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)
Add to solve later
Sponsored Links
Contents
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.
Add to solve later
Sponsored Links
2 Responses
[…] post contains the first three problems. Check out Part 2 and Part 3 for the rest of the exam […]
[…] post is Part 3 and contains Problem 7, 8, and 9. Check out Part 1 and Part 2 for the rest of the exam […]