WEBVTT
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Consider the equation the determinant of the three-by-three matrix π₯ minus one, zero, zero, zero, π₯ squared plus π₯ plus one, zero, zero, zero, one is equal to two.
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Determine the value of π₯ to the sixth power.
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In this question, weβre given an equation involving the determinant of a three-by-three matrix which has a variable π₯.
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We need to use this to determine the value of π₯ to the sixth power.
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To do this, we need to find an expression for the determinant of this matrix.
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We could do this by using the definition of a determinant, expanding over one of the rows or columns.
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However, we can also notice that every element not on the main diagonal of this matrix is equal to zero.
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In other words, this three-by-three matrix is a diagonal matrix.
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And we can use this to find the determinant of this matrix because every diagonal matrix is an upper and lower triangular matrix.
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For example, every entry below the main diagonal of our matrix is zero.
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So, this is an example of an upper triangular matrix.
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Now, we just need to recall that the determinant of any square triangular matrix is the product of the elements on its main diagonal.
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And itβs worth noting since all diagonal matrices are both upper and lower triangular matrices, this property will also hold for any square diagonal matrix.
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Therefore, we can evaluate the determinant of this matrix by finding the product of its leading diagonal, π₯ minus one times π₯ squared plus π₯ plus one multiplied by one.
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We can then distribute over our parentheses.
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We get π₯ cubed plus π₯ squared plus π₯ minus π₯ squared minus π₯ minus one.
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And if we simplify this expression, we see itβs equal to π₯ cubed minus one.
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And remember, weβre told in the question this determinant is equal to two.
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Therefore, our expression for the determinant is equal to two.
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π₯ cubed minus one is equal to two.
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We want to use this to find the value of π₯ to the sixth power.
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So, letβs start by adding one to both sides of the equation to give us that π₯ cubed is equal to three.
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And then, we can find an expression for π₯ to the sixth power by squaring both sides of our equation, giving us that π₯ to the sixth power is equal to nine, which is our final answer.
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Therefore, weβve shown if the determinant of the three-by-three matrix π₯ minus one, zero, zero, zero, π₯ squared plus π₯ plus one, zero, zero, zero, one is equal to two, then the value of π₯ to the sixth power is nine.