# 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)

## 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.

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

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