WEBVTT
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Find the determinant of the matrix one, two, three; three, two, two; zero, nine, eight.
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So in this problem, what weβre asked to do is find the determinant of a three-by-three matrix.
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Now to find the determinant of a three-by-three matrix, weβve got a general rule.
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So if we take a look here, weβve got our three-by-three matrix π, π, π; π, π, π; π, β, π.
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So these vertical lines tell us that weβre looking at the determinant.
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These mean the determinant of the matrix π, π, π; π, π, π; π, β, π.
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And to find the determinant, itβs equal to π β thatβs because itβs the first term in the first row β multiplied by the determinant of the submatrix that is formed when we remove the row and column that π is in.
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So it would be the submatrix π, π, β, π, so π multiplied by the term in this submatrix.
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Then minus π.
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And thatβs because we have a pattern with our coefficients.
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So we choose each of the first terms in the first row.
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But they follow the pattern positive, negative, positive.
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So therefore, as I said, itβs minus π multiplied by the two-by-two submatrix formed if you remove the column and row that π is in.
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Plus π multiplied by the two-by-two submatrix, the determinant of this, which is π, π, π, β.
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So, great, now we know how to work it out.
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Letβs go on and work out the determinant of our matrix.
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So therefore, we can say, using our rule, the determinant of the matrix one, two, three; three, two, two; zero, nine, eight can be calculated as one multiplied by the two-by-two submatrix determinant two, two, nine, eight minus two multiplied by the determinant of the two-by-two submatrix three, two, zero, eight plus three multiplied by the determinant of the submatrix three, two, zero, nine.
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And Iβm just gonna look at the third section just to remind us for how we found that.
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So we chose the three because thatβs the third term in the top row.
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And itβs positive because our pattern says that the first column is positive, second column is negative, third column is positive, et cetera.
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And we found the submatrix three, two, zero, nine by deleting the row and column that the three was in.
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And then we had three, two, zero, nine left.
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Okay, great.
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So now letβs find out the value of this.
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So to calculate the value of our determinant of a three-by-three matrix, weβre gonna need to find out the determinants of our two-by-two submatrices.
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The way we do that is by using this general rule.
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If we have the determinant of the matrix π, π, π, π, this is equal to π multiplied by π minus π multiplied by π.
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So weβre gonna use that to find the value of the determinant of our three-by-three matrix.
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So first of all, weβre gonna have one multiplied by.
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Then weβve got two multiplied by eight β thatβs cause itβs our π and our π β minus two multiplied by nine, our π and our π.
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And then weβre gonna have minus two multiplied by three multiplied by eight minus two multiplied by zero and then finally plus three multiplied by three multiplied by nine minus two multiplied by zero.
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Okay, great.
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So weβve now got in a form where we can now just carry on and work out the value of our determinant.
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So this is gonna give us negative two.
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And thatβs because two multiplied by eight is 16, 16 minus 18, because two multiplied by nine is 18, gives us negative two.
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And this multiplied by one is just negative two.
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So itβs negative two minus 48 plus 81, which is equal to 31.
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So therefore, we can say that the determinant of the matrix one, two, three; three, two, two; zero, nine, eight is equal to 31.