# 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)

## 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.

### 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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##### 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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