How do you solve #Ax=B# given #A=((2, 0, 0), (-1, 2, 0), (-2, 4, 1))# and #B=((4), (10), (11))#?

1 Answer
Jul 30, 2016

Answer:

Column vector #x=[(x),(y),(z)]=[(2),(6),(50)]#

Explanation:

You may regain the original system of linear equations represented in the given matrix form by applying the definition of matrix multiplication using the Euclidean inner product in #RR^3#, together with the definition of matrix equality to obtain :

#2x+0y+0z=4# .......1.
#-x+2y+0z=10# .......2.
#-2x+4y+z=11# ........3.

where #x# is the column vector #x=[(x),(y),(z)]#.

Solving equation 1 we get #x=2#.

Substituting this into equation 2, we get #y=6#.

Substituting this into equation3 we get #z=50#.