Answer
See answers below
Work Step by Step
a) We are given:
$A_1=\begin{bmatrix}
1 & 1\\
0 & 1
\end{bmatrix}$
$A_2=\begin{bmatrix}
1 & 3\\
-1 & 0
\end{bmatrix}$
$A_3=\begin{bmatrix}
1 & 0\\
1 & 2
\end{bmatrix}$
$A_4=\begin{bmatrix}
0 & -1\\
2 & 3
\end{bmatrix}$
The matrixes must be
$c_1A_1+c_2A_2+c_3A_3+c_4A_4=0$
We can obtain:
$-c_1+c_2+c_3=0$
$c_1+3c_2+-c_4=0$
$-c_2+c_3+2c_4=0$
$c_1+2c_3+3c_4=0$
$\rightarrow \begin{bmatrix}
-1 & 1 & 1 & 0 | 0 \\\
1 & 3 &0 & -1 | 0\\
0 & -1 & 1 & 2 | 0\\
1 & 0 & 2 & 3 | 0
\end{bmatrix} \approx \begin{bmatrix}
-1 & 1 & 1 & 0 | 0\\\
0 & 4 &1 & -1 | 0\\
0 & -1 & 1 & 2 | 0\\
0 & 1 & 3 & 3 | 0
\end{bmatrix} \approx \begin{bmatrix}
-1 & 1 & 1 & 0 | 0 \\\
0 & 4 &1 & -1 | 0 \\
0 & 0 & 5 & 7 | 0 \\
0 & 0 & 4 & 5 | 0
\end{bmatrix} \approx \begin{bmatrix}
-1 & 1 & 1 & 0 | 0\\
0 & 4 &1 & -1 | 0\\
0 & 0 & 5 & 7 | 0\\
0 & 0 & 0 & 3 | 0
\end{bmatrix} $
$\rightarrow rank (A)=4 \rightarrow c_1=c_2=c_3=c_4 $
Hence $\{A_1,A_2,A_3,A_4\}$ is a basis for $M_2(R)$
b) Assume
$A_5=\begin{bmatrix}
5 & 6\\
7 & 8
\end{bmatrix}$
as a linear combination of the basis $\{A_1,A_2,A_3,A_4\}$
Let $c_1A_1+c_2A_2+c_3A_3+c_4A_4=A_5$
We obtain:
$\begin{bmatrix}
-1 & 1 & 1 & 0 | 5\\
1 & 3 & 0 & -1 |6 \\
0 & -1 & 1 & 2 | 7 \\
1 & 0 & 2 & 3 | 8
\end{bmatrix} \approx \begin{bmatrix}
-1 & 1 & 1 & 0 | 5\\
0 & 4 & 1 & -1 |11 \\
0 & -1 & 1 & 2 | 7\\
0 & 1 & 3 & 3 | 13
\end{bmatrix} \approx \begin{bmatrix}
-1 & 1 & 1 & 0 | 5\\
0 & 4 & 1 & -1 |11 \\
0 & 0 & 5 & 7 | 39\\
0 & 0 & 4 & 5| 20
\end{bmatrix} \approx \begin{bmatrix}
-1 & 1 & 1 & 0 | 5\\
0 & 4 & 1 & -1 |11 \\
0 & 0 & 5 & 7 | 39\\
0 & 0 & 0 & 3| 56
\end{bmatrix} $
We have:
$c_4=\frac{56}{3}$
$c_3=39-7c_4=39-7(\frac{56}{3})=\frac{-55}{3}$
$c_2=\frac{11+\frac{56}{3}-(-\frac{55}{3})}{4}=12$
$c_1=-[5-(-\frac{55}{3})-12]=\frac{-34}{3}$
Therefore, we can express the vector $\begin{bmatrix}
5 & 6\\
7 & 8
\end{bmatrix}$ as $\frac{-34}{3}A_1+12A_2-\frac{55}{3}A_3+\frac{56}{3}A_4$
