00:02
Using gaussian elimination we will find a row echelon form not reduced row echelon form of the augmented matrix for the following system and then use it to determine for which value of a the following system has infinitely many solutions.
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The system is x -2y plus 6z equals 1 and x plus 5y plus z equals negative 13 and 2x minus 3y plus 8 times z equals 0.
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So we write down the augmented matrix for this linear system and that is the matrix where we put together side by side the confusion matrix on the right hand side forming a unique matrix.
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So the confusion matrix is 1, negative 2, 6, 1, 5, 1, 2, negative 3, 8 and the last column of this augmented matrix is the right hand side vector 1, negative 13, 0.
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And this is a 3 by 4 matrix so it is a unique matrix but we recognize that the first 3 rows and 3 columns represent the confusion matrix while the last column of the matrix is the right hand side vector of the linear system.
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And we draw typically this dot line to separate visually the two elements, the two parts, confusion matrix on the left and right hand side on the right.
01:49
Good, so now we apply elementary row operations in order to transform this augmented matrix into a row echelon form.
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And those elementary operations that can be performed to get that form are linear combinations of the rows or shifting of rows.
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Let's say the first thing here would be to put zeros under the element on the main diagonal in position 1, 1.
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That is we want to put a zero here and a zero here.
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But before all that we also wanted to put a 1 in position 1, 1.
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We have it already in this particular example but if it was not the case, or if it were not the case, then we should divide by that number in position 1, 1 in order to put it equal to zero.
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Of course it is a row operation, we have to modify the whole row doing that.
02:52
Ok, so in order to put a zero here, all we have to do is to subtract row 2 minus row 1.
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And with that operation we get a zero right here.
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So let's say row 2 is going to be modified as what it has now minus row 1.
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And that row operation will lead us to the following augmented matrix.
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Since we are going to modify the second row, the first and third rows are the same.
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So 1, negative 2, 6, 1 and 2, negative 3, 8, 0.
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And we can calculate the second row using this elementary row operation.
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So we subtract the second row minus the first row and put the result in the second row.
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So 1 minus 1 is 0, 5 minus negative 2 is 7, 1 minus 6 is negative 5, and negative 13 minus 1 is negative 14.
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And we put a dot line here, a separation line.
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Good, this is the second augmented matrix which is equivalent to the first one.
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In the sense that if we write down the linear system associated to this new augmented matrix, the solutions of the two linear systems are the same.
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And that's a key thing in order to use the row -ational form of the augmented matrix.
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Good, so now we have to put a zero here in the element 2 which is the third row, first column.
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And the operation will be the row 3 has to be modified by using what it has now, minus 2 times row 1.
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You can see that we have used row 1 to put zeros under element 1, 1.
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So row 1 in this case, for these two row operations, is what we call the pivot row.
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Okay, so because we're going to modify the third row, the first and second are the same.
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So it's 1, negative 2, 6, 1, 0, 7, negative 5, negative 14.
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Look that we are using the last augmented matrix.
05:35
That is, we proceed ahead in order to modify the augmented matrix.
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We always use the last one to apply the row operation.
05:46
Okay, so the third row will be what it has now, minus 2 times the first row.
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So 2 minus 2 times 1 is 0.
05:57
Negative 3 minus 2 times negative 2 is negative 3 plus 4.
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That is 1.
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And then here we have a, so it's going to be indicated as an operation.
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A minus 2 times 6 is a minus 12.
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And i put it inside parentheses, not to be confused around the numbers, the entries.
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And then finally, 0 minus 2 times 1 is negative 2.
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And sorry here, the line is going to be not a straight line, but not to write it again.
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Okay, so we have this new augmented matrix.
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And over here, we have already zeros below the element at position 1, 1.
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Now we need to put a 1 right here.
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And we can do that several ways in this case.
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That is, we can divide the second row by 7.
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But we have a 1 here, just below where we want to put a 1.
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So we can shift the second and third rows to have already a 1 in position 2, 2.
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So we will do that.
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So the elementary row operation at this point will be row 2 interchanges with row 3.
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That is, row 2 becomes row 3 and vice versa.
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So we get the first row is the same, 1, negative 2, 6, 1.
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And the second row now is what was the third row here.
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That is 0, 1, a minus 12, and negative 2.
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And the third row is what was the second row here.
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That is 0, 7, negative 5, and negative 14.
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Okay, so we are done with the lines in a straight line.
08:14
Okay, we have that.
08:19
And now we have at position 2, 2, we have a 1.
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And that 1 helps us to put a 0 at position 3, 2.
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That is below the 2, 2 element.
08:34
Okay, so we do another row operation.
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This case will be the row 3 is going to be modified by using what it has now.
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A minus 7 times row 2.
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And we use row 2 because we have a 1 here and we know this operation, this row operation here, will produce a 0 at this position.
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But the question is what happens with the 0 we had already put on the element 1, 1? well, this row operation does not modify those 0s.
09:10
Let's see that.
09:11
So we are going to modify the third row.
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So the first and second are the same.
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So i put them down already.
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1, negative 2, 6, 1, and 0, 1, a, minus 12, negative 2.
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And now i do this row operation.
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Let's see what happens here with these two 0s here.
09:39
We are going to do 0 minus 7 times 0.
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That is 0 minus 0, so we get 0 again.
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So this operation here does not modify the 0s we had already put under the element 1, 1.
09:58
Okay, so we keep on doing the modifications.
10:03
Now we do 7 minus 7 times 1 is 7 minus 7 is 0.
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Now negative 5 here.
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I am going to put the operation right here, do some simplifications and put the result right here.
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So it is negative 5, what we have now on row 3, at position 3, 3.
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And we subtract 7 times what we have in row 2 at that same position, that is a, minus 12.
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And now we need to simplify this expression.
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We get negative 5 minus 7a plus 84.
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And now 84 minus 5 is 79, so we get 79 minus 7a.
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That is the result.
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So we put a parenthesis, 79 minus 7 times a.
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And that came from the row operation at that position.
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And finally, we have negative 14 minus 7 times negative 2.
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We have negative 14 plus 14, so we get 0 right here.
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And here we have a separator line.
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And good, we have done this other row operation.
11:39
Now, normally here, to be precise and have a row echelon form, we should put a 1 here.
11:49
In order to have zeros below all diagonal elements, which we already have zeros here, you see...