Solving Matrix question AX=B to find the possible solutions of B Question


  Find $b_1$ and $b_2$ so that the equation $Ax = b$ has solutions where
  $$A = 
\begin{bmatrix}
1 &amp; 2 \\
0 &amp; 1\\
-1 &amp; 2
\end{bmatrix};\quad 
b = 
\begin{bmatrix}
b_1 \\
b_2 \\
0
\end{bmatrix}.$$
  Can this equation have a unique solution? Why or why not?


Work

$$\begin{bmatrix}1&amp;2 \\0&amp;1\\-1&amp;2\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix} = \begin{bmatrix}b_1\\b_2\\0\end{bmatrix}$$

$$\begin{bmatrix}
x_1 + 2x_2 \\
x_2 \\
-x_1 + 2x_2
\end{bmatrix} = 
\begin{bmatrix}
b_1 \\
b_2 \\
0 
\end{bmatrix}$$

$$\left.\begin{array}{ll}
x_1 + 2x_2 = b_1 \\
x_2 = b_2 \\
2x_2 = x_1 
\end{array}\right\} \Rightarrow \left.\begin{array}{ll}x_1 = b_1/2 \\
x_2 = b_1/4 \end{array}\right\} \Rightarrow b_1/4 = b_2 \Rightarrow b_1 = 4b_2$$
Hint: the equation $Ax=b$ has solutions $ \iff rank(A|b)=rank(A)$.
Hint: the equation $Ax=b$ has solutions $ \iff rank(A|b)=rank(A)$.