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Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 23/Solution 2

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Alternatively, we can also use row reduction.

Form the augmented matrix [A|I]= [213100101010012001]

Interchange R2 and R1 : [101010213100012001]

R2−2R1→R2 : [1010100111−20012001]

R3−R2→R3 : [1010100111−20001−121]

R1−R3→R1 : [1001−1−10111−20001−121]

R2−R3→R2 : [1001−1−10102−4−1001−121]

Therefore,

A−1=[1−1−12−4−1−121].