Question Video: Solving a Matrix Equation by Finding the Inverse Mathematics • 10th Grade

Video Transcript

Given that 𝐵 is equal to negative two, negative five, negative six, negative 10, 𝐴 multiplied by 𝐵 is equal to 𝐼, where 𝐼 is the identity matrix, find the matrix 𝐴.

Let’s begin by using the equation 𝐴 multiplied by 𝐵 is equal to the identity matrix to form an equation for which 𝐴 is the subject. To make 𝐴 the subject, we’ll need to multiply both sides by the inverse of 𝐵. Since the inverse of 𝐵 multiplied by 𝐵 is simply the identity matrix, multiplying both sides by the inverse of 𝐵 gives 𝐴 is equal to the inverse of 𝐵 multiplied by 𝐼. Since 𝐼 is simply the identity matrix, 𝐴 must be equal to the inverse of 𝐵.

So we’ll need to work out the multiplicative inverse of 𝐵. For two-by-two matrix 𝐵, where 𝐵 is given by 𝑎, 𝑏, 𝑐, 𝑑, its inverse is given by the formula one over the determinant of 𝐵 multiplied by 𝑑, negative 𝑏, negative 𝑐, 𝑎, where the determinant is 𝑎, 𝑑 minus 𝑏, 𝑐.

Notice this means that if the determinant of the matrix 𝐵 is zero, the inverse does not exist since one over the determinant of 𝐵 is one over zero, which is undefined. Let’s begin then by substituting each of the individual elements of 𝐵 into our formula for the determinant.

𝑎 multiplied by 𝑑 is negative two multiplied by negative 10. And 𝑏 multiplied by 𝑐 is negative five multiplied by negative six. Negative two multiplied by negative 10 is 20 and negative five multiplied by negative six is 30. 20 minus 30 is negative 10. So the determinant of our matrix 𝐵 is negative 10.

Now, we’ll substitute all of this into the formula for the inverse of 𝐵. Remember we switch the elements 𝑎 and 𝑑. Those are the elements in the top-left- and bottom-right-hand corner of our two-by-two matrix. We switched negative 10 with a negative two. We multiply both 𝑏 and 𝑐, which are the elements in the top-right-hand corner and the bottom left by negative one, essentially, changing the sign.

The inverse of our matrix becomes negative one tenth all multiplied by negative 10, five, six, negative two. Finally, we can multiply every element of this matrix by negative one tenth to give us one, negative a half, negative three-fifths, and one-fifth.

Earlier, we said that 𝐴 was equal to the inverse of 𝐵. So our matrix 𝐴 then is given by one, negative a half, negative three-fifths, one-fifth.