If I have a matrix $A$, find $b \in \Bbb R^4$ so $Ax=b$ has no solution

Question

If I have a matrix $A= \begin{pmatrix} 0 & 1 & 2 \\ 0 & 2 & 4 \\ 0 & 3 & 6 \\ 1 & 4 & 8 \end{pmatrix}$, find $b \in \Bbb R^4$ so that $Ax=b$ has no solution.

Answer 1

Note that the first $3$ rows $R_i$ of your matrix are linearly dependent: $6R_1 = 3 R_2 = 2 R_1$; and they are clearly linearly independent of the fourth row; hence the rank of your matrix is $2$. (Or, conversely, for the columns you find that $2C_2 = C_3$).

Now let us pick $x = (x_1, x_2, x_3)^T$ and $b = (b_1,b_2,b_3,b_4)^T$. Then $Ax = b$ thranslates to $$\begin{pmatrix} 0 & 1 & 2 \\ 0 & 2 & 4 \\ 0 & 3 & 6 \\ 1 & 4 & 8 \end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} = \begin{pmatrix}x_2 + 2x_3\\2x_2 + 4x_3\\3x_2 + 6x_3\\x_1 + 4x_2 + 8x_3\\\end{pmatrix}= \begin{pmatrix}b_1\\b_2\\b_3\\b_4\end{pmatrix}.$$ One of the (many!) things you learn from this, is that $$b_2 = 2x_2+4x_3 = 2(x_2 + 2x_3) = 2b_1.$$ In other words: for a solution $x$ to exist, $b$ must be such that $b_2 = 2b_1$. So if you pick any vector $b \in \mathbb{R}^4$ with $b_2 \neq 2b_1$, no solution will exist.