For what b does Ax = b have infinitely many solutions

Question

\begin{bmatrix}2&1\\-2&-1\end{bmatrix}

I reduced it to

\begin{bmatrix}2&1\\0&0\end{bmatrix}

But I don't understand where to go from here the hint says that it is a linear combination of the columns of Ax of columns of A

Answer 1

Try writing it as an augmented matrix with a general vector $b=(b_1,b_2)$. So apply your row reduction steps to $$\begin{bmatrix}2&1&|&b_1\\-2&-1&|&b_2\end{bmatrix}$$then you can find a relationship with $b_1,b_2$ such that the system has infinitely many solutions.

Answer 2

Say $b=0$...

Then $2x_1+x_2=0$, where $x$ is $\begin{bmatrix} x_1\\ x_2 \end{bmatrix}$

Choose any $x_2$, and then $2x_1=-x_2\implies x_1=\frac{-x_2}2$.

So $\begin{bmatrix} \frac{-x_2}2\\ x_2\end{bmatrix}$ is a solution for any $x_2$.

So infinitely many solutions.

More generally, if $b_1=-b_2$, where $b={\begin{bmatrix} b_1\\ b_2 \end{bmatrix}}$ then $\begin{bmatrix} \frac{-x_2+b_1}2\\ x_2\end{bmatrix}$ is a solution for any $x_2$... To see this row reduce the general augmented matrix...

Answer 3

This is the general answer. We denote $[A\,|\, b]$ the augmented matrix:

An $n\times n$ linear system $Ax = b$ has

• One solution if the matrix $A$ has maximal rank ($n$);
• An infinity of solutions if $A$ has rank $<n$ AND $\DeclareMathOperator{\rk}{rank}\rk [A\,|\, b]=\rk A$;
• No solution if $A$ has rank $<n$ AND $\rk [A\,|\,b] >\rk A$.

Answer 4

Hint: Observe that if $x = \begin{bmatrix}x_1 \\ x_2\end{bmatrix}$ then $Ax = \underbrace{x_1\begin{bmatrix}2 \\ -2\end{bmatrix} + x_2\begin{bmatrix}1 \\ -1\end{bmatrix}}_{\text{linear combination}}= (2x_1+x_2)\begin{bmatrix}1 \\ -1\end{bmatrix}$.