Question

by using matrices

Answer 1

Using Kramer's rule (with determinants and all) ?

Answer 2

but det A is cmg zero

Answer 3

cmg?

Answer 4

coming

Answer 5

\[\left[\begin{matrix}2 & 1 & 1 \\1 & -2 & -1\\3&4&3\end{matrix}\right]\]

Answer 6

You're right, its determinant is zero... that means it'd have infinitely many solutions, right?

Answer 7

It's linear dependent matrix, you cannot use Cramer's rule to solve, Just rref,
I got \[x= \frac{ 7-z }{ 5 } ; y = \frac{ 1-3z }{ 5 } \]

Answer 8

z is free variable.