# Use a software program or a graphing utility with matrix capabilities to find the transition matrix from B to B' B=left{(1,2,7),(-1,2,0),(2,4,0)right} , B'=left{(0,2,1),(-2,1,0),(1,1,1)right}

Use a software program or a graphing utility with matrix capabilities to find the transition matrix from B to B
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Clara Reese

Step 1
Consider the given matrices:
$B=\left\{\left(1,2,7\right),\left(-1,2,0\right),\left(2,4,0\right)\right\},{B}^{\prime }=\left\{\left(0,2,1\right),\left(-2,1,0\right),\left(1,1,1\right)\right\}$
The objective is to find the transition matrix from B to B′.
Step 2
Consider the given matrices:
$B=\left\{\left(1,2,7\right),\left(-1,2,0\right),\left(2,4,0\right)\right\},{B}^{\prime }=\left\{\left(0,2,1\right),\left(-2,1,0\right),\left(1,1,1\right)\right\}$
Now,
$\left[{B}^{\prime }|B\right]=\left[\begin{array}{cccccc}0& -2& 1& 1& -1& 2\\ 2& 1& 1& 2& 2& 4\\ 1& 0& 1& 7& 0& 0\end{array}\right]$
Now, reducing it into Echelon’s form:
Interchange the first row and second row:
$\left[{B}^{\prime }|B\right]=\left[\begin{array}{cccccc}2& 1& 1& 2& 2& 4\\ 0& -2& 1& 1& -1& 2\\ 1& 0& 1& 7& 0& 0\end{array}\right]$
Multiplying the first row by $\frac{1}{2}$
$\left[{B}^{\prime }|B\right]=\left[\begin{array}{cccccc}1& \frac{1}{2}& \frac{1}{2}& 1& 1& 2\\ 0& -2& 1& 1& -1& 2\\ 1& 0& 1& 7& 0& 0\end{array}\right]$
Add -1 times the first row to the third row:
$\left[{B}^{\prime }|B\right]=\left[\begin{array}{cccccc}1& \frac{1}{2}& \frac{1}{2}& 1& 1& 2\\ 0& -2& 1& 1& -1& 2\\ 0& \frac{-1}{2}& \frac{1}{2}& 6& -1& -2\end{array}\right]$
Multiply the second row by $\frac{-1}{2}$
$\left[{B}^{\prime }|B\right]=\left[\begin{array}{cccccc}1& \frac{1}{2}& \frac{1}{2}& 1& 1& 2\\ 0& 1& -\frac{1}{2}& -\frac{1}{2}& \frac{1}{2}& -1\\ 0& \frac{-1}{2}& \frac{1}{2}& 6& -1& -2\end{array}\right]$
Add $\frac{1}{2}$-times the second row to the third row:
$\left[{B}^{\prime }|B\right]=\left[\begin{array}{cccccc}1& \frac{1}{2}& \frac{1}{2}& 1& 1& 2\\ 0& 1& -\frac{1}{2}& -\frac{1}{2}& \frac{1}{2}& -1\\ 0& 0& \frac{1}{4}& \frac{23}{4}& \frac{-3}{4}& \frac{-5}{4}\end{array}\right]$
Multiply the third row by 4 and add $\frac{1}{2}$-times the third row to the second row:
$\left[{B}^{\prime }|B\right]=\left[\begin{array}{cccccc}1& \frac{1}{2}& \frac{1}{2}& 1& 1& 2\\ 0& 1& 0& 11& -1& 6\\ 0& 0& 1& 23& -3& -5\end{array}\right]$
Add -$\frac{1}{2}$ times the third row to the first row:

Jeffrey Jordon