Question Video: Simplifying a Matrix Expression Mathematics

Video Transcript

Given that 𝐴 and 𝐵 are invertible matrices, which of the following is the value of 𝐵 times the inverse of 𝐴𝐵? Is it option (A) matrix 𝐴, option (B) the inverse of 𝐴, option (C) the inverse of 𝐵, option (D) the transpose of 𝐴? Or is it option (E) the transpose of 𝐵?

In this question, we’re given two invertible matrices, matrix 𝐴 and matrix 𝐵. We need to use the fact that these two matrices are invertible to find an expression for 𝐵 times the inverse of 𝐴𝐵. We can do this by noting that since 𝐴 and 𝐵 are invertible matrices, we can rewrite 𝐴𝐵 inverse. We recall if we have two invertible matrices 𝐴 and 𝐵 of the same order, then the inverse of 𝐴𝐵 is equal to 𝐵 inverse times 𝐴 inverse. We can apply this to the expression given to us in the question. We get 𝐵 multiplied by 𝐵 inverse times 𝐴 inverse.

We can now notice something interesting. We have 𝐵 in our expression. And to the right of matrix 𝐵, we’re multiplying by 𝐵 inverse. So, we’re going to want to simplify this. First, we’re going to need to use the fact that matrix multiplication is associative. We recall this tells us we can evaluate the multiplication of three or more matrices in any order. In other words, for three matrices, 𝑀 one, 𝑀 two, and 𝑀 three, whose orders match up so that we know the product of these three matrices make sense, we know that 𝑀 sub one times the product of 𝑀 sub two and 𝑀 sub three will be equal to the product of 𝑀 sub one and 𝑀 sub two multiplied by 𝑀 sub three.

And we can then apply this to our expression to evaluate 𝐵 multiplied by 𝐵 inverse first. And this is very useful because we know any matrix multiplied by its inverse gives us the identity matrix of the same order. So, this leaves us with the identity matrix of order 𝑛 multiplied by 𝐴 inverse, where it’s worth noting 𝑛 is the order of the square matrix 𝐵.

And there is something worth pointing out here. We’re given in the question that we’re multiplying 𝐴 and 𝐵. And since matrices 𝐴 and 𝐵 are both square and we’re multiplying them together, we can just assume they’re square matrices of the same order. So 𝑛 is also the order of matrix 𝐴 inverse.

We can then recall multiplying a matrix by the identity matrix of the same order gives us the same matrix. So, the identity matrix multiplied by 𝐴 inverse is just 𝐴 inverse, which is our final answer, which we can see is given as option (B).

Therefore, we were able to show if 𝐴 and 𝐵 are invertible matrices, then 𝐵 times the inverse of 𝐴𝐵 is just equal to 𝐴 inverse, which was option (B).