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In Exercises 29–32, find the elementary row operation that transforms the first matrix into the second, and then find the reverse row operation that transforms the second matrix into the first.$$\left[\begin{array}{rrrr}{1} & {-2} & {1} & {0} \\ {0} & {5} & {-2} & {8} \\ {4} & {-1} & {3} & {-6}\end{array}\right],\left[\begin{array}{rrrr}{1} & {-2} & {1} & {0} \\ {0} & {5} & {-2} & {8} \\ {0} & {7} & {-1} & {-6}\end{array}\right]$$

Add $-4 R 1$ to $R 3$ to convert the 1 st matrix to the 2 ndAdd $4 R 1$ to $R 3$ to convert the 2 nd matrix to the 1 st.

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 1

Systems of Linear Equations

Introduction to Matrices

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So for this problem, were asked to look at two major sees and find what transformation would take. One. The left matrix to the right, in the right matrix to the left. So going from left to right, we can see that we have this row here or this column here that's 100 And that might be something with really desirable for us and the way that we could have gotten that would have been taking this third row and subtract King four times the first row. And indeed, if we do that, we'll see that That would also give us a negative one. Plus four times two is eat will give us seven, uh, three AA minus four times one would give us negative one and a minus six minus four times zero would weave us at negative six. So in order to go the other direction, we would just do the backwards of that. So we would add four times the first row to the fourth row, and that would give us before back, and it would give us a negative one back. You guys are three back and once again that negative six is unchanged because we have a zero above it

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