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Evaluate the determinant of the three-by-three matrix one, negative nine, negative six, negative eight, four, one, two, negative one, nine.
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Now when we’re looking to evaluate a determinant of a three-by-three matrix, then what we do if we take a look at the first row, and we use the first row as coefficients.
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And what we do is we multiply in turn each of these by the submatrix, so the two-by-two submatrix that’s formed when you delete the row and column that that value is in.
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It’s also worth noting that our coefficients have to follow a pattern.
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So, our first column is positive, so the coefficient is multiplied by positive one.
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The second column is negative, so multiplied by negative one.
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Third column, positive.
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So, we’re gonna use that in a second when we put it altogether.
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So, the first coefficient we’re looking at is the top-left term.
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So, it’s one because it’s in the first row.
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And we’re gonna multiply this by the determinant of this submatrix, which is formed, so the two-by-two submatrix when we delete the row and column that the one is in.
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So, it’s gonna give us one multiplied by the determinant of the two-by-two submatrix four, one, negative one, nine.
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Then, next, we’re gonna have minus negative nine.
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And that’s cause, as we said, the second column has to be negative.
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And we’ve already got negative nine as our coefficient.
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Then, multiplied by the two-by-two submatrix, again, formed when you delete the row and column that the negative nine is in.
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So, it’s the determinant of the submatrix negative eight, one, two, nine.
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Then, finally, we’re gonna have minus six multiplied by the determinant of the two-by-two submatrix negative eight, four, two, negative one.
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And it stayed as negative six as the coefficient because, as we said, this third column is positive.
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So, the sign stays the same.
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So, now the next stage is to know how to deal with the two-by-two determinant.
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Well, to find the two-by-two determinant, what we do is if we’ve got the two-by-two determinant of the matrix 𝑎, 𝑏, 𝑐, 𝑑, you multiply 𝑎 by 𝑑, so we diagonally multiply, and then subtract 𝑏 multiplied by 𝑐.
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So, first of all, we’re gonna have one multiplied by then we’ve got four multiplied by nine minus one multiplied by negative one.
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And then, we’ve got plus nine multiplied by negative eight multiplied by nine minus one multiplied by two.
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And we get that because we had minus negative nine.
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And if we subtract a negative, it turns positive.
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And then, finally, you’ve got minus six multiplied by negative eight multiplied by negative one minus four multiplied by two.
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So, if we evaluate this, we’re gonna get 37.
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And that’s because we had one multiplied by then you’ve got 36 minus negative one, which is 36, add one, which is 37.
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Then, we’re gonna get minus 666.
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And that’s because we had nine multiplied by negative 74.
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That’s cause we had negative 72 minus two, which is negative 74, which gives us negative 666.
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And then, minus zero.
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And we get that because we have negative eight multiplied by negative one, which is eight.
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Four multiplied by two, which is eight.
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Eight minus eight is just zero.
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So, this gives us a final answer of negative 629.
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So, therefore, we can say that if we evaluate the determinant of the three-by-three matrix one, negative nine, negative six, negative eight, four, one, two, negative one, nine, then the result is negative 629.