frac{1}{2}\left[\begin{array}{ccc}27-22\\12-8\\-9+8\end{array}\right]$

$X=\left[\begin{array}{ccc}5/2\\2\\-1/2\end{array}\right]$

$\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}5/2\\2\\-1/2\end{array}\right]$

We get

x=5/2

y=2

z=-1/2