# If a Matrix $A$ is Singular, then Exists Nonzero $B$ such that $AB$ is the Zero Matrix

## Problem 301

Let $A$ be a $3\times 3$ singular matrix.

Then show that there exists a nonzero $3\times 3$ matrix $B$ such that
$AB=O,$ where $O$ is the $3\times 3$ zero matrix.

## Proof.

Since $A$ is singular, the equation $A\mathbf{x}=\mathbf{0}$ has a nonzero solution.
Let $\mathbf{x}_1$ be a nonzero solution of $A\mathbf{x}=\mathbf{0}$.
We define the $3\times 3$ matrix $B$ to be
$B=[\mathbf{x}_1, \mathbf{0}, \mathbf{0}],$ that is, the first column of $B$ is the vector $x_1$ and the second and the third column vectors are $3$-dimensional zero vectors.

Then since $x_1\neq \mathbf{0}$, the matrix $B$ is not the zero matrix.
We have
\begin{align*}
AB&=A[\mathbf{x}_1, \mathbf{0}, \mathbf{0}]\\
&=[A\mathbf{x}_1, A\mathbf{0}, A\mathbf{0}]\\
&=[\mathbf{0}, \mathbf{0}, \mathbf{0}]=O,
\end{align*}
since $A\mathbf{x}_1=\mathbf{0}$.

Thus we have found the nonzero matrix $B$ such that the product $AB=O$.

## Comment.

This is one of the midterm exam 1 problems of linear algebra (Math 2568) at the Ohio State University.

In the exam the following hint was given:

Hint: Let $B=[\mathbf{x}_1, \mathbf{x}_2, \mathbf{x}_3]$, where $\mathbf{x}_i$ is the $i$-th column vector of $B$ for $i=1,2,3$. Then $AB=[A\mathbf{x}_1, A\mathbf{x}_2, A\mathbf{x}_3]$.

### Common Mistake

Several students wrote “Since $A$ is singular, $A$ has a solution”.
This does not make sense. A solution of what? You need to remember the definition correctly.

A matrix $A$ is singular if the equation $A\mathbf{x}=\mathbf{0}$ has a nonzero solution $\mathbf{x}$.

Also, note that $\mathbf{x}$ is a vector, not a number.

## Midterm 1 problems and solutions

The following list is the problems and solutions/proofs of midterm exam 1 of linear algebra at the Ohio State University in Spring 2017.

1. Problem 1 and its solution: Possibilities for the solution set of a system of linear equations
2. Problem 2 and its solution: The vector form of the general solution of a system
3. Problem 3 and its solution: Matrix operations (transpose and inverse matrices)
4. Problem 4 and its solution: Linear combination
5. Problem 5 and its solution: Inverse matrix
6. Problem 6 and its solution: Nonsingular matrix satisfying a relation
7. Problem 7 and its solution: Solve a system by the inverse matrix
8. Problem 8 and its solution (The current page): A proof problem about nonsingular matrix

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### 9 Responses

1. Michael Braun says:

There is an error. The problem says A is singular. The proof says A is nonsingular.

• Yu says:

Dear Michael,
Thank you for pointing out the typo. I fixed it.

1. 02/13/2017

[…] Problem 8 and its solution:A proof problem about nonsingular matrix […]

2. 02/13/2017

[…] Problem 8 and its solution:A proof problem about nonsingular matrix […]

3. 02/13/2017

[…] Problem 8 and its solution:A proof problem about nonsingular matrix […]

4. 02/13/2017

[…] Problem 8 and its solution:A proof problem about nonsingular matrix […]

5. 02/13/2017

[…] Problem 8 and its solution:A proof problem about nonsingular matrix […]

6. 04/21/2017

[…] Problem 8 and its solution:A proof problem about nonsingular matrix […]

7. 07/27/2017

[…] Problem 8 and its solution:A proof problem about nonsingular matrix […]

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##### Solve a System by the Inverse Matrix and Compute $A^{2017}\mathbf{x}$

Let $A$ be the coefficient matrix of the system of linear equations \begin{align*} -x_1-2x_2&=1\\ 2x_1+3x_2&=-1. \end{align*} (a) Solve the system...

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