Matrix Transformation Examples and Practice Problems

Find a nonzero vector in Nul A and a nonzero vector in Col A for the matrix A below.
$A=\left[\begin{array}{cccc}2& 3& 5& -9\\ -8& -9& -11& 21\\ 4& -3& -17& 27\end{array}\right]$
In Nul A, locate a nonzero vector.
$A=\left[\begin{array}{c}-3\\ 2\\ 0\\ 1\end{array}\right]$

Determine if the columns of the matrix form a linearly independent set. Justify each answer.
$\left[\begin{array}{cccc}1& 4& -3& 0\\ -2& -7& 5& 1\\ -4& -5& 7& 5\end{array}\right]$

Determine the value(s) of h such that the matrix is the augmented matrix of a consistent linear system.
$\left[\begin{array}{ccc}1& 3& -2\\ -4& h& 8\end{array}\right]$

Using T defined by T(x)=Ax, find a vector x whose image under T is b, and determine whether x is unique.
$A\left[\begin{array}{ccc}1& -5& -7\\ -3& 7& 5\end{array}\right],B=\left[\begin{array}{c}-2\\ -2\end{array}\right]$

T must be a linear transformation, we assume. Identify the T standard matrix. 𝟚𝟚${\displaystyle T:{\mathbb{R}}^{\mathbb{2}}\to {\mathbb{R}}^{\mathbb{2}}}$ is a vertical shear transformation that maps $e_{1}into{e}_1-2e_2$ but leaves the vector ${\displaystyle e_2}$ unchanged.

Let W be the subspace spanned by the given vectors. Find a basis for
$w_1=\left[\begin{array}{c}1\\ -1\\ 3\\ -2\end{array}\right],w_2=\left[\begin{array}{c}0\\ 1\\ -2\\ 1\end{array}\right]$

What is the area of the parallelogram whose vertices are listed? (0,0), (5,2), (6,4), (11,6)

The reduced row echelon form of a system of linear equations is given. Write the system of equations corresponding to the given matrix. Use x, y;x,y; or x, y, z;x,y,z; or ${\displaystyle x_1,x_2,x_3,x_4}$ as variables. Determine whether the system is consistent or inconsistent. If it is consistent, give the solution.
$\left[\begin{array}{cccccc}1& 0& 0& 0& |& 1\\ 0& 1& 0& 0& |& 2\\ 0& 0& 1& 2& |& 3\end{array}\right]$

Consider the linear system
${\overrightarrow{y}}^{\prime }=\left[\begin{array}{cc}-3& -2\\ 6& 4\end{array}\right]\overrightarrow{y}$
a) Find the eigenvalues and eigenvectors for the coefficient matrix
${\lambda }_{1}1,{\overrightarrow{v}}_1=\left[\begin{array}{c}-1\\ 2\end{array}\right]$ and ${\lambda }_2=0,{\overrightarrow{v}}_2=\left[\begin{array}{c}-2\\ 3\end{array}\right]$
b) For each eigenpair in the previos part, form a solution of ${\displaystyle {\overrightarrow{y}}^{\prime }=A\overrightarrow{y}}$. Use t as the independent variable in your answers.
${\overrightarrow{y}}_1(t)=\left[\begin{array}{cc}& \\ & \end{array}\right]$ and ${\overrightarrow{y}}_2(t)=\left[\begin{array}{c}-2\\ 3\end{array}\right]$
c) Does the set of solutions you found form a fundamental set (i.e., linearly independent set) of solution? No, it is not a fundamental set.

Find an orthogonal basis for the column space of each matrix.
$\left[\begin{array}{ccc}3& -5& 1\\ 1& 1& 1\\ -1& 5& -2\\ 3& -7& -8\end{array}\right]$

Use the definition of Ax to write the matrix equation as a vector equation, or vice versa.
$\left[\begin{array}{cc}7& -3\\ 2& 1\\ 9& -6\\ -3& 2\end{array}\right]\left[\begin{array}{c}-2\\ -5\end{array}\right]=\left[\begin{array}{c}1\\ -9\\ 12\\ -4\end{array}\right]$

I need to find a unique description of Nul A, namely by listing the vectors that measure the null space.
$A=\left[\begin{array}{ccccc}1& 5& -4& -3& 1\\ 0& 1& -2& 1& 0\\ 0& 0& 0& 0& 0\end{array}\right]$

Let A be an ${\displaystyle m\times n}$ matrix, and ${\displaystyle C=AB}$.
Show that:
a) $Null(B)$ is a subspace of $Null(C)$.
b) $Null(C{)}^{\mathrm{\perp }}$ is a subspace of $Null(B{)}^{\mathrm{\perp }}$ and, consequently, ${\displaystyle Col\left(CT\right)}$ is a subspace of ${\displaystyle Col\left(BT\right)}$.

T must be a linear transformation, we assume. Can u find the T standard matrix.${\displaystyle T:{\mathbb{R}}^2\to {\mathbb{R}}^4,T\left(e_1\right)=(3,1,3,1)\text{and}T\left(e_2\right)=(-5,2,0,0),where{e}_1=(1,0)\text{and}{e}_2=(0,1)}$.

Suppose that A is row equivalent to B. Find bases for the null space of A and the column space of A.
$A=\left[\begin{array}{ccccc}1& 2& -5& 11& -3\\ 2& 4& -5& 15& 2\\ 1& 2& 0& 4& 5\\ 3& 6& -5& 19& -2\end{array}\right]$
$B=\left[\begin{array}{ccccc}1& 2& 0& 4& 5\\ 0& 0& 5& -7& 8\\ 0& 0& 0& 0& -9\\ 0& 0& 0& 0& 0\end{array}\right]$