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

Ohio State University exam problems and solutions in mathematics

Problem 571

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 2 and contains Problem 4, 5, and 6.
Check out Part 1 and Part 3 for the rest of the exam problems.


Problem 4. Let
\[\mathbf{a}_1=\begin{bmatrix}
1 \\
2 \\
3
\end{bmatrix}, \mathbf{a}_2=\begin{bmatrix}
2 \\
-1 \\
4
\end{bmatrix}, \mathbf{b}=\begin{bmatrix}
0 \\
a \\
2
\end{bmatrix}.\]

Find all the values for $a$ so that the vector $\mathbf{b}$ is a linear combination of vectors $\mathbf{a}_1$ and $\mathbf{a}_2$.


Problem 5.
Find the inverse matrix of
\[A=\begin{bmatrix}
0 & 0 & 2 & 0 \\
0 &1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
1 & 0 & 0 & 1
\end{bmatrix}\] if it exists. If you think there is no inverse matrix of $A$, then give a reason.


Problem 6.
Consider the system of linear equations
\begin{align*}
3x_1+2x_2&=1\\
5x_1+3x_2&=2.
\end{align*}

(a) Find the coefficient matrix $A$ of the system.

(b) Find the inverse matrix of the coefficient matrix $A$.

(c) Using the inverse matrix of $A$, find the solution of the system.

(Linear Algebra Midterm Exam 1, the Ohio State University)
 
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Solution of Problem 4.

To be able to express the vector $\mathbf{b}$ as a linear combination of $\mathbf{a}_1$ and $\mathbf{a}_2$, there must be scalars $c_1, c_2$ such that
\[c_1\mathbf{a}+c_2\mathbf{a}_2=\mathbf{b}.\] This is equivalent to the matrix equation
\[A\begin{bmatrix}
c_1 \\
c_2
\end{bmatrix}=\mathbf{b},\] where
\[A=[\mathbf{a}_1, \mathbf{a}_2]=\begin{bmatrix}
1 & 2 \\
2 & -1 \\
3 &4
\end{bmatrix}.\] Thus, the vector $\mathbf{b}$ is a linear combination of $\mathbf{a}_1$ and $\mathbf{a}_2$ if and only if the system $A\mathbf{x}=\mathbf{b}$ is consistent.

So let us consider the augmented matrix of the system and reduce it by elementary row operations.
We have
\begin{align*}
[A\mid \mathbf{b}]&=
\left[\begin{array}{rr|r}
1 & 2 & 0 \\
2 &-1 &a \\
3 & 4 & 2
\end{array}\right] \xrightarrow[R_3-3R_1]{R_2-2R_1}
\left[\begin{array}{rr|r}
1 & 2 & 0 \\
0 &-5 &a \\
0 & -2 & 2
\end{array}\right]\\[6pt] &\xrightarrow{-\frac{1}{2}R_3}
\left[\begin{array}{rr|r}
1 & 2 & 0 \\
0 &-5 &a \\
0 & 1 & -1
\end{array}\right] \xrightarrow{R_2 \leftrightarrow R_3}
\left[\begin{array}{rr|r}
1 & 2 & 0 \\
0 & 1 & -1\\
0 &-5 & a
\end{array}\right]\\[6pt] &\xrightarrow[R_3+5R_2]{R_1-2R_2}
\left[\begin{array}{rr|r}
1 & 0 & 2 \\
0 &1 &-1 \\
0 & 0 & a-5
\end{array}\right].
\end{align*}
If $a-5=0$, then we obtain the solution $x_1=2, x_2=-1$. Thus the system is consistent when $a=5$.

On the other hand, if $a-5 \neq 0$, then we divide the third row by $a-5$ and then the third row becomes $ \left[\begin{array}{rr|r}
0 & 0 & 1
\end{array}\right]$, which implies that the system is inconsistent (as we have $0=1$.)

Therefore, the only possible value for $a$ is $a=5$.

Note that when $a=5$, we have
\[2\mathbf{a}_1-\mathbf{a}_2=\mathbf{b}.\]

Solution of Problem 5.

We form the augmented matrix $[A\mid I]$, where $I$ is the $4\times 4$ identity matrix and apply elementary row operations as follows. We have
\begin{align*}
&[A\mid I]= \left[\begin{array}{rrrr|rrrr}
0 & 0 & 2 & 0 &1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
1 &0 & 0 & 1 & 0 & 0 & 0 &1 \\
\end{array} \right]\\[6pt] &\xrightarrow{R_1\leftrightarrow R_3}
\left[\begin{array}{rrrr|rrrr}
1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 2 & 0 &1 & 0 & 0 & 0 \\
1 &0 & 0 & 1 & 0 & 0 & 0 &1 \\
\end{array} \right] \xrightarrow[\frac{1}{2}R_3]{R_4-R_1}
\left[\begin{array}{rrrr|rrrr}
1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 &1/2 & 0 & 0 & 0 \\
0 &0 & 0 & 1 & 0 & 0 & -1 &1 \\
\end{array} \right].
\end{align*}
The left $4\times 4$ part became the identity matrix.
This means that the matrix $A$ is invertible and the inverse matrix of $A$ is given by the right $4\times 4$ part.
Hence
\[A^{-1}=\begin{bmatrix}
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
1/2 & 0 & 0 & 0 \\
0 & 0 & -1 &1 \\
\end{bmatrix}.\]

Solution of Problem 6.

(a) Find the coefficient matrix $A$ of the system.

The coefficient matrix of the system is
\[A=\begin{bmatrix}
3 & 2\\
5& 3
\end{bmatrix}.\]

(b) Find the inverse matrix of the coefficient matrix $A$.

The determinant of $A$ is given by $\det(A)=3\cdot 3 -2\cdot 5=-1\neq 0$. Thus $A$ is invertible.
Using the formula for the inverse matrix of a $2\times 2$ matrix, we obtain
\[A^{-1}=\frac{1}{-1}\begin{bmatrix}
3 & -2\\
-5& 3
\end{bmatrix}=\begin{bmatrix}
-3 & 2\\
5& -3
\end{bmatrix}.\]

(c) Using $A^{-1}$, find the solution of the system.

The system can be written as
\[A\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}=\begin{bmatrix}
1 \\
2
\end{bmatrix}.\] Multiplying it by $A^{-1}$ on right and using the identity $A^{-1}A=I$, we obtain
\begin{align*}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}&=A^{-1}\begin{bmatrix}
1 \\
2
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
-3 & 2\\
5& -3
\end{bmatrix}\begin{bmatrix}
1 \\
2
\end{bmatrix}
=\begin{bmatrix}
1 \\
-1
\end{bmatrix}.
\end{align*}
Hence the solution of the system is $x_1=1, x_2=-1$.

Go to Part 1 and Part 3

Go to Part 1 for Problem 1, 2, and 3.

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


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2 Responses

  1. 09/25/2017

    […] post contains the first three problems. Check out Part 2 and Part 3 for the rest of the exam […]

  2. 09/25/2017

    […] post is Part 3 and contains Problem 7, 8, and 9. Check out Part 1 and Part 2 for the rest of the exam […]

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Ohio State University exam problems and solutions in mathematics
Linear Algebra Midterm 1 at the Ohio State University (1/3)

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

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