# Solve a System by the Inverse Matrix and Compute $A^{2017}\mathbf{x}$

## Problem 300

Let $A$ be the coefficient matrix of the system of linear equations

\begin{align*}

-x_1-2x_2&=1\\

2x_1+3x_2&=-1.

\end{align*}

**(a)** Solve the system by finding the inverse matrix $A^{-1}$.

**(b)** Let $\mathbf{x}=\begin{bmatrix}

x_1 \\

x_2

\end{bmatrix}$ be the solution of the system obtained in part (a).

Calculate and simplify

\[A^{2017}\mathbf{x}.\]

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

Add to solve later

Contents

## Solution.

### (a) Solve the system by finding the inverse matrix $A^{-1}$.

The coefficient matrix $A$ of the system is

\[A=\begin{bmatrix}

-1 & -2\\

2& 3

\end{bmatrix}.\]
The determinant of the matrix $A$ is $\det(A)=(-1)(3)-(-2)(2)=1\neq 0$, and thus $A$ is invertible and the inverse matrix can be found by the $2\times 2$ inverse matrix formula

\[A^{-1}=\frac{1}{ad-bc}\begin{bmatrix}

d & -b\\

-c& a

\end{bmatrix}\]
if $A=\begin{bmatrix}

a & b\\

c& d

\end{bmatrix}$ and $\det(A)=ad-bc\neq 0$.

Thus

\[A^{-1}=\begin{bmatrix}

3 & 2\\

-2& -1

\end{bmatrix}.\]
(Or you could form the augmented matrix $[A\mid I]$ and find $A^{-1}$.)

Now the given system of linear equations can be written as

\[A\begin{bmatrix}

x_1 \\

x_2

\end{bmatrix}=\begin{bmatrix}

1 \\

-1

\end{bmatrix}.\]
Multiplying by $A^{-1}$ on the left, we obtain

\[\begin{bmatrix}

x_1 \\

x_2

\end{bmatrix}=A^{-1}\begin{bmatrix}

1 \\

-1

\end{bmatrix}=\begin{bmatrix}

3 & 2\\

-2& -1

\end{bmatrix}\begin{bmatrix}

1 \\

-1

\end{bmatrix}=\begin{bmatrix}

1 \\

-1

\end{bmatrix}.\]
Thus the solution of the system is $x_1=1, x_2=-1$, or in the vector form

\[\begin{bmatrix}

x_1 \\

x_2

\end{bmatrix}=\begin{bmatrix}

1 \\

-1

\end{bmatrix}.\]

### (b) Calculate and simplify $A^{2017}\mathbf{x}$

Since $\mathbf{x}=\begin{bmatrix}

x_1 \\

x_2

\end{bmatrix}=\begin{bmatrix}

1 \\

-1

\end{bmatrix}$ is the solution of the system $A\mathbf{x}=\begin{bmatrix}

1 \\

-1

\end{bmatrix}$, we see that

\[A\mathbf{x}=\mathbf{x}.\]
From this, we have

\begin{align*}

A^2\mathbf{x}&=AA\mathbf{x}=A\mathbf{x}=\mathbf{x}\\

A^3\mathbf{x}&=AA^2\mathbf{x}=A\mathbf{x}=\mathbf{x}\\

A^4\mathbf{x}&=AA^3\mathbf{x}=A\mathbf{x}=\mathbf{x}\\

\vdots &

\end{align*}

Repeating this, we see that

\[A^{2017}\mathbf{x}=\mathbf{x}.\]
(To be more precise, you can prove that $A^n\mathbf{x}=\mathbf{x}$ for any positive integer $n$ by mathematical induction on $n$.)

Thus, we have

\[A^{2017}\mathbf{x}=\begin{bmatrix}

1 \\

-1

\end{bmatrix}.\]

## Comment.

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

## 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: 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 (The current page): Solve a system by the inverse matrix
- Problem 8 and its solution:A proof problem about nonsingular matrix

Add to solve later

Sponsored Links

## 5 Responses

[…] Problem 7 and its solution: Solve a system by the inverse matrix […]

[…] Problem 7 and its solution: Solve a system by the inverse matrix […]

[…] Problem 7 and its solution: Solve a system by the inverse matrix […]

[…] Problem 7 and its solution: Solve a system by the inverse matrix […]

[…] Problem 7 and its solution: Solve a system by the inverse matrix […]