# 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)

## 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.

### 2 Responses

1. 09/25/2017

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