Linear Algebra (Math 220)
Midterm Test Solutions

March 18, 2008

1. Find the reduced row echelon forms of the following matrices: $$\text{(a)}\left(\begin{array}{cc}\hfill 4& \hfill -3\\ \hfill 3& \hfill -2\end{array}\right)\text{(b)}\left(\begin{array}{ccc}\hfill 2& \hfill -6& \hfill 0\\ \hfill -3& \hfill 9& \hfill 1\end{array}\right)\text{.}$$

   Response. $$\text{(a)}{R}_1\to R_1-R_2\text{:}\left(\begin{array}{cc}\hfill 1& \hfill -1\\ \hfill 3& \hfill -2\end{array}\right)R_2\to R_2-3R_1\text{:}\left(\begin{array}{cc}\hfill 1& \hfill -1\\ \hfill 0& \hfill 1\end{array}\right)R_1\to R_1+R_2\text{:}\left(\begin{array}{cc}\hfill 1& \hfill 0\\ \hfill 0& \hfill 1\end{array}\right)$$ $$\text{(b)}{R}_1\to \left(1⁄2\right)R_1\text{:}\left(\begin{array}{ccc}\hfill 1& \hfill -3& \hfill 0\\ \hfill -3& \hfill 9& \hfill 1\end{array}\right)R_2\to R_2+3R_1\text{:}\left(\begin{array}{ccc}\hfill 1& \hfill -3& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1\end{array}\right)$$

2. Let $f$ be the linear function from ${\mathbf{R}}^4$ to ${\mathbf{R}}^4$ given by $f\left(x\right)=Mx$ where $M$ is the $4\times 4$ matrix $$M=\left(\begin{array}{cccc}\hfill 1& \hfill 0& \hfill 1& \hfill 0\\ \hfill -2& \hfill 3& \hfill 0& \hfill -1\\ \hfill 0& \hfill 1& \hfill 0& \hfill -1\\ \hfill 1& \hfill 0& \hfill 1& \hfill 0\end{array}\right)\text{.}$$ Find $f\left(x\right)$ when $x$ is: $$\text{(a)}\left(\begin{array}{c}\hfill 1\\ \hfill -1\\ \hfill 0\\ \hfill 0\end{array}\right)\text{(b)}\left(\begin{array}{c}\hfill 0\\ \hfill 1\\ \hfill -1\\ \hfill 1\end{array}\right)\text{.}$$

   Response. In each case multiply $M$ by $x$ to get: $$\text{(a)}\left(\begin{array}{c}\hfill 1\\ \hfill -5\\ \hfill -1\\ \hfill 1\end{array}\right)\text{(b)}\left(\begin{array}{c}\hfill -1\\ \hfill 2\\ \hfill 0\\ \hfill -1\end{array}\right)$$

3. Let $M$ be the $3\times 3$ matrix $$\left(\begin{array}{ccc}\hfill 0& \hfill 2& \hfill 4\\ \hfill 1& \hfill 0& \hfill 1\\ \hfill 3& \hfill 1& \hfill 0\end{array}\right)\text{.}$$ Find the inverse of $M\text{.}$

   Response. $$\left(\begin{array}{cccccc}\hfill 0& \hfill 2& \hfill 4& \hfill 1& \hfill 0& \hfill 0\\ \hfill 1& \hfill 0& \hfill 1& \hfill 0& \hfill 1& \hfill 0\\ \hfill 3& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 1\end{array}\right)R_1\leftrightarrow R_2\text{:}\left(\begin{array}{cccccc}\hfill 1& \hfill 0& \hfill 1& \hfill 0& \hfill 1& \hfill 0\\ \hfill 0& \hfill 2& \hfill 4& \hfill 1& \hfill 0& \hfill 0\\ \hfill 3& \hfill 1& \hfill 0& \hfill 0& \hfill 0& \hfill 1\end{array}\right)$$ $$\left.\begin{array}{c}\hfill R_2\to \left(1⁄2\right)R_2\hfill \\ \hfill R_3\to R_3-3R_1\hfill \end{array}\right\}\text{:}\left(\begin{array}{cccccc}\hfill 1& \hfill 0& \hfill 1& \hfill 0& \hfill 1& \hfill 0\\ \hfill 0& \hfill 1& \hfill 2& \hfill 1⁄2& \hfill 0& \hfill 0\\ \hfill 0& \hfill 1& \hfill -3& \hfill 0& \hfill -3& \hfill 1\end{array}\right)R_3\to R_3-R_2\text{:}\left(\begin{array}{cccccc}\hfill 1& \hfill 0& \hfill 1& \hfill 0& \hfill 1& \hfill 0\\ \hfill 0& \hfill 1& \hfill 2& \hfill 1⁄2& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill -5& \hfill -1⁄2& \hfill -3& \hfill 1\end{array}\right)$$ $$R_3\to -\left(1⁄5\right)R_3\text{:}\left(\begin{array}{cccccc}\hfill 1& \hfill 0& \hfill 1& \hfill 0& \hfill 1& \hfill 0\\ \hfill 0& \hfill 1& \hfill 2& \hfill 1⁄2& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1& \hfill 1⁄10& \hfill 3⁄5& \hfill -1⁄5\end{array}\right)\left.\begin{array}{c}\hfill R_1\to R_1-R_3\hfill \\ \hfill R_2\to R_2-2R_3\hfill \end{array}\right\}\text{:}\left(\begin{array}{cccccc}\hfill 1& \hfill 0& \hfill 0& \hfill -1⁄10& \hfill 2⁄5& \hfill 1⁄5\\ \hfill 0& \hfill 1& \hfill 0& \hfill 3⁄10& \hfill -6⁄5& \hfill 2⁄5\\ \hfill 0& \hfill 0& \hfill 1& \hfill 1⁄10& \hfill 3⁄5& \hfill -1⁄5\end{array}\right)$$ Hence, $$M^{-1}=\left(\begin{array}{ccc}\hfill -1⁄10& \hfill 2⁄5& \hfill 1⁄5\\ \hfill 3⁄10& \hfill -6⁄5& \hfill 2⁄5\\ \hfill 1⁄10& \hfill 3⁄5& \hfill -1⁄5\end{array}\right)\text{.}$$

