Find an elementary matrix E such that EA = B, where A = [[3, -4], [2, 5]] and B = [[3, -4], [-4, 13]].

Steps to Solve

1. Identify Matrix A and B

The given matrices are:

[ A = \begin{bmatrix} 3 & -4 \ 2 & 5 \end{bmatrix}, \quad B = \begin{bmatrix} 3 & -4 \ -4 & 13 \end{bmatrix} ]

2. Determine the Transformation

We need to find an elementary matrix ( E ) such that ( E A = B ). This can involve either row operations or scaling. Let's analyze what changes from ( A ) to ( B ).

3. Set Up the Equation

Using the matrix equation,

[ E A = B ]

we substitute for ( E ):

[ E = B A^{-1} ]

4. Calculate the Inverse of A

Now, we find the inverse of ( A ).

The formula for the inverse of a 2x2 matrix ( \begin{bmatrix} a & b \ c & d \end{bmatrix} ) is:

[ \text{Inverse} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} ]

For ( A ), ( a = 3, b = -4, c = 2, d = 5 ):

[ ad - bc = 3 \cdot 5 - (-4) \cdot 2 = 15 + 8 = 23 ]

Thus, the inverse of ( A ) is:

[ A^{-1} = \frac{1}{23} \begin{bmatrix} 5 & 4 \ -2 & 3 \end{bmatrix} = \begin{bmatrix} \frac{5}{23} & \frac{4}{23} \ -\frac{2}{23} & \frac{3}{23} \end{bmatrix} ]

5. Compute E

Now we compute ( E ):

[ E = B A^{-1} = \begin{bmatrix} 3 & -4 \ -4 & 13 \end{bmatrix} \begin{bmatrix} \frac{5}{23} & \frac{4}{23} \ -\frac{2}{23} & \frac{3}{23} \end{bmatrix} ]

Perform the multiplication:

First row:

[ \begin{bmatrix} 3 \cdot \frac{5}{23} + (-4) \cdot \left(-\frac{2}{23}\right) & 3 \cdot \frac{4}{23} + (-4) \cdot \frac{3}{23} \end{bmatrix} ] [ = \begin{bmatrix} \frac{15}{23} + \frac{8}{23} & \frac{12}{23} - \frac{12}{23} \end{bmatrix} = \begin{bmatrix} \frac{23}{23} & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \end{bmatrix} ]

Second row:

[ \begin{bmatrix} -4 \cdot \frac{5}{23} + 13 \cdot \left(-\frac{2}{23}\right) & -4 \cdot \frac{4}{23} + 13 \cdot \frac{3}{23} \end{bmatrix} ] [ = \begin{bmatrix} -\frac{20}{23} - \frac{26}{23} & -\frac{16}{23} + \frac{39}{23} \end{bmatrix} = \begin{bmatrix} -\frac{46}{23} & \frac{23}{23} \end{bmatrix} = \begin{bmatrix} -2 & 1 \end{bmatrix} ]

Thus,

[ E = \begin{bmatrix} 1 & 0 \ -2 & 1 \end{bmatrix} ]