Question

Consider the linear system
x + 2y + 3z = a
x + 3y + 8z = b
x + 2y + 2z = c
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where a, b and c are arbitrary constants. Find all solutions of this system.

Answer 1

x + 2y + 3z = a

x + 3y + 8z = b

x + 2y + 2z = c

factorizing

$\begin{bmatrix} 1 & 2&3 \\ 1 & 3&8\\ 1 &2&2 \end{bmatrix}$ $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ =$\begin{bmatrix} a \\ b\\ c \end{bmatrix}$

writing the argument matrix

$\begin{bmatrix} 1 & 2&3&a \\ 1 & 3&8&b\\ 1&2&2&c \end{bmatrix}$

reducing to row echelon

R₁ to R₁, R₂-R₁, R₃-R₁ we obtain

$\begin{bmatrix} 1 & 2&3&a \\ 0 & 1&5&b-a\\ 0&0&-1&c-a \end{bmatrix}$

replacing the variables to form system of equations

x+2y+3Z=a

y+5z=b-a

-z=c-a

taking back substitution

• z=a-c

y+5(a-c)=b-a

y=b-a-5a+5c

• y=b-6a+5c

x+2(b-6a+5c)+3(a-c)=a

x+2b-12a+10c+3a-3c=a

x+2b-9a+7c=a

• x=10a-2b-7c