Video Transcript
Consider the equation the
determinant of the three-by-three matrix 𝑥 minus one, zero, zero, zero, 𝑥 squared
plus 𝑥 plus one, zero, zero, zero, one is equal to two. Determine the value of 𝑥 to the
sixth power.
In this question, we’re given an
equation involving the determinant of a three-by-three matrix which has a variable
𝑥. We need to use this to determine
the value of 𝑥 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 it’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, 𝑥 minus
one times 𝑥 squared plus 𝑥 plus one multiplied by one. We can then distribute over our
parentheses. We get 𝑥 cubed plus 𝑥 squared
plus 𝑥 minus 𝑥 squared minus 𝑥 minus one. And if we simplify this expression,
we see it’s equal to 𝑥 cubed minus one. And remember, we’re told in the
question this determinant is equal to two.
Therefore, our expression for the
determinant is equal to two. 𝑥 cubed minus one is equal to
two. We want to use this to find the
value of 𝑥 to the sixth power. So, let’s start by adding one to
both sides of the equation to give us that 𝑥 cubed is equal to three. And then, we can find an expression
for 𝑥 to the sixth power by squaring both sides of our equation, giving us that 𝑥
to the sixth power is equal to nine, which is our final answer.
Therefore, we’ve shown if the
determinant of the three-by-three matrix 𝑥 minus one, zero, zero, zero, 𝑥 squared
plus 𝑥 plus one, zero, zero, zero, one is equal to two, then the value of 𝑥 to the
sixth power is nine.
