# 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. Add to solve later

## 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

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###### More in Linear Algebra ##### Linear Algebra Midterm 1 at the Ohio State University (2/3)

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

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