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

Ohio State University exam problems and solutions in mathematics

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)
 
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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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