Video: Finding the Inverse of a Matrix

Given that 𝐴 = π‘₯Β³, βˆ’π‘₯Β³ 𝑦³ and 0, 𝑦³, find 𝐴⁻¹.

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Video Transcript

Given that 𝐴 is equal to π‘₯ cubed, negative π‘₯ cubed 𝑦 cubed, zero, and 𝑦 cubed, find the inverse of 𝐴.

For a two-by-two matrix 𝐴, which is equal to π‘Ž, 𝑏, 𝑐, 𝑑, its inverse is given by the formula one over the determinant of π‘Ž multiplied by 𝑑, negative 𝑏, negative 𝑐, π‘Ž, where the determinant of 𝐴 is given by π‘Žπ‘‘ minus 𝑏𝑐.

Notice, this means that if the determinant of the matrix is zero, the inverse does not exist, since one over the determinant of 𝐴 is one over zero, which is undefined. There is no multiplicative inverse for 𝐴 if the determinant of 𝐴 is zero.

So let’s first substitute the values of π‘Ž, 𝑏, 𝑐, and 𝑑 into our formula for the determinant of 𝐴. π‘Ž multiplied by 𝑑 is π‘₯ cubed multiplied by 𝑦 cubed, and 𝑏 multiplied by 𝑐 is negative π‘₯ cubed 𝑦 cubed multiplied by zero. Negative π‘₯ cubed 𝑦 cubed multiplied by zero is simply zero, so the determinant of 𝐴 is π‘₯ cubed 𝑦 cubed.

Now that we know the determinant of 𝐴 and we know it’s 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 𝑦 cubed, and we change the signs of the negative π‘₯ cubed 𝑦 cubed and zero. The inverse of 𝐴 then becomes one over π‘₯ cubed 𝑦 cubed all multiplied by 𝑦 cubed, π‘₯ cubed 𝑦 cubed, zero, and π‘₯ cubed.

We’ll now multiply each element in this matrix by one over π‘₯ cubed 𝑦 cubed. One over π‘₯ cubed 𝑦 cubed multiplied by 𝑦 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 𝑦 cubed multiplied by π‘₯ cubed is one over 𝑦 cubed. The inverse of 𝐴 then is one over π‘₯ cubed, one, zero, and one over 𝑦 cubed.

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