1
Question
Using matrices, solve the following system of linear equation:
x−y+2z=7
3x+4y−5z=−5
2x−y+3z=12
x−y+2z=7
3x+4y−5z=−5
2x−y+3z=12
Open in App
Solution
The given system of equation can be expressed can be represented in matrix form as AX = B, where
A=∣∣
∣∣1−1234−52−13∣∣
∣∣X=∣∣
∣∣xyz∣∣
∣∣,B=∣∣
∣∣7−512∣∣
∣∣
Now |A|=∣∣
∣∣1−1234−52−13∣∣
∣∣=1(12−5)+1(9+10)+2(−3−8)⇒7+19−22=4≠0
Hence A−1 exist and system have unique solution.
C11=(−1)1+1∣∣∣4−5−13∣∣∣=12−5=7
C12=(−1)1+2∣∣∣3−523∣∣∣=−(9+10)=−19
C13=(−1)1+3∣∣∣342−1∣∣∣=(−3−8)=−11
C21=(−1)2+1∣∣∣−1213∣∣∣=−(−3+2)=1
C22=(−1)2+2∣∣∣1223∣∣∣=3−4=−1
C23=(−1)2+3∣∣∣1−12−1∣∣∣=−(−1+2)=−1
C31=(−1)3+1∣∣∣−124−5∣∣∣=5−8=−3
C32=(−1)3+2∣∣∣123−5∣∣∣=4+3=7
C33=(−1)3+3∣∣∣1−134∣∣∣=4+3=7
AdjA=∣∣
∣∣71−19−111−1−1−3117∣∣
∣∣T=∣∣
∣∣71−3−19−111−11−17∣∣
∣∣
A−1=1|A|adjA=14∣∣
∣∣71−3−19−111−11−17∣∣
∣∣
AX=B⇒X=A−1B
Therefore, ⎡⎢⎣xyz⎤⎥⎦=14∣∣
∣∣71−3−19−111−11−17∣∣
∣∣∣∣
∣∣7−512∣∣
∣∣
=14⎡⎢⎣49−5−36−133+5+132−77+5+84⎤⎥⎦
=14⎡⎢⎣8412⎤⎥⎦
∣∣
∣∣xyz∣∣
∣∣=∣∣
∣∣213∣∣
∣∣
Equating the corresponding elements we get
x = 2, y = 1, z = 3.
A=∣∣ ∣∣1−1234−52−13∣∣ ∣∣X=∣∣ ∣∣xyz∣∣ ∣∣,B=∣∣ ∣∣7−512∣∣ ∣∣
Now |A|=∣∣ ∣∣1−1234−52−13∣∣ ∣∣=1(12−5)+1(9+10)+2(−3−8)⇒7+19−22=4≠0
Hence A−1 exist and system have unique solution.
C11=(−1)1+1∣∣∣4−5−13∣∣∣=12−5=7
C12=(−1)1+2∣∣∣3−523∣∣∣=−(9+10)=−19
C13=(−1)1+3∣∣∣342−1∣∣∣=(−3−8)=−11
C21=(−1)2+1∣∣∣−1213∣∣∣=−(−3+2)=1
C22=(−1)2+2∣∣∣1223∣∣∣=3−4=−1
C23=(−1)2+3∣∣∣1−12−1∣∣∣=−(−1+2)=−1
C31=(−1)3+1∣∣∣−124−5∣∣∣=5−8=−3
C32=(−1)3+2∣∣∣123−5∣∣∣=4+3=7
C33=(−1)3+3∣∣∣1−134∣∣∣=4+3=7
AdjA=∣∣ ∣∣71−19−111−1−1−3117∣∣ ∣∣T=∣∣ ∣∣71−3−19−111−11−17∣∣ ∣∣
A−1=1|A|adjA=14∣∣ ∣∣71−3−19−111−11−17∣∣ ∣∣
AX=B⇒X=A−1B
Therefore, ⎡⎢⎣xyz⎤⎥⎦=14∣∣ ∣∣71−3−19−111−11−17∣∣ ∣∣∣∣ ∣∣7−512∣∣ ∣∣
=14⎡⎢⎣49−5−36−133+5+132−77+5+84⎤⎥⎦
=14⎡⎢⎣8412⎤⎥⎦
∣∣ ∣∣xyz∣∣ ∣∣=∣∣ ∣∣213∣∣ ∣∣
Equating the corresponding elements we get
x = 2, y = 1, z = 3.
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
