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

## Problem 570

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 1 and contains the first three problems.

Check out Part 2 and Part 3 for the rest of the exam problems.

**Problem 1.** Determine all possibilities for the number of solutions of each of the systems of linear equations described below.

**(a)** A consistent system of $5$ equations in $3$ unknowns and the rank of the system is $1$.

**(b)** A homogeneous system of $5$ equations in $4$ unknowns and it has a solution $x_1=1$, $x_2=2$, $x_3=3$, $x_4=4$.

**Problem 2.** Consider the homogeneous system of linear equations whose coefficient matrix is given by the following matrix $A$. Find the vector form for the general solution of the system.

\[A=\begin{bmatrix}

1 & 0 & -1 & -2 \\

2 &1 & -2 & -7 \\

3 & 0 & -3 & -6 \\

0 & 1 & 0 & -3

\end{bmatrix}.\]

**Problem 3.** Let $A$ be the following invertible matrix.

\[A=\begin{bmatrix}

-1 & 2 & 3 & 4 & 5\\

6 & -7 & 8& 9& 10\\

11 & 12 & -13 & 14 & 15\\

16 & 17 & 18& -19 & 20\\

21 & 22 & 23 & 24 & -25

\end{bmatrix}

\]
Let $I$ be the $5\times 5$ identity matrix and let $B$ be a $5\times 5$ matrix.

Suppose that $ABA^{-1}=I$.

Then determine the matrix $B$.

(*Linear Algebra Midterm Exam 1, the Ohio State University*)

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## Solution of Problem 1.

**(a)** As the system is consistent, we first narrow down the possibilities to either a unique solution or infinitely many solutions.

Since the number of unknowns minus the rank is the number of free variables, the system has $3-1=2$ free variables.

It follows that the system must have infinitely many solutions.

**(b)** A homogeneous system has a trivial (zero) solution. As we know the homogeneous system has one nontrivial solution $x_1=1$, $x_2=2$, $x_3=3$, $x_4=4$, there must be infinitely many solutions.

## Solution of Problem 2.

The homogeneous system is given by

\[A\begin{bmatrix}

x_1 \\

x_2 \\

x_3 \\

x_4

\end{bmatrix}=\begin{bmatrix}

0 \\

0 \\

0 \\

0

\end{bmatrix}.\]
We apply elementary row operations to the augmented matrix of the system as follows.

\begin{align*}

[A\mid \mathbf{0}]&= \left[\begin{array}{rrrr|r}

1 & 0 & -1 & -2 & 0 \\

2 & 1 & -2 & -7 & 0 \\

3 & 0 & -3 & -6 & 0 \\

0 & 1 & 0 & -3 & 0

\end{array} \right]
\xrightarrow[R_3-3R_1]{R_2-2R_1}

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

1 & 0 & -1 & -2 & 0 \\

0 & 1 & 0 & -3 & 0 \\

0 & 0 & 0 & 0 & 0 \\

0 & 1 & 0 & -3 & 0 \\

\end{array} \right]\\[6pt]
&\xrightarrow{R_4-R_2}

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

1 & 0 & -1 & -2 & 0 \\

0 & 1 & 0 & -3 & 0 \\

0 & 0 & 0 & 0 & 0 \\

0 & 0 & 0 & 0 & 0 \\

\end{array} \right].

\end{align*}

It follows that the general solution is given by

\begin{align*}

x_1&=x_3+2x_4\\

x_2&=3x_4,

\end{align*}

where $x_3$ and $x_4$ are free variables.

The vector form of the general solution is

\[\mathbf{x}=\begin{bmatrix}

x_1 \\

x_2 \\

x_3 \\

x_4

\end{bmatrix}=\begin{bmatrix}

x_3+2x_4 \\

3x_4 \\

x_3 \\

x_4

\end{bmatrix}

=

x_3\begin{bmatrix}

1 \\

0 \\

1 \\

0

\end{bmatrix}+x_4\begin{bmatrix}

2 \\

3 \\

0 \\

1

\end{bmatrix}.\]

## Solution of Problem 3.

Multiplying $ABA^{-1}=I$ by $A$ on right, we obtain

\begin{align*}

(ABA^{-1})A&=IA\\

\Leftrightarrow AB(A^{-1}A)&=A && \text{by associativity and $IA=A$}\\

\Leftrightarrow ABI&=A && \text{as $A^{-1}A=I$}\\

\Leftrightarrow AB&=A && \text{as $BI=B$}.

\end{align*}

Next, we multiply $AB=A$ by $A^{-1}$ on left and get

\begin{align*}

A^{-1}(AB)&=A^{-1}A\\

\Leftrightarrow (A^{-1}A)B&=I && \text{by associativity and $A^{-1}A=I$}\\

\Leftrightarrow IB&=I && \text{as $A^{-1}A=I$}\\

\Leftrightarrow B&=I && \text{as $IB=I$}.

\end{align*}

Therefore, the matrix $B$ must be the $5\times 5$ identity matrix $I$.

### Another Solution of Problem 3

We could have combined the above two steps as follows.

We have

\begin{align*}

I&=A^{-1}A\\

&=A^{-1}(ABA^{-1})A && \text{by assumption $ABA^{-1}=I$}\\

&=(A^{-1}A)B(A^{-1}A)=IBI=B.

\end{align*}

### A Common Mistake

A common mistake of this problem is that one changes the order of matrix products:

\begin{align*}

I=ABA^{-1}=AA^{-1}B=IB=B &&\text{This is WRONG!}.

\end{align*}

## Go to Part 2 and Part 3

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

Go to Part 3 for Problem 7, 8, and 9.

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