4. Let $g$ be the linear function from ${\mathbf{R}}^3$ to ${\mathbf{R}}^3$ that is defined by $$g\left(x\right)=\left(\begin{array}{ccc}\hfill 1& \hfill 10& \hfill 22\\ \hfill 1& \hfill -2& \hfill -4\\ \hfill 2& \hfill 2& \hfill 5\end{array}\right)\left(\begin{array}{c}\hfill x_1\\ \hfill x_2\\ \hfill x_3\end{array}\right)\text{.}$$

   1. Find a parametric representation of, or a basis for, the kernel of $g\text{.}$

   2. Find one or more equations in three variables that characterize the image of $g\text{.}$

   Response. Manuever a generic augmented matrix so that its first 3 columns are brought to reduced row echelon form: $$\left.\begin{array}{c}\hfill R_2\to R_2-R_1\hfill \\ \hfill R_3\to R_3-2R_1\hfill \end{array}\right\}\text{:}\left(\begin{array}{cccc}\hfill 1& \hfill 10& \hfill 22& \hfill y_1\\ \hfill 0& \hfill -12& \hfill -26& \hfill y_2-y_1\\ \hfill 0& \hfill -18& \hfill -39& \hfill y_3-2y_1\end{array}\right)R_2\to -\left(1⁄12\right)R_2\text{:}\left(\begin{array}{cccc}\hfill 1& \hfill 10& \hfill 22& \hfill y_1\\ \hfill 0& \hfill 1& \hfill 13⁄6& \hfill \left(y_1-y_2\right)⁄12\\ \hfill 0& \hfill -18& \hfill -39& \hfill y_3-2y_1\end{array}\right)$$ $$R_3\to R_3+18R_2\text{:}\left(\begin{array}{cccc}\hfill 1& \hfill 10& \hfill 22& \hfill y_1\\ \hfill 0& \hfill 1& \hfill 13⁄6& \hfill \left(y_1-y_2\right)⁄12\\ \hfill 0& \hfill 0& \hfill 0& \hfill \left(2y_3-3y_2-y_1\right)⁄2\end{array}\right)R_1\to R_1-10R_2\text{:}\left(\begin{array}{cccc}\hfill 1& \hfill 0& \hfill 1⁄3& \hfill \left(y_1+5y_2\right)⁄6\\ \hfill 0& \hfill 1& \hfill 13⁄6& \hfill \left(y_1-y_2\right)⁄12\\ \hfill 0& \hfill 0& \hfill 0& \hfill \left(2y_3-3y_2-y_1\right)⁄2\end{array}\right)$$

   1. The matrix has rank $2$. $y_3$ may be used as a parameter for its null space (the kernel of $g$): $$t\left(\begin{array}{c}\hfill -1⁄3\\ \hfill -13⁄6\\ \hfill 1\end{array}\right)$$

   2. The image of $g$ is the column space of the matrix. An equation for it is: $$y_1=2y_3-3y_2$$

5. Let ${\mathcal{P}}_2$ be the vector space of all polynomials $a{t}^2+bt+c$ having degree at most $2$ in the variable $t\text{.}$ Define ${\mathcal{P}}_2\stackrel{\varphi }{\to }{\mathcal{P}}_2$ by $$\varphi \left(f\left(t\right)\right)=f^{\prime \prime }\left(t\right)-3f^{\prime }\left(t\right)+2f\left(t\right)\text{.}$$ Find the matrix of $\varphi$ relative to the basis $\mathbf{v}$ of ${\mathcal{P}}_2$ (playing the role of basis for both the domain and the target of $\varphi$) given by $$\mathbf{v}=\left\{1,t,t^2\right\}\text{.}$$

   Response. One computes $\varphi$ at each of the three polynomials in $\mathbf{v}$: $$\begin{array}{cccc}& \hfill \varphi \left(1\right)& \hfill =\hfill & 2\hfill \\ & \hfill \varphi \left(t\right)& \hfill =\hfill & -3+2t\hfill \\ & \hfill \varphi \left(t^2\right)& \hfill =\hfill & 2-6t+2t^2\hfill \end{array}$$ The coefficient vectors of these $\varphi$ values relative to the basis $\mathbf{v}$ (in its role as basis of the target) are: $$\left(\begin{array}{c}\hfill 2\\ \hfill 0\\ \hfill 0\end{array}\right),\left(\begin{array}{c}\hfill -3\\ \hfill 2\\ \hfill 0\end{array}\right),\left(\begin{array}{c}\hfill 2\\ \hfill -6\\ \hfill 2\end{array}\right)\text{.}$$ Hence, the matrix of $\varphi$ with respect to $\mathbf{v}$ is: $$\left(\begin{array}{ccc}\hfill 2& \hfill -3& \hfill 2\\ \hfill 0& \hfill 2& \hfill -6\\ \hfill 0& \hfill 0& \hfill 2\end{array}\right)\text{.}$$