Question

How do you solve `3a + 6b + c + 4d = 2`, `a - 2b + 3c - d = -3`, `2a - 4c + d= -15` and `a + 8b + 2c + d = 6` using matrices?

Answer 1

The solution is `((a),(b),(c),(d))=((-5),(1/2),(2),(3))`

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

`A=((1,8,2,1,|,6),(1,-2,3,-1,|,-3),(3,6,1,4,|,2),(2,0,-4,1,|,-15))`

I have written the equations not in the sequence as in the question in order to get `1` as pivot.

Perform the folowing operations on the rows of the matrix

`R2larrR2-R1` ; `R3larrR3-3R1` ; `R4-2R1`

`A=((1,8,2,1,|,6),(0,-10,1,-2,|,-9),(0,-18,-5,1,|,-16),(0,-16,-8,-1,|,-27))`

`R3larr(R3)/(-10)`

`A=((1,8,2,1,|,6),(0,1,-1/10,1/5,|,9/10),(0,-18,-5,1,|,-16),(0,-16,-8,-1,|,-27))`

`R3larr(R3+18R2)` , `R4larr (R4+16R2)`

`A=((1,8,2,1,|,6),(0,1,-1/10,1/5,|,9/10),(0,0,-34/5,23/5,|,1/5),(0,0,-48/5,11/5,|,-63/5))`

`R3larr(R3xx(-5/34))`

`A=((1,8,2,1,|,6),(0,1,-1/10,1/5,|,9/10),(0,0,1,-23/34,|,-1/34),(0,0,-48/5,11/5,|,-63/5))`

`R4larr(R4+(48/5)*R3)`

`A=((1,8,2,1,|,6),(0,1,-1/10,1/5,|,9/10),(0,0,1,-23/34,|,-1/34),(0,0,0,-73/17,|,-219/17))`

`R4larr(R4*(-17/73))`

`A=((1,8,2,1,|,6),(0,1,-1/10,1/5,|,9/10),(0,0,1,-23/34,|,-1/34),(0,0,0,1,|,3))`

`R3larr(R3+23/34R4)`

`A=((1,8,2,1,|,6),(0,1,-1/10,1/5,|,9/10),(0,0,1,0,|,2),(0,0,0,1,|,3))`

`R2larr(R2+1/10R3)` , `R2larrR2-1/5R3`

`A=((1,8,2,1,|,6),(0,1,0,0,|,1/2),(0,0,1,0,|,2),(0,0,0,1,|,3))`

`R1larr(R1-8R2)` , `R1larrR1-2R3` ,; `R1larrR1-R4`

`A=((1,0,0,0,|,-5),(0,1,0,0,|,1/2),(0,0,1,0,|,2),(0,0,0,1,|,3))`