# Question Video: Solving Equations by Finding the Determinant of a Diagonal Matrix Mathematics

Consider the equation |π₯ β 1, 0, 0 and 0, π₯Β² + π₯ + 1, 0 and 0, 0, 1| = 2. Determine the value of π₯βΆ.

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### Video Transcript

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. Determine the value of π₯ to the sixth power.

In this question, weβre given an equation involving the determinant of a three-by-three matrix which has a variable π₯. We need to use this to determine the value of π₯ to the sixth power. To do this, we need to find an expression for the determinant of this matrix. We could do this by using the definition of a determinant, expanding over one of the rows or columns. However, we can also notice that every element not on the main diagonal of this matrix is equal to zero. In other words, this three-by-three matrix is a diagonal matrix. And we can use this to find the determinant of this matrix because every diagonal matrix is an upper and lower triangular matrix. For example, every entry below the main diagonal of our matrix is zero. So, this is an example of an upper triangular matrix.

Now, we just need to recall that the determinant of any square triangular matrix is the product of the elements on its main diagonal. And itβs worth noting since all diagonal matrices are both upper and lower triangular matrices, this property will also hold for any square diagonal matrix. 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. We can then distribute over our parentheses. We get π₯ cubed plus π₯ squared plus π₯ minus π₯ squared minus π₯ minus one. And if we simplify this expression, we see itβs equal to π₯ cubed minus one. And remember, weβre told in the question this determinant is equal to two.

Therefore, our expression for the determinant is equal to two. π₯ cubed minus one is equal to two. We want to use this to find the value of π₯ to the sixth power. So, letβs start by adding one to both sides of the equation to give us that π₯ cubed is equal to three. 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.

Therefore, weβve 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.