Question

Use the inverse matrix method to solve the system of equations

2y + 2z = 3
x + 3z = 1
4x - 3y + 8z = 0​

Answer 1

x=5/2

y=2

z=-1/2

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)=(1)/(|A|)adj A`

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

Now,

`X=(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=(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