# Linear Algebra Midterm 1 at the Ohio State University (3/3)

## Problem 572

The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.

There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).

The time limit was 55 minutes.

This post is Part 3 and contains Problem 7, 8, and 9.

Check out Part 1 and Part 2 for the rest of the exam problems.

**Problem 7**. Let $A=\begin{bmatrix}

-3 & -4\\

8& 9

\end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix}

-1 \\

2

\end{bmatrix}$.

**(a)** Calculate $A\mathbf{v}$ and find the number $\lambda$ such that $A\mathbf{v}=\lambda \mathbf{v}$.

**(b)** Without forming $A^3$, calculate the vector $A^3\mathbf{v}$.

**Problem 8**. Prove that if $A$ and $B$ are $n\times n$ nonsingular matrices, then the product $AB$ is also nonsingular.

**Problem 9**.

Determine whether each of the following sentences is true or false.

**(a)** There is a $3\times 3$ homogeneous system that has exactly three solutions.

**(b)** If $A$ and $B$ are $n\times n$ symmetric matrices, then the sum $A+B$ is also symmetric.

**(c)** If $n$-dimensional vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent, then the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ is also linearly dependent for any $n$-dimensional vector $\mathbf{v}_4$.

**(d)** If the coefficient matrix of a system of linear equations is singular, then the system is inconsistent.

**(e)** The vectors

\[\mathbf{v}_1=\begin{bmatrix}

1 \\

0 \\

1

\end{bmatrix}, \mathbf{v}_2=\begin{bmatrix}

0 \\

1 \\

0

\end{bmatrix}, \mathbf{v}_3=\begin{bmatrix}

0 \\

0 \\

1

\end{bmatrix}\]
are linearly independent.

Sponsored Links

Contents

- Problem 572
- Solution of Problem 7.
- Proof of Problem 8.
- Solution of Problem 9.
- True or False: (a) There is a $3\times 3$ homogeneous system that has exactly three solutions.
- True or False: (b) If $A$ and $B$ are $n\times n$ symmetric matrices, then the sum $A+B$ is also symmetric.
- True or False: (c) If $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent, then $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ is linearly dependent.
- True or False: (d) If the coefficient matrix of a system of linear equations is singular, then the system is inconsistent.
- True or False: (e) The vectors \[\mathbf{v}_1=\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \mathbf{v}_2=\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \mathbf{v}_3=\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\] are linearly independent.

- Go to Part 1 and Part 2

## Solution of Problem 7.

### (a) Calculate $A\mathbf{v}$.

We calculate

\[A\mathbf{v}=\begin{bmatrix}

-3 & -4\\

8& 9

\end{bmatrix}\begin{bmatrix}

-1 \\

2

\end{bmatrix}=\begin{bmatrix}

-5 \\

10

\end{bmatrix}=5\begin{bmatrix}

-1 \\

2

\end{bmatrix}.\]
Thus we see that $\lambda=5$.

### (b) Without forming $A^3$, calculate the vector $A^3\mathbf{v}$.

From part (a), we know that $A\mathbf{v}=5\mathbf{v}$.

Using the properties of matrix multiplication, we compute

\begin{align*}

A^3\mathbf{v}&=A^2(A\mathbf{v})=A^2(5\mathbf{v})=5A^2\mathbf{v}=5A(A\mathbf{v})=5A(5\mathbf{v})\\

&=5^2A\mathbf{v}=5^2(5\mathbf{v})=5^3\mathbf{v}\\

&=5^3\begin{bmatrix}

-1 \\

2

\end{bmatrix}=\begin{bmatrix}

-125 \\

250

\end{bmatrix}.

\end{align*}

Hence we obtain $A^3\mathbf{v}=\begin{bmatrix}

-125 \\

250

\end{bmatrix}$.

## Proof of Problem 8.

Let $A$ and $B$ be nonsingular matrices.

Suppose that $(AB)\mathbf{v}=\mathbf{0}$ for some $n$-dimensional vector $\mathbf{v}$.

Then we have

\[A(B\mathbf{v})=\mathbf{0}.\]
It follows that the vector $B\mathbf{v}$ is a solution of $A\mathbf{x}=\mathbf{0}$.

As the matrix $A$ is nonsingular, any solution must be the zero vector.

Hence we have $B\mathbf{v}=\mathbf{0}$.

