# Quiz 2. The Vector Form For the General Solution / Transpose Matrices. Math 2568 Spring 2017.

## Problem 273

(a) The given matrix is the augmented matrix for a system of linear equations.
Give the vector form for the general solution.
$\left[\begin{array}{rrrrr|r} 1 & 0 & -1 & 0 &-2 & 0 \\ 0 & 1 & 2 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 \\ \end{array} \right].$

(b) Let
$A=\begin{bmatrix} 1 & 2 & 3 \\ 4 &5 &6 \end{bmatrix}, B=\begin{bmatrix} 1 & 0 & 1 \\ 0 &1 &0 \end{bmatrix}, C=\begin{bmatrix} 1 & 2\\ 0& 6 \end{bmatrix}, \mathbf{v}=\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}.$ Then compute and simplify the following expression.
$\mathbf{v}^{\trans}\left( A^{\trans}-(A-B)^{\trans}\right)C.$

## Solution.

### (a) Give the vector form for the general solution

The system of linear equation represented by the augmented matrix is
\begin{align*}
x_1-x_3-2x_5&=0\\
x_2+2x_3-x_5&=0\\
x_4+x_5=0.
\end{align*}
Solving the preceding system, we obtain
\begin{align*}
x_1&=x_3+2x_5\\
x_2&=-2x_3+x_5\\
x_4&=-x_5.
\end{align*}
Here $x_2, x_5$ are free variables and the rest are dependent variables.

Hence the general solution $\mathbf{x}$ can be expressed as
\begin{align*}
\mathbf{x}&=\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5
\end{bmatrix}=\begin{bmatrix}
x_3+2x_5 \\
-2x_3+x_5 \\
x_3 \\
-x_5 \\
x_5
\end{bmatrix}\10pt] &=\begin{bmatrix} x_3 \\ -2x_3 \\ x_3 \\ 0 \\ 0 \end{bmatrix}+\begin{bmatrix} 2x_5 \\ x_5 \\ 0 \\ -x_5 \\ x_5 \end{bmatrix}=x_3\begin{bmatrix} 1 \\ -2 \\ 1 \\ 0 \\ 0 \end{bmatrix}+x_5\begin{bmatrix} 2 \\ 1 \\ 0 \\ -1\\ 1 \end{bmatrix}. \end{align*} Therefore, the vector form for the general solution is \[\mathbf{x}=x_3\begin{bmatrix} 1 \\ -2 \\ 1 \\ 0 \\ 0 \end{bmatrix}+x_5\begin{bmatrix} 2 \\ 1 \\ 0 \\ -1\\ 1 \end{bmatrix}.

### (b) Matrix product and transpose

First of all, note that by the property of the transpose we have
$(A-B)^{\trans}=A^{\trans}-B^{\trans}.$ Hence we can simply the middle part:
\begin{align*}
A^{\trans}-(A-B)^{\trans}&= A^{\trans}-(A^{\trans}-B^{\trans})\\
&=A^{\trans}-A^{\trans}+B^{\trans}=B^{\trans}.
\end{align*}
Thus the expression becomes
$\mathbf{v}^{\trans}\left( A^{\trans}-(A-B)^{\trans}\right)C=\mathbf{v}^{\trans}B^{\trans}C.$ The transposes of $\mathbf{v}$ and $B$ are
$\mathbf{v}^{\trans}=\begin{bmatrix} 0 & 1 & 0 \\ \end{bmatrix} \text{ and } B^{\trans}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 1 &0 \end{bmatrix}.$ Thus we have
\begin{align*}
&\mathbf{v}^{\trans}\left( A^{\trans}-(A-B)^{\trans}\right)C=\mathbf{v}^{\trans}B^{\trans}C\\
&=\begin{bmatrix}
0 & 1 & 0 \\
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\
0 & 1 \\
1 &0
\end{bmatrix}
\begin{bmatrix}
1 & 2\\
0& 6
\end{bmatrix}\\
&=\begin{bmatrix}
0 & 1
\end{bmatrix}\begin{bmatrix}
1 & 2\\
0& 6
\end{bmatrix}
=\begin{bmatrix}
0 & 6
\end{bmatrix}.
\end{align*}

In conclusion, we have obtained
$\mathbf{v}^{\trans}\left( A^{\trans}-(A-B)^{\trans}\right)C=\begin{bmatrix} 0 & 6 \end{bmatrix}.$

## Comment.

These are Quiz 2 problems for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017.

### List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017

There were 13 weekly quizzes. Here is the list of links to the quiz problems and solutions.

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