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Question

Using matrices, solve the following system of equations:
x+yz=3;2x+3y+z=10;3xy7z=1.

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Solution

The given system of equation can be expressed can be represented in matrix form as AX = B, where


A=∣ ∣111231317∣ ∣,X=∣ ∣xyz∣ ∣,B=∣ ∣3101∣ ∣


Now |A|=∣ ∣111231317∣ ∣=1(21+1)1(143)1(29)20+17+11=80

C11=(1)1+13117=21+1=20
C12=(1)1+22137=(143)=17
C13=(1)1+32331=29=11
C21=(1)2+11117=(71)=8
C22=(1)2+21137=7+3=4
C23=(1)2+31131=(13)=4
C31=(1)3+11131=1+3=4
C32=(1)3+21121=(1+2)=3
C33=(1)3+31123=32=1
AdjA=∣ ∣201711844431∣ ∣T=∣ ∣208417431141∣ ∣
A1=1|A|adjA=18∣ ∣208417431141∣ ∣
AX=BX=A1B
Therefore, xyz=18∣ ∣208417431141∣ ∣∣ ∣3101∣ ∣
=1860+80+45140333+40+1
=182488
∣ ∣xyz∣ ∣=∣ ∣311∣ ∣
Equating the corresponding elements we get
x=3,y=1,z=1.

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