5
Question
By the method of matrix inversion, solve the system.
⎛⎜⎝11125721−1⎤⎥⎦⎛⎜⎝x1y1x2y2x3y3⎤⎥⎦=⎛⎜⎝9252150−1⎤⎥⎦
By the method of matrix inversion, solve the system.
⎛⎜⎝11125721−1⎤⎥⎦⎛⎜⎝x1y1x2y2x3y3⎤⎥⎦=⎛⎜⎝9252150−1⎤⎥⎦
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Solution
The correct option is C x1=1,x2=3,x3=5y1=−1,y2=2,y3=1
We have⎛⎜⎝11125721−1⎤⎥⎦⎛⎜⎝x1y1x2y2x3y3⎤⎥⎦=⎛⎜⎝9252150−1⎤⎥⎦⇒AX=B −(1)Here (A|=−4≠0. Therefore,adj A=⎛⎜⎝−1216−82−31213⎤⎥⎦T=⎛⎜⎝−122216−3−5−813⎤⎥⎦∴A−1=adj 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 equation (1), we getX=A−1B⇒⎛⎜⎝x1y1x2y2x3y3⎤⎥⎦=⎛⎜⎝1−13251⎤⎥⎦⇒x1=1, x2=3, x3=5⇒y1=−1, y2=2, y3=1option (c) is correct.
We have⎛⎜⎝11125721−1⎤⎥⎦⎛⎜⎝x1y1x2y2x3y3⎤⎥⎦=⎛⎜⎝9252150−1⎤⎥⎦⇒AX=B −(1)Here (A|=−4≠0. Therefore,adj A=⎛⎜⎝−1216−82−31213⎤⎥⎦T=⎛⎜⎝−122216−3−5−813⎤⎥⎦∴A−1=adj 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 equation (1), we getX=A−1B⇒⎛⎜⎝x1y1x2y2x3y3⎤⎥⎦=⎛⎜⎝1−13251⎤⎥⎦⇒x1=1, x2=3, x3=5⇒y1=−1, y2=2, y3=1option (c) is correct.
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