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Hello.
00:01
So here we have this system of equations.
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We have x minus y plus z equal to negative 4, 2x minus 3y plus 4 z equals negative 15, and 5x plus y minus 2 z equal to 12.
00:15
So to use kramer's rule, you want to find the determinant of the coefficient matrix.
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So d is equal to the determinant.
00:23
We just take the coefficients on all of our variables.
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So we have the determinant of this matrix here.
00:28
Now we have a determinant of a 3x3 matrix.
00:30
So we can go ahead and expand along the first column.
00:34
So we just have one times the determinant of its minor.
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We just cross out the row and column where one is in and look at the corresponding 2x2 matrix.
00:44
And then we have minus, well, the next entry times the determinant of its minor, but the next entry is negative.
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So we have minus a negative 1, which is a plus 1 times its determinant, and then plus one times its minor times its determinant of its minor here.
01:04
Okay, so what we have here, this is going to be equal to, well, one times we get negative three times negative two, which is a negative or positive six minus one times four.
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So that's just six minus four.
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And then we get a plus one times while two times negative two is a negative four.
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And then minus five times four.
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So that's minus 20.
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And then plus one times two times one is two, minus five times a negative three.
01:40
So that's minus a negative 15, which becomes plus 15.
01:44
Okay, so this becomes what? it becomes just equal to one times six minus four.
01:49
That's just one times two, which is two.
01:52
And then we get a plus, well, one times minus four, minus 20 is minus 24.
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So we have two minus 24.
01:59
And then we get a plus one times 17.
02:03
So we get a plus 17.
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And this works out to be equal to negative 5.
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So d here is negative 5, which is not equal to 0.
02:13
So since the determinant here is not equal to 0, yes, we can use kramer's rule.
02:18
Kramer's rule just says that, well, x is equal to d subx over d.
02:27
And then every variable, y is equal to d sub y over d, and in this case, z is equal to d sub z over d...