# Compute and Simplify the Matrix Expression Including Transpose and Inverse Matrices

## Problem 297

Let $A, B, C$ be the following $3\times 3$ matrices.
$A=\begin{bmatrix} 1 & 2 & 3 \\ 4 &5 &6 \\ 7 & 8 & 9 \end{bmatrix}, B=\begin{bmatrix} 1 & 0 & 1 \\ 0 &3 &0 \\ 1 & 0 & 5 \end{bmatrix}, C=\begin{bmatrix} -1 & 0\ & 1 \\ 0 &5 &6 \\ 3 & 0 & 1 \end{bmatrix}.$ Then compute and simplify the following expression.
$(A^{\trans}-B)^{\trans}+C(B^{-1}C)^{-1}.$

(The Ohio State University, Linear Algebra Midterm Exam Problem)

## Solution.

We first use the properties of transpose matrices and inverse matrices and simplify the expression.
Note that we have
$(A^{\trans}-B)^{\trans}=(A^{\trans})^{\trans}-B^{\trans}=A-B$ since the double transpose $(A^{\trans})^{\trans}=A$ and $B$ is a symmetric matrix.

Also, note that we have
$(B^{-1}C)^{-1}=C^{-1}(B^{-1})^{-1}=C^{-1}B$ since $(B^{-1})^{-1}=B$. Care must be taken when you distribute the inverse sign in the first equality. We needed to switch the order of the product.
Then we have
$C(B^{-1}C)^{-1}=CC^{-1}B=IB=B,$ where $I$ is the $3\times 3$ identity matrix.

Therefore, the given expression can be simplified into
$(A^{\trans}-B)^{\trans}+C(B^{-1}C)^{-1}=A-B+B=A.$ Hence we have
$(A^{\trans}-B)^{\trans}+C(B^{-1}C)^{-1}=A=\begin{bmatrix} 1 & 2 & 3 \\ 4 &5 &6 \\ 7 & 8 & 9 \end{bmatrix}.$

## Comment.

This is one of the midterm exam 1 problems of linear algebra (Math 2568) at the Ohio State University.

### Common Mistakes

Here are two common mistakes.

The first one is
$(B^{-1}C)^{-1}=(B^{-1})^{-1}C^{-1} \quad \text{(This is wrong!!)}.$ Note that in general, we have
$(AB)^{-1}=B^{-1}A^{-1} \quad \text{(Note the order of products)}.$

The second common mistake is that
$CBC^{-1}=CC^{-1}B=B \quad \text{(The first equality is wrong!!)}.$ Recall that in general matrix multiplication is not commutative, meaning that
$AB\neq BA.$

And surprisingly, if you combine these two mistakes, you get the right answer.
$C(B^{-1}C)^{-1}=CBC^{-1}=B. \quad \text{(This is wrong!!)}.$

However, these mistakes show that you didn’t understand matrix operations including transpose and inverse matrices.

## Midterm 1 problems and solutions

The following list is the problems and solutions/proofs of midterm exam 1 of linear algebra at the Ohio State University in Spring 2017.

1. Problem 1 and its solution: Possibilities for the solution set of a system of linear equations
2. Problem 2 and its solution: The vector form of the general solution of a system
3. Problem 3 and its solution (The current page): Matrix operations (transpose and inverse matrices)
4. Problem 4 and its solution: Linear combination
5. Problem 5 and its solution: Inverse matrix
6. Problem 6 and its solution: Nonsingular matrix satisfying a relation
7. Problem 7 and its solution: Solve a system by the inverse matrix
8. Problem 8 and its solution:A proof problem about nonsingular matrix

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