# 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*)

Add to solve later

## 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.

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

Add to solve later

## 7 Responses

[…] Problem 3 and its solution: Matrix operations (transpose and inverse matrices) […]

[…] Problem 3 and its solution: Matrix operations (transpose and inverse matrices) […]

[…] Problem 3 and its solution: Matrix operations (transpose and inverse matrices) […]

[…] Problem 3 and its solution: Matrix operations (transpose and inverse matrices) […]

[…] Problem 3 and its solution: Matrix operations (transpose and inverse matrices) […]

[…] Problem 3 and its solution: Matrix operations (transpose and inverse matrices) […]

[…] Problem 3 and its solution: Matrix operations (transpose and inverse matrices) […]