How do you solve #3x + 2y + z = 8#, #2x -3y + 2z = -16#, and #x + 4y -z = 20# using matrices?

1 Answer

Answer:

#x=1# and #y=4# and #z=-3#

Explanation:

from the given
#3x+2y+z=8#
#2x-3y+2z=-16#
#x+4y-z=20#

For the matrix

#D=((a_11, a_12,a_13),(a_21,a_22,a_23),(a_31,a_32, a_33))=((3,2,1),(2, -3, 2), (1, 4, -1))#

#C=((c_1),(c_2),(c_3))=((8),(-16),(20))#

We will use the formula

#D=(a_11* a_22*a_33+a_12* a_23*a_31+a_13* a_21*a_32#
#-a_31* a_22*a_13-a_32* a_23*a_11-a_33* a_21*a_12)#

For x:

#x=(((c_1, a_12,a_13),(c_2,a_22,a_23),(c_3,a_32, a_33)))/D#

For y:

#y=(((a_11, c_1,a_13),(a_21,c_2,a_23),(a_31,c_3, a_33)))/D#

For z:

#z=(((a_11, a_12,c_1),(a_21,a_22,c_2),(a_31,a_32, c_3)))/D#

We form the matrix

#D=((3,2,1),(2, -3, 2), (1, 4, -1))=4#

For x:

#x=(((c_1, a_12,a_13),(c_2,a_22,a_23),(c_3,a_32, a_33)))/D=(((8, 2,1),(-16,-3,2),(20,4, -1)))/4=1#

For y:

#y=(((a_11, c_1,a_13),(a_21,c_2,a_23),(a_31,c_3, a_33)))/D=(((3, 8,1),(2,-16,2),(1,20, -1)))/4=4#

For z:

#z=(((a_11, a_12,c_1),(a_21,a_22,c_2),(a_31,a_32, c_3)))/D=(((3,2,8),(2,-3,-16),(1,4, 20)))/4=-3#

solution set #x=1# and #y=4# and #z=-3#

God bless...I hope the explanation is useful.