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

Ohio State University exam problems and solutions in mathematics

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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2 Responses

  1. 09/25/2017

    […] post contains Problem 4, 5, and 6. Check out Part 1 and Part 3 for the rest of the exam […]

  2. 09/25/2017

    […] post contains Problem 7, 8, and 9. Check out Part 1 and Part 2 for the rest of the exam […]

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