Let $A$ and $B$ be $3\times 3$ matrices and let $C=A-2B$.
If
\[A\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}=B\begin{bmatrix}
2 \\
6 \\
10
\end{bmatrix},\]
then is the matrix $C$ nonsingular? If so, prove it. Otherwise, explain why not.

Using the given equality, we have
\begin{align*}
\mathbf{0}&=A\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}-B\begin{bmatrix}
2 \\
6 \\
10
\end{bmatrix}\\[6pt]
&=A\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}-2B\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}\\[6pt]
&=(A-2B)\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}\\[6pt]
&=C\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}.
\end{align*}
Thus, we obtain
\[C\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}=\mathbf{0}.\]
This implies that the vector $\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}$ is a nonzero solution to the equation $C\mathbf{x}=\mathbf{0}$.
Therefore, the matrix $C$ is singular.

Common Mistake

This is a midterm exam problem of Lienar Algebra at the Ohio State University.

One common mistake is that once you have
\[A\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}=2B\begin{bmatrix}
1 \\
3\\
5
\end{bmatrix},\]
then you conclude that $A=2B$, and hence $C$ is the zero matrix.

This is WRONG!! You cannot cancel matrices, like numbers!

Another common mistake is to state something like “Because $C$ has a nonzero solution…”
There is no “solution” to a matrix. We can talk about the solution of a system, for example.

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