Answer on Question #82857 - Math - Linear Algebra
1. We present the equation in the form of a 3 by 3 matrix:
⎣⎡2131−2−1−123∣∣∣11−25⎦⎤
2. For convenience, we change the 1st and second line:
⎣⎡123−21−12−13∣∣∣−2115⎦⎤
3. Multiply the 1st row by -2 and -3 and add to the second and third row respectively:
⎣⎡100−2552−5−3∣∣∣−21511⎦⎤
4. Multiply the 2nd row by -1 and add to the 3rd. Divide the 2nd line by 5:
⎣⎡100−2102−12∣∣∣−23−4⎦⎤
5. Divide the 3rd line by 2:
⎣⎡100−2102−11∣∣∣−23−2⎦⎤
6. Add the third line to the second. Multiply the 3rd row by -2 and add to 1st:
⎣⎡100−21000121−2⎦⎤
7. Multiply the 2nd row by 2 and add to 1st:
⎣⎡10001000141−2⎦⎤
Answer:
⎩⎨⎧x=4y=1z=−2
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