This equation says that the vector $\mathbf{v}$ is a solution of $B\mathbf{x}=\mathbf{0}$.

As the matrix $B$ is nonsingular, any solution must be the zero vector.

This implies that $\mathbf{v}=\mathbf{0}$.

This proves that if $(AB)\mathbf{v}=\mathbf{0}$, then $\mathbf{v}=\mathbf{0}$.

This is equivalent to say that the matrix $AB$ is nonsingular by definition.

## Solution of Problem 9.

### True or False: (a) There is a $3\times 3$ homogeneous system that has exactly three solutions.

**False**. Every system of linear equations has no solution at all or it has either one or infinitely many solutions.

### True or False: (b) If $A$ and $B$ are $n\times n$ symmetric matrices, then the sum $A+B$ is also symmetric.

** True**. Since $A$ and $B$ are symmetric, we have $A^{\trans}=A$ and $B^{\trans}=B$.

It follows that

\[(A+B)^{\trans}=A^{\trans}+B^{\trans}=A+B.\]
Thus the sum $A+B$ is symmetric.

### True or False: (c) If $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent, then $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ is linearly dependent.

**True**. Since the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent, there exists scalars $c_1, c_2, c_3$, not all of them are zero, such that

\[c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3=\mathbf{0}.\]
Let $\mathbf{v}_4$ be any $n$-dimensional vector.

Then we have the linear combination

\[c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3+0\mathbf{v}_4=\mathbf{0}\]
whose coefficient is not trivial as at least one of $c_1, c_2, c_3$ is nonzero.

This implies that the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ is also linearly dependent.

### True or False: (d) If the coefficient matrix of a system of linear equations is singular, then the system is inconsistent.

**False**. The system could be consistent even though the coefficient matrix is singular.

For example, consider the system

\begin{align*}

x_1+2x_2&=3\\

2x_1+4x_2&=6.

\end{align*}

The coefficient matrix of the system is $A=\begin{bmatrix}

1 & 2\\

2& 4

\end{bmatrix}$.

This is singular since, for example, it is row equivalent to $\begin{bmatrix}

1 & 2\\

0& 0

\end{bmatrix}$.

However, the system has a solution, for example, $x_1=1, x_2=1$.

Hence the system is consistent even though its coefficient matrix is singular.

### True or False: (e) The vectors

\[\mathbf{v}_1=\begin{bmatrix}

1 \\

0 \\

1

\end{bmatrix}, \mathbf{v}_2=\begin{bmatrix}

0 \\

1 \\

0

\end{bmatrix}, \mathbf{v}_3=\begin{bmatrix}

0 \\

0 \\

1

\end{bmatrix}\]
are linearly independent.

** True**. Consider the linear combination

\[c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3=\mathbf{0}\]
for some scalars $c_1, c_2, c_3$.

Then this can be written as

\[A\begin{bmatrix}

c_1 \\

c_2 \\

c_3

\end{bmatrix}=\begin{bmatrix}

0 \\

0 \\

0

\end{bmatrix},\]
where

\[A=[\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3]=\begin{bmatrix}

1 & 0 & 0 \\

0 &1 &0 \\

1 & 0 & 1

\end{bmatrix}.\]
So the scalars $c_1, c_2, c_3$ is a solution of the system $A\mathbf{x}=\mathbf{0}$.

The augmented matrix of the system is

\begin{align*}

\left[\begin{array}{rrr|r}

1 & 0 & 0 & 0 \\

0 &1 & 0 & 0 \\

1 & 0 & 1 & 0

\end{array} \right] \xrightarrow{R_3-R_1}

\left[\begin{array}{rrr|r}

1 & 0 & 0 & 0 \\

0 &1 & 0 & 0 \\

0 & 0 & 1 & 0

\end{array} \right].

\end{align*}

Thus the solution is $c_1=0, c_2=0, c_3=0$.

It follows that the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly independent.

## Go to Part 1 and Part 2

Go to Part 1 for Problem 1, 2, and 3.

Go to Part 2 for Problem 4, 5, and 6.

Add to solve later

Sponsored Links

## 2 Responses

[…] post contains the first three problems. Check out Part 2 and Part 3 for the rest of the exam […]

[…] post is Part 1 and contains the first three problems. Check out Part 2 and Part 3 for the rest of the exam […]