I am trying to figure out this math problem.
For what values $a,b$ does the linear system have infinitely many solutions?
This is the matrix
$$ \left[ \begin{array}{ccc|c} 1&-2&3&-4\\ 2&1&1&2\\ 1&a&2&-b \end{array} \right] $$
I know the solution. $a = -1$ and $b = 2,$ but this is my problem. In the solutions it is written like this: $a=-1, b=2, x=t[-1\ 1\ 1]^{T}+[0\ 2\ 0]^{T}$.
I don't understand the solution as it is given; can somebody please explain it to me?
Thank you very much!!!
Using Gaussian elimination, we arrive at
$$\begin{bmatrix} 1&0&1&0\\ 0&1&-1&2\\ 0&0&a+1&-b-2a\\ \end{bmatrix}$$
As you know, when $a = -1$ and $b=2$, the linear system of equations represented by the matrix above will have infinitely many solutions.
So assuming $a = -1,\;b = 2$, we have
$$\begin{bmatrix} 1&0&1&0\\ 0&1&-1&2\\ 0&0&0&0\\ \end{bmatrix}$$
Since $0x_3 = 0$ is true *for all real values $x_3$, we use $x_3 = t$ with $t\in \mathbb R$ to denote this.
Then we solve for $x_2$ using our matrix: $x_2 - x_3 = 2 \iff x_2 = x_3 + 2 = t + 2$.
Solving for $x_1$ using our matrix, we have $x_1 + x_3 = 0 \iff x_1 = -x_3 = -t$.
This gives us the (infinite number of) solutions of the form$$x = \begin{bmatrix} -t\\ t+2\\ t \end{bmatrix} = \begin{bmatrix} -t \\ t \\ t\end{bmatrix} + \begin{bmatrix} 0 \\ 2 \\ 0\end{bmatrix} = t \begin{bmatrix} -1 \\1\\1\end{bmatrix} + \begin{bmatrix} 0 \\ 2\\0\end{bmatrix},\quad t \in \mathbb R$$