Question

How do you solve `x-y+2z = -5`, `-x +3z = 0`, and `2x+ y = 1` using matrices?

Answer 1

The answer is `((x),(y),(z))=((-12/11),(35/11),(-4/11))`

Explanation:

Rewrite the equation in matrix form

`((1,-1,2),(-1,0,3),(2,1,0))((x),(y),(z))=((-5),(0),(1))`

Let `A=((1,-1,2),(-1,0,3),(2,1,0))`

We must find the inverse of matrix `A`

Let's calculate the

`Det A` `= | (1,-1,2), (-1,0,3), (2,1,0) |`

`=1* | (0,3), (1,0) |+1* | (-1,3), (2,0) | +2* | (-1,0), (2,1) |`

`=-3-6-2=-11`

As `Det A!=0`, matrix A is invertible

Now, we calculte the matrix of co-factors

`C=(( | (0,3), (1,0) | , -| (-1,3), (2,0) | , | (-1,0), (2,1) | ),( -| (-1,2), (1,0) | , | (1,2), (2,0) | , -| (1,-1), (2,1) | ),( | (-1,2), (0,3) | , -| (1,2), (-1,3) | , | (1,-1), (-1,0) | ))`

`=((-3,6,-1),(2,-4,-3),(-3,-5,-1))`

We calculate the transpose of matrix `C`

`C^T=((-3,2,-3),(6,-4,-5),(-1,-3,-1))`

The inverse is

`A^-1=C^T/detA=-1/11*((-3,2,-3),(6,-4,-5),(-1,-3,-1))`

`=((3/11,-2/11,3/11),(-6/11,4/11,5/11),(1/11,3/11,1/11))`

Verification, by doing `A*A^-1`

`A*A^-1=((1,-1,2),(-1,0,3),(2,1,0))*((3/11,-2/11,3/11),(-6/11,4/11,5/11),(1/11,3/11,1/11))`

`=((1,0,0),(0,1,0),(0,0,1))=I`

Now, we can solve our equation

`((x),(y),(z))=((3/11,-2/11,3/11),(-6/11,4/11,5/11),(1/11,3/11,1/11))*((-5),(0),(1))`

`=((-12/11),(35/11),(-4/11))`