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Find the inverses of the matrices in Exercises $29-32,$ if they exist. Use the algorithm introduced in this section. $$\left[\begin{array}{rrr}{1} & {0} & {-2} \\ {-3} & {1} & {4} \\ {2} & {-3} & {4}\end{array}\right]$$

$A ^ { - 1 } = \left[ \begin{array} { c c c } { 8 } & { 3 } & { 1 } \\ { 10 } & { 4 } & { 1 } \\ { 7 / 2 } & { 3 / 2 } & { 1 / 2 } \end{array} \right]$

Algebra

Chapter 2

Matrix Algebra

Section 2

The Inverse of a Matrix

Introduction to Matrices

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Lectures

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for this example, we have a three by three matrix A that's provided here for this matrix. We want to be able to determine what it's in verses. If it does indeed have one. And one way we can tell if there is an inverse is reduce A until we get to the identity matrix. This will tell us that this is an in vertical matrix. If we never make it to the identity matrix, then A is not in vertical. So here's our trouble, though with this approach we want to find the inverse so fits equal to the identity matrix. Then the next step is to find that in verse, which is a lot of trouble. So we're going to take an alternate approach. What we'll do is take a matrix a then augment with a three by three identity matrix. So if we do Rover operations and we get the identity matrix here, then we have an inverse, and it turns out the Arian verse will be in this position. So let's begin row reduction. First, this matrix will be equal to all place A, which is one negative. 32 01 negative, three negative 244 and now we begin augmenting. So the augments break will be here. And for the I three identity matrix, we have 100 010 and 001 So this is the matrix we must reduce. This matrix is of size three bike six. So it feels a little bit daunting, but it's not too bad. Really? Unless it had something like four rows to consider. First, we'll begin with this pivot position. Eliminate the negative three and the positive two below. So let me re copy Row one of this matrix 10 negative too. 100 and multiply. Row one by three. Added to row two. We'll obtain 01 Then we have a four or excuse me. Negative too. A three. A one and a zero for a next operation. Multiply row one by negative too. Added to row three. Then we'll obtain zero negative. Three eight Negative too zero and then a one. So for this pivot position were all set because we have zeros below for this divot position. We need to eliminate this entry and then call him too will be sorted out. So will say that this is row equivalent to the following first I'm going to copy Rose one end to so 10 negative too. 100 Then Row two is 01 negative. 231 and zero. And this is what our operations going to be. Multiply row two by three Add the results to row three. The result is going to be zero that we have a zero A to seven, a three and a one. So now looking at the pivots Columns one and columns to are the 1st 2 columns of the three by three identity matrix. But we still have some trouble here for this pivot position. We need to eliminate the entries above. And now, ordinarily, I would take the step of dividing this to here by two to make it a one. But that would give us, ah, whole bunch of fractions to work with. So let's make that our final step in a row operations. So if we copy row three, which is 002731 then our operations are going to be the same. Add row three to row to and will have 01 Then we have obtained zero 10 four and a one. Then add row three to Row one, the same operation and we get one 008 three and one. Now we have one final roll operation. We need to take this row and divide it by two so that we have the first portion of this matrix row equivalent to the identity matrix. So the last step is that this is roadkill in equivalent to 00 one seven halves, three halves and one halves for our third row. Then I'm just going to copy the rest. So 010 10 41 makes Row two and Ra one is 100 83 and one. So here we have the three by three Identity matrix. And don't forget we have arguments. Break here. This matrix here is the main matrix we want. Let's pause for a moment, though we've really shown two things in this row operations that a here is raw equivalent to the three by three identity matrix. This proves a inverse exists. And the second thing we've shown with these row operations is that a inverse is this matrix 831 10 41 seven halves, three halves and 1/2. So this is the method for calculating the inverse of a three by three matrix. There are other methods, but doing by hand computations. Typically, this is a fastest route.

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