1
Question
By the method of matrix inversion, solve the system
⎡⎢⎣11125721−1⎤⎥⎦⎡⎢⎣xuyvzw⎤⎥⎦=⎡⎢⎣9252150−1⎤⎥⎦
and find the value of xyz−uvw
⎡⎢⎣11125721−1⎤⎥⎦⎡⎢⎣xuyvzw⎤⎥⎦=⎡⎢⎣9252150−1⎤⎥⎦
and find the value of xyz−uvw
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Solution
We have
⎡⎢⎣11125721−1⎤⎥⎦⎡⎢⎣xuyvzw⎤⎥⎦=⎡⎢⎣9252150−1⎤⎥⎦
or AX=BorX=A−1B...(1)
Where
A=⎡⎢⎣11125721−1⎤⎥⎦, X=⎡⎢⎣xuyvzw⎤⎥⎦ and B=⎡⎢⎣9252150−1⎤⎥⎦
∴|A|=1(−5−7)−1(−2−14)+1(2−10)=−12+16−8=−4≠0
Let C be the matrix of cofactors of elements of |A|
∴C=⎡⎢⎣C11C12C13C21C22C23C31C32C33⎤⎥⎦
=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣∣∣∣571−1∣∣∣−∣∣∣272−1∣∣∣∣∣∣2521∣∣∣−∣∣∣111−1∣∣∣∣∣∣112−1∣∣∣−∣∣∣1121∣∣∣∣∣∣1157∣∣∣−∣∣∣1127∣∣∣∣∣∣1125∣∣∣⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦=⎡⎢⎣−1216−82−312−53⎤⎥⎦
∴Adj A=C′=⎡⎢⎣−122216−3−5−813⎤⎥⎦
∴A−1=Adj.fat A|A|=−14⎡⎢⎣−122216−3−5−813⎤⎥⎦
Now, A−1B=−14⎡⎢⎣−122216−3−5−813⎤⎥⎦×⎡⎢⎣9252150−1⎤⎥⎦=−14⎡⎢⎣−44−12−8−20−4⎤⎥⎦
=⎡⎢⎣1−13251⎤⎥⎦
From (1)
X=A−1B
⇒⎡⎢⎣xuyvzw⎤⎥⎦=⎡⎢⎣1−13251⎤⎥⎦
On equating the corresponding elements, we have
x=1,u=−1y=3,v=2z=5,w=1
⎡⎢⎣11125721−1⎤⎥⎦⎡⎢⎣xuyvzw⎤⎥⎦=⎡⎢⎣9252150−1⎤⎥⎦
or AX=BorX=A−1B...(1)
Where
A=⎡⎢⎣11125721−1⎤⎥⎦, X=⎡⎢⎣xuyvzw⎤⎥⎦ and B=⎡⎢⎣9252150−1⎤⎥⎦
∴|A|=1(−5−7)−1(−2−14)+1(2−10)=−12+16−8=−4≠0
Let C be the matrix of cofactors of elements of |A|
∴C=⎡⎢⎣C11C12C13C21C22C23C31C32C33⎤⎥⎦
=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣∣∣∣571−1∣∣∣−∣∣∣272−1∣∣∣∣∣∣2521∣∣∣−∣∣∣111−1∣∣∣∣∣∣112−1∣∣∣−∣∣∣1121∣∣∣∣∣∣1157∣∣∣−∣∣∣1127∣∣∣∣∣∣1125∣∣∣⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦=⎡⎢⎣−1216−82−312−53⎤⎥⎦
∴Adj A=C′=⎡⎢⎣−122216−3−5−813⎤⎥⎦
∴A−1=Adj.fat A|A|=−14⎡⎢⎣−122216−3−5−813⎤⎥⎦
Now, A−1B=−14⎡⎢⎣−122216−3−5−813⎤⎥⎦×⎡⎢⎣9252150−1⎤⎥⎦=−14⎡⎢⎣−44−12−8−20−4⎤⎥⎦
=⎡⎢⎣1−13251⎤⎥⎦
From (1)
X=A−1B
⇒⎡⎢⎣xuyvzw⎤⎥⎦=⎡⎢⎣1−13251⎤⎥⎦
On equating the corresponding elements, we have
x=1,u=−1y=3,v=2z=5,w=1
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