Q. x1+x2+2x3=−1x1−2x2+x3=−53x1+x2+x3=3Find all solutions by using the Gaussian elimination & Gauss-Jordan Reduction.
Write Augmented Matrix: Write the augmented matrix for the system of equations.⎣⎡113amp;1amp;−2amp;1amp;2amp;1amp;1amp;∣amp;∣amp;∣amp;−1amp;−5amp;3⎦⎤
Leading 1 in First Row: Perform row operations to get a leading 1 in the first row, first column (R1 is already set).No changes needed for R1.
Eliminate Below Leading 1: Make the elements below the leading 1 in the first column zero, using R2 - R1 → R2 and R3 - 3R1 → R3.⎣⎡100amp;1amp;−3amp;−2amp;2amp;−1amp;−5amp;∣amp;∣amp;∣amp;−1amp;−4amp;6⎦⎤
Leading 1 in Second Row: Get a leading 1 in the second row, second column by dividing R2 by −3.⎣⎡100amp;1amp;1amp;−2amp;2amp;1/3amp;−5amp;∣amp;∣amp;∣amp;−1amp;4/3amp;6⎦⎤
Eliminate Above and Below: Eliminate the entries above and below the leading 1 in the second column. Add R2 to R1 and add 2R2 to R3.⎣⎡100amp;0amp;1amp;0amp;7/3amp;1/3amp;−7/3amp;∣amp;∣amp;∣amp;1/3amp;4/3amp;14/3⎦⎤
Leading 1 in Third Row: Get a leading 1 in the third row, third column by dividing R3 by −7/3.⎣⎡100amp;0amp;1amp;0amp;7/3amp;1/3amp;1amp;∣amp;∣amp;∣amp;1/3amp;4/3amp;−2⎦⎤
Eliminate Above Leading 1: Eliminate the entries above the leading 1 in the third column. Subtract (7/3)R3 from R1 and subtract (1/3)R3 from R2.⎣⎡100amp;0amp;1amp;0amp;0amp;0amp;1amp;∣amp;∣amp;∣amp;5amp;5amp;−2⎦⎤