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

1 Answer
Jan 11, 2017

Answer:

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))#