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Byju's Answer
Standard XII
Mathematics
Special Determinants
Let A = いbegi...
Question

Let A=⎡⎢⎣4−144023−24⎤⎥⎦
Find A−1 by using elementery row transformations.

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Solution

Lets take the given matrix
A=⎡⎢⎣4−144023−24⎤⎥⎦

|A|=∣∣ ∣∣4−144023−24∣∣ ∣∣

|A|=4(0+4)+1(16−6)+4(−8−0)
⇒|A|=−6
⇒|A|≠0
∴A−1 exists.

We know that, AA−1=I
⎡⎢⎣4−144023−24⎤⎥⎦A−1=⎡⎢⎣100010001⎤⎥⎦

Perform R1→14R1
⎡⎢ ⎢ ⎢⎣1−1414023−24⎤⎥ ⎥ ⎥⎦A−1=⎡⎢ ⎢ ⎢⎣1400010001⎤⎥ ⎥ ⎥⎦

Perform R2→R2−4R1 and R3→R3−3R1
⎡⎢ ⎢ ⎢ ⎢ ⎢⎣1−14101−20−541⎤⎥ ⎥ ⎥ ⎥ ⎥⎦A−1=⎡⎢ ⎢ ⎢ ⎢ ⎢⎣1400−110−3401⎤⎥ ⎥ ⎥ ⎥ ⎥⎦

Perform R3→R3+54R2 and R1→R1+14R2
⎡⎢ ⎢ ⎢ ⎢ ⎢⎣101201−200−32⎤⎥ ⎥ ⎥ ⎥ ⎥⎦A−1=⎡⎢ ⎢ ⎢ ⎢ ⎢⎣0140−110−2541⎤⎥ ⎥ ⎥ ⎥ ⎥⎦

Perform R3→−23R3
⎡⎢ ⎢ ⎢⎣101201−2001⎤⎥ ⎥ ⎥⎦A−1=⎡⎢ ⎢ ⎢ ⎢ ⎢⎣0140−11043−56−23⎤⎥ ⎥ ⎥ ⎥ ⎥⎦

Perform R2→R2+2R3 and R1→R1−12R3

⎡⎢⎣100010001⎤⎥⎦A−1=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−23231353−46−4343−56−23⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦

⇒I⋅A−1=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−23231353−23−4343−56−23⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦



∴A−1=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣−23231353−23−4343−56−23⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦

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