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\noindent
{\Large Math 211 \quad  Homework 4 \hfill Name: \hskip 1.5in \ ~}

\medskip
\centerline{Five points for each question unless specified otherwise.}

\medskip
\noindent{\bf 1.}
Compute $AB$ if 
$$A=\left[\matrix{4 & -3\cr -3 & 5\cr 0 & 1}\right],  \quad 
B=\left[\matrix{1&4\cr 3&-2}\right].$$

\vskip .5in


\medskip
\noindent{\bf 2.}
Construct nonzero $3\times 3$ matrices $A, B, C$ such that $AB = AC$ and $B \ne C$.

[Hint: You can construct $A, B, C$ that only has one nonzero entry in each of the matrices].



\medskip
\noindent{\bf 3.}
Consider system of equations
$$7x_1+3x_2=-9$$
$$-6x_1-3x_2=4$$

(a) Create a matrix equation of the form $A{\bf x}={\bf b}$ 
by specifying $A$ and ${\bf b}$.

(b) Compute $A^{-1}$ using the algorithm in Section 2.3.

(c) Compute ${\bf x}_0 = A^{-1}{\bf b}$ and verify that 
$A{\bf x}_0 = {\bf b}$.

\medskip 
\noindent{\bf 4.} Let $A$ be a $3\times 3$ matrix.
Determine all possible reduced row echelon form of the augmented 

matrix $[A \ I_3]$ 
and decide whether $A$ is invertible in each of the following cases.

(a) The matrix $A$ has 1 pivoting column.

(b) The matrix $A$ has 2 pivoting columns.

(c) The matrix $A$ has 3 pivoting columns.

\medskip\noindent
{\bf 5.} Suppose $A$ is invertible, and $AB = AC$. Show that $B = C$.

[Hint: Show that $A(B-C) = 0$ ensures that $B-C= 0$ if $A$ is invertible.]

\vskip .5in

\medskip
\noindent{\bf 6.}
Let 
$A = \left[\matrix{1 & -2 & 1 \cr 4 & -7 & 3 \cr -2 & 6 & -3\cr}\right]$.
Apply row reduction to the matrix  $\left[\matrix{A & I_3}\right]$ 
until we have the 

reduced row echelon form $\left[\matrix{I_3& B }\right]$,
and verify that $AB = I_3$.

\vskip .5in 

\medskip
\noindent{\bf 7.}
Let $A = \left[\matrix{1 & 1 & 1 & 0 \cr 0 & 1 & 1 & 1\cr}\right]$.


(a) Find a $4\times 2$ matrix $B$ such that $AB = I_2$.

\qquad [Hint: Solve 
${\bf b_1}, {\bf b_2} \in \IR^4$ such that $A{\bf b}_1 = {\bf e}_1 \in \IR^2$ and
$A{\bf b}_2 = {\bf e}_2 \in \IR^2$.

\smallskip
(b) Explain why there is no $4\times 2$ matrix $C$ such that $CA = I_4$.

\vskip .7in

\medskip
\noindent{\bf 8.}
Suppose $A$ is an $n\times n$ matrix such that 
the columns of $A^2$ spans $\mathbb{R}^n$. 

Show that $A$ has linearly independent columns.




\medskip
\noindent{\bf 9.} (10 points)
Let $T: \mathbb{R}^2\rightarrow \mathbb{R}^2$ be given by
$$T(x_1, x_2)=(2x_1-8x_2,-2x_1+7x_2).$$

(a) Determine $A$ such that 
$T(x_1,x_2) = (b_1,b_2)$ holds if 
$A\left[\matrix{x_1\cr x_2\cr}\right] = \left[\matrix{b_1\cr b_2\cr}\right]$,
and compute $A^{-1}$.

(b) Define $R: \mathbb{R}^2\rightarrow \mathbb{R}^2$  by
$R(x_1,x_2) = (b_1,b_2)$ holds if 
$A^{-1}\left[\matrix{x_1\cr x_2\cr}\right] = \left[\matrix{b_1\cr b_2\cr}\right]$.

(c) Verify that $R\circ T(x_1,x_2) = (x_1,x_2)$ for all $x_1, x_2 \in {\mathbb R}$.

(d)  Verify that $T\circ R(x_1,x_2) = (x_1,x_2)$ for all $x_1, x_2 \in {\mathbb R}$.
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