1
Question
Find the inverse of the matrix ⎡⎢⎣121−10221−3⎤⎥⎦ by elementary row transformation. Hence solve the system of equations x+2y+z=4,−x+2z=0,2x+y−3z=0
Open in App
Solution
Let A=⎡⎢⎣121−10221−3⎤⎥⎦AA−1=I⟹⎡⎢⎣121−10221−3⎤⎥⎦A−1=⎡⎢⎣100010001⎤⎥⎦
R2→R2+R12,R3→R3−2R1⟹⎡⎢
⎢
⎢⎣12101320−3−5⎤⎥
⎥
⎥⎦A−1=⎡⎢
⎢
⎢⎣10012120−201⎤⎥
⎥
⎥⎦
R1→R1−2R2,R3→R3+3R2⟹⎡⎢
⎢
⎢
⎢
⎢⎣10−2013200−12⎤⎥
⎥
⎥
⎥
⎥⎦A−1=⎡⎢
⎢
⎢
⎢
⎢⎣0−1012120−12321⎤⎥
⎥
⎥
⎥
⎥⎦
R1→R1−4R3,R2→R2+3R3,R3→−2R3⟹⎡⎢⎣100010001⎤⎥⎦A−1=⎡⎢⎣2−7−4−1501−3−2⎤⎥⎦
⟹A−1=⎡⎢⎣2−7−4−1501−3−2⎤⎥⎦Given equations are x+2y+z=4,−x+2z=0,2x+y−3z=0 which can be represented in matrix is as follows
⎡⎢⎣121−10221−3⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣400⎤⎥⎦⟹⎡⎢⎣xyz⎤⎥⎦=A−1⎡⎢⎣400⎤⎥⎦⟹⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣2−7−4−1501−3−2⎤⎥⎦⎡⎢⎣400⎤⎥⎦ ⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣8−44⎤⎥⎦x=8,y=−4,z=4
AA−1=I⟹⎡⎢⎣121−10221−3⎤⎥⎦A−1=⎡⎢⎣100010001⎤⎥⎦
R2→R2+R12,R3→R3−2R1⟹⎡⎢
⎢
⎢⎣12101320−3−5⎤⎥
⎥
⎥⎦A−1=⎡⎢
⎢
⎢⎣10012120−201⎤⎥
⎥
⎥⎦
R1→R1−2R2,R3→R3+3R2⟹⎡⎢
⎢
⎢
⎢
⎢⎣10−2013200−12⎤⎥
⎥
⎥
⎥
⎥⎦A−1=⎡⎢
⎢
⎢
⎢
⎢⎣0−1012120−12321⎤⎥
⎥
⎥
⎥
⎥⎦
R1→R1−4R3,R2→R2+3R3,R3→−2R3⟹⎡⎢⎣100010001⎤⎥⎦A−1=⎡⎢⎣2−7−4−1501−3−2⎤⎥⎦
⟹A−1=⎡⎢⎣2−7−4−1501−3−2⎤⎥⎦
Given equations are x+2y+z=4,−x+2z=0,2x+y−3z=0 which can be represented in matrix is as follows
⎡⎢⎣121−10221−3⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣400⎤⎥⎦⟹⎡⎢⎣xyz⎤⎥⎦=A−1⎡⎢⎣400⎤⎥⎦⟹⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣2−7−4−1501−3−2⎤⎥⎦⎡⎢⎣400⎤⎥⎦
⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣8−44⎤⎥⎦
x=8,y=−4,z=4
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
