# Determine whether the Matrix is Nonsingular from the Given Relation ## Problem 698

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. Add to solve later

## Solution.

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. Add to solve later

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