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Evaluate the determinant of the matrix given by seven, zero, π plus one, zero, one, nine π, negative π plus one, negative four π, and negative 10.
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Finding the determinant of a three-by-three matrix is a little more complicated than finding the determinant for a two-by-two.
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So we need to be really careful carrying out each step.
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The determinant for a three-by-three matrix is found using determinants of two-by-two matrices as shown.
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This may look a little complicated.
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However, it is simply the elements of the top row multiplied by the determinant of their minors, that is, the two-by-two matrix thatβs left when we eliminate the row and the column weβre looking at.
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Letβs substitute the values from our matrix into this formula.
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Next, we need to find the determinants of the two-by-two matrices.
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Remember, we work this out by finding the product of the top-left and bottom-right elements and subtracting the product of the top-right and bottom-left elements.
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The determinant of our first two-by-two matrix is, therefore, one multiplied by negative 10 minus 9π multiplied by negative four π.
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Remember, π is the square root of negative one.
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So π squared is just negative one.
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Our expression becomes negative 10 minus 36, which is negative 46.
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We can substitute this back into our original formula.
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Now at this stage, we could work out the determinant for the second two-by-two matrix.
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However, ultimately, weβre just going to multiply it by zero, which will give a value of zero.
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Instead, we can go straight on to finding the determinant of our final two-by-two matrix.
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The determinant here is zero multiplied by negative four π minus one multiplied by negative π plus one.
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This simplifies to π minus one.
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Again, letβs substitute this back into our formula.
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And our determinant is going to be calculated by working out the product of seven and negative 46, subtracting zero, then adding the product of π plus one and π minus one.
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Seven multiplied by negative 46 is negative 322.
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And π plus one multiplied by π minus one can be calculated pretty quickly using the rule of the difference of two squares to get π squared minus one.
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Once again, π squared gives us a value of negative one.
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So our expression becomes negative 322 minus one minus one.
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This gives us negative 324.
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The determinant of this three-by-three matrix is negative 324.