# Please solve this matrix by putting it into echelon form.a+b+c+d=6 2a-b+c-d=-2 b+2c+d=8 3a-2d=-6

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We can construct the augmented matrix:

`((1,1,1,1,6),(2,1,1,-1,-2),(0,1,2,1,8),(3,0,0,-2,-6))` Note the use of 0 as a coefficient.

We can now use row-reduction rules: we can add any two rows together, we can multiply a row by a constant, and we can interchange rows.

`((1,1,1,1,6),(0,-1,-1,-3,-14),(0,1,2,1,8),(0,-3,-3,-5,-24))` Leave row 1 alone as the pivot is 1; now replace row 2 with (-2R1)+(R2); leave row 3 alone, and replace row 4 with (-3R1)+(R3)

`==>((1,0,0,-2,-8),(0,1,1,3,14),(0,0,1,-2,-6),(0,0,0,4,18))` We now concentrate on the second column.

Replace R1 with (R1)+(R2);multiply R2 by (-1);replace R3 with R2+R3;replace R4 with (-3R2)+(R4)

`==>((1,0,0,-2,-8),(0,1,0,5,20),(0,0,1,-2,-6),(0,0,0,1,9/2))` Leave row 1 alone;replace R2 with (R2)-(R3);leave R3 alone;replace R4 with `1/4` R4

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**This is in echelon form. From here you could use back substitution to solve: `d=9/2` ; `c-9=-6==>c=3` ;`b+45/2=20==>b=-5/2` , and `a-9=-8==>a=1` .**

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You could continue and put in reduced row-echelon form:

`==>((1,0,0,0,1),(0,1,0,0,-5/2),(0,0,1,0,3),(0,0,0,1,9/2))` Replace R1 with R1+2R4; replace R2 with R2-5R4;replace R3 with R3+2R4;leave R4 alone. Then the solution can be read from the matrix.

Checking we get `1-5/2+3+9/2=6;2-5/2+3-9/2=-2;-5/2+6+9/2=8;3-9=-6` as required.

**Sources:**