Video Transcript
Given that π΄ is equal to π₯ cubed,
negative π₯ cubed π¦ cubed, zero, and π¦ cubed, find the inverse of π΄.
For a two-by-two matrix π΄, which is
equal to π, π, π, π, its inverse is given by the formula one over the
determinant of π multiplied by π, negative π, negative π, π, where the
determinant of π΄ is given by ππ minus ππ.
Notice, this means that if the
determinant of the matrix is zero, the inverse does not exist, since one over the
determinant of π΄ is one over zero, which is undefined. There is no multiplicative inverse
for π΄ if the determinant of π΄ is zero.
So letβs first substitute the
values of π, π, π, and π into our formula for the determinant of π΄. π multiplied by π is π₯ cubed
multiplied by π¦ cubed, and π multiplied by π is negative π₯ cubed π¦ cubed
multiplied by zero. Negative π₯ cubed π¦ cubed
multiplied by zero is simply zero, so the determinant of π΄ is π₯ cubed π¦ cubed.
Now that we know the determinant of
π΄ and we know itβs not equal to zero, we can substitute everything into the formula
for the inverse of a two-by-two matrix. We swap the positions of the π₯
cubed and the π¦ cubed, and we change the signs of the negative π₯ cubed π¦ cubed
and zero. The inverse of π΄ then becomes one
over π₯ cubed π¦ cubed all multiplied by π¦ cubed, π₯ cubed π¦ cubed, zero, and π₯
cubed.
Weβll now multiply each element in
this matrix by one over π₯ cubed π¦ cubed. One over π₯ cubed π¦ cubed
multiplied by π¦ cubed is one over π₯ cubed. Multiplying one over the
determinant by the element on the first row and the second column gives us one. In the second row, on the first
column, it gives us zero. And one over π₯ cubed π¦ cubed
multiplied by π₯ cubed is one over π¦ cubed. The inverse of π΄ then is one over
π₯ cubed, one, zero, and one over π¦ cubed.