# Let T : M 2 x 2 </mrow> </msub> &#x2192;<!-- \rightarrow

Let $T:{M}_{2x2}\to {\mathbb{R}}^{3}$ have matrix $\left[T{\right]}_{B,A}=\left[\begin{array}{cccc}1& 2& 0& 1\\ 0& 1& 1& 0\\ 1& 1& -1& -1\end{array}\right]$ relative to $A=\left\{\left[\begin{array}{cc}2& 0\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 3\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 5& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 0& 6\end{array}\right]\right\}$ and $\beta =\left\{\left(1,1,1\right),\left(1,2,3\right),\left(1,4,9\right)\right\}$. Find the matrix of T relative to the bases ${A}^{\prime }=\left\{\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 4\\ 0& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 2& 0\end{array}\right],\left[\begin{array}{cc}0& 0\\ 0& 7\end{array}\right]\right\}$ and ${\beta }^{\prime }=\left\{\left(1,1,1\right),\left(1,0,0\right),\left(1,1,0\right)\right\}$
You can still ask an expert for help

## Want to know more about Matrix transformations?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Kaya Kemp
First of all, notice that ${M}_{2x2}\cong {R}^{4}$ with $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\to \left(a,b,c,d\right)$.
Your first step is the right way to go, but I would be careful with the notation. Let $\alpha$ be the standard basis auf ${\mathbb{R}}^{3}$, then
$T\left(\left[\begin{array}{cc}2& 0\\ 0& 0\end{array}\right]\right)=\left(1,0,1{\right)}_{B}=\left(2,5,10{\right)}_{\alpha }=\left(10,-3,-5{\right)}_{{B}^{\prime }}$
$T\left(\left[\begin{array}{cc}0& 3\\ 0& 0\end{array}\right]\right)=\left(2,1,1{\right)}_{B}=\left(4,8,9{\right)}_{\alpha }=\left(9,-4,-1{\right)}_{{B}^{\prime }}$
$T\left(\left[\begin{array}{cc}0& 0\\ 5& 0\end{array}\right]\right)=\left(0,1,-1{\right)}_{B}=\left(2,5,12{\right)}_{\alpha }=\left(12,-3,-7{\right)}_{{B}^{\prime }}$
$T\left(\left[\begin{array}{cc}0& 0\\ 0& 6\end{array}\right]\right)=\left(1,0,-1{\right)}_{B}=\left(0,-3,-8{\right)}_{\alpha }=\left(-8,3,5{\right)}_{{B}^{\prime }}$
Thus, we have
$\left[T{\right]}_{{B}^{\prime },A}=\left[\begin{array}{cccc}10& 9& 12& -8\\ -3& -4& -3& 3\\ -5& -1& -7& 5\end{array}\right]$
Now we are almost done. The new basis ${A}^{\prime }$ is good to handle because every new basis vector is a scalar multiple of a vector of $A$. For example the first basis vector of ${A}^{\prime }$ is half of its counterpart of $A$ (So $\left(2,0,0,0{\right)}_{{A}^{\prime }}=\left(1,0,0,0{\right)}_{A}$). So if you map $\left(1,0,0,0{\right)}_{{A}^{\prime }}$ with our wanted matrix, it will have the same result as if we map $\left(\frac{1}{2},0,0,0{\right)}_{A}$ into $\left[T{\right]}_{{B}^{\prime },A}$ (because they both map into the vector space generated by ${B}^{\prime }$) Thus, the first column of $\left[T{\right]}_{{B}^{\prime },{A}^{\prime }}$. You can get the other columns with the same method. The solution is:
$\left[T{\right]}_{{B}^{\prime },{A}^{\prime }}==\left[\begin{array}{cccc}5& 9\frac{4}{3}& 12\frac{2}{5}& -8\frac{7}{6}\\ -\frac{3}{2}& -4\frac{4}{3}& -3\frac{2}{5}& 3\frac{7}{6}\\ \frac{5}{2}& -1\frac{4}{3}& -7\frac{2}{5}& 5\frac{7}{6}\end{array}\right]$