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) Add to solve later

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. Add to solve later

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 […]

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Linear Algebra An Example of Matrices $A$, $B$ such that $\mathrm{rref}(AB)\neq \mathrm{rref}(A) \mathrm{rref}(B)$

For an $m\times n$ matrix $A$, we denote by $\mathrm{rref}(A)$ the matrix in reduced row echelon form that is row...

Close