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

## 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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