1
Question
If A= ⎡⎢⎣2−11−12−11−12⎤⎥⎦
Find A−1 and solve following system of linear equations.
2x−y+z=3
−x+2y−z=−4
x−y+2z=1
Find A−1 and solve following system of linear equations.
2x−y+z=3
−x+2y−z=−4
x−y+2z=1
Open in App
Solution
Lets find the A−1 by adjoint method,
|A| = 2(4−1)+1(−2+1)+1(1−2)
= 6−1−1
= 4
Cofactors of elements of matrix A are given as
C11 = (−1)1+1[2−1−12] = 1(4−1)=3
C12 = (−1)1+2[−1−112] = (−1)(−2+1)=1
C13 = (−1)1+3[−121−1] = (1−2)=−1
C21 = (−1)2+1[−11−12] = (−1)(1−2)=1
C22 = (−1)2+2[2112] = (4−1)=3
C23 = (−1)2+3[2−11−1] = (−1)(1−2)=1
C31 = (−1)3+1[−112−1] = (1−2)=−1
C32 = (−1)3+2[21−1−1] = (−1)(−2+1)=1
C33 = (−1)3+3[2−1−12] = (4−1)=3
So, cofactor matrix, [C] = ⎡⎢⎣31−1131−113⎤⎥⎦
Adj. A = [C]T = ⎡⎢⎣31−1131−113⎤⎥⎦
A−1= Adj.A|A|= 14⎡⎢⎣31−1131−113⎤⎥⎦
The given system of equations can be expressed as
⎡⎢⎣2−11−12−11−12⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣3−41⎤⎥⎦
⟹[A][X]=[B]
Where [A]=⎡⎢⎣2−11−12−11−12⎤⎥⎦, [X]=⎡⎢⎣xyz⎤⎥⎦ and [B] = ⎡⎢⎣3−41⎤⎥⎦
[X] = A−1 [B]
= 14⎡⎢⎣31−1131−113⎤⎥⎦ ⎡⎢⎣3−41⎤⎥⎦
=14⎡⎢⎣4−8−4⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦ = ⎡⎢⎣1−2−1⎤⎥⎦
⇒x=1;y=−2;z=−1
|A| = 2(4−1)+1(−2+1)+1(1−2)
= 6−1−1
= 4
Cofactors of elements of matrix A are given as
C11 = (−1)1+1[2−1−12] = 1(4−1)=3
C12 = (−1)1+2[−1−112] = (−1)(−2+1)=1
C13 = (−1)1+3[−121−1] = (1−2)=−1
C21 = (−1)2+1[−11−12] = (−1)(1−2)=1
C22 = (−1)2+2[2112] = (4−1)=3
C23 = (−1)2+3[2−11−1] = (−1)(1−2)=1
C31 = (−1)3+1[−112−1] = (1−2)=−1
C32 = (−1)3+2[21−1−1] = (−1)(−2+1)=1
C33 = (−1)3+3[2−1−12] = (4−1)=3
So, cofactor matrix, [C] = ⎡⎢⎣31−1131−113⎤⎥⎦
Adj. A = [C]T = ⎡⎢⎣31−1131−113⎤⎥⎦
A−1= Adj.A|A|= 14⎡⎢⎣31−1131−113⎤⎥⎦
The given system of equations can be expressed as
⎡⎢⎣2−11−12−11−12⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣3−41⎤⎥⎦
⟹[A][X]=[B]
Where [A]=⎡⎢⎣2−11−12−11−12⎤⎥⎦, [X]=⎡⎢⎣xyz⎤⎥⎦ and [B] = ⎡⎢⎣3−41⎤⎥⎦
[X] = A−1 [B]
= 14⎡⎢⎣31−1131−113⎤⎥⎦ ⎡⎢⎣3−41⎤⎥⎦
=14⎡⎢⎣4−8−4⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦ = ⎡⎢⎣1−2−1⎤⎥⎦
⇒x=1;y=−2;z=−1
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
