# Find the Inverse Matrix of a Matrix With Fractions

## Problem 214

Find the inverse matrix of the matrix
$A=\begin{bmatrix} \frac{2}{7} & \frac{3}{7} & \frac{6}{7} \\[6 pt] \frac{6}{7} &\frac{2}{7} &-\frac{3}{7} \\[6pt] -\frac{3}{7} & \frac{6}{7} & -\frac{2}{7} \end{bmatrix}.$

Contents

## Hint.

You may use the augmented matrix method to find the inverse matrix.
Here we give an alternative way to find the inverse matrix by noting that $A$ is an orthogonal matrix.

Recall that a matrix $B$ is orthogonal if $B^{\trans}B=B^{\trans}B=I$.
Thus, once we know $B$ is an orthogonal matrix, then the inverse matrix $B^{-1}$ is just the transpose matrix $B^{\trans}$.

Also, recall that a matrix $B$ is orthogonal if and only if the column vectors of $B$ form an orthonormal set.

## Solution.

We first show that $A$ is an orthogonal matrix.
To do this, it suffices to that the column vectors form an orthonormal set.

Let
$\mathbf{v}_1= \begin{bmatrix} \frac{2}{7} \\[6 pt] \frac{6}{7} \\[6pt] -\frac{3}{7} \end{bmatrix}, \mathbf{v}_2=\begin{bmatrix} \frac{3}{7} \\[6 pt] \frac{2}{7} \\[6pt] \frac{6}{7} \end{bmatrix}, \mathbf{v}_3=\begin{bmatrix} \frac{6}{7} \\[6 pt] -\frac{3}{7} \\[6pt] -\frac{2}{7} \end{bmatrix}$ be the column vectors of $A$.

Then the length of the vector $\mathbf{v}_1$ is
$||\mathbf{v}_1||=\sqrt{(2/7)^2+(6/7)^2+(-3/7)^2}=1.$ Similarly, we have $||\mathbf{v}_2||=||\mathbf{v}_3||=1$.
Thus, column vectors are unit vectors.

The dot (inner) product of the vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ is
$\mathbf{v}_1\cdot \mathbf{v}_2=\frac{2}{7}\cdot \frac{3}{7}+\frac{6}{7}\cdot \frac{2}{7}+\left( -\frac{3}{7}\right) \cdot \frac{6}{7}=0.$ Similarly, we have
$\mathbf{v}_1\cdot \mathbf{v}_3=0, \quad \mathbf{v}_2\cdot \mathbf{v}_3=0.$

Therefore, the column vectors are orthogonal.
Hence the column vectors of $A$ are orthonormal, and this implies that $A$ is an orthogonal matrix. Namely, $A^{\trans}=A^{-1}$.
Thus the inverse matrix of $A$ is
$A^{-1}=\begin{bmatrix} \frac{2}{7} & \frac{6}{7} & -\frac{3}{7} \\[6 pt] \frac{3}{7} &\frac{2}{7} &\frac{6}{7} \\[6pt] \frac{6}{7} & -\frac{3}{7} & -\frac{2}{7} \end{bmatrix}.$

### More from my site

• Rotation Matrix in Space and its Determinant and Eigenvalues For a real number $0\leq \theta \leq \pi$, we define the real $3\times 3$ matrix $A$ by $A=\begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta &\cos\theta &0 \\ 0 & 0 & 1 \end{bmatrix}.$ (a) Find the determinant of the matrix $A$. (b) Show that $A$ is an […]
• Find the Distance Between Two Vectors if the Lengths and the Dot Product are Given Let $\mathbf{a}$ and $\mathbf{b}$ be vectors in $\R^n$ such that their length are $\|\mathbf{a}\|=\|\mathbf{b}\|=1$ and the inner product $\mathbf{a}\cdot \mathbf{b}=\mathbf{a}^{\trans}\mathbf{b}=-\frac{1}{2}.$ Then determine the length $\|\mathbf{a}-\mathbf{b}\|$. (Note […]
• Construction of a Symmetric Matrix whose Inverse Matrix is Itself Let $\mathbf{v}$ be a nonzero vector in $\R^n$. Then the dot product $\mathbf{v}\cdot \mathbf{v}=\mathbf{v}^{\trans}\mathbf{v}\neq 0$. Set $a:=\frac{2}{\mathbf{v}^{\trans}\mathbf{v}}$ and define the $n\times n$ matrix $A$ by $A=I-a\mathbf{v}\mathbf{v}^{\trans},$ where […]
• Inner Products, Lengths, and Distances of 3-Dimensional Real Vectors For this problem, use the real vectors $\mathbf{v}_1 = \begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix} , \mathbf{v}_2 = \begin{bmatrix} 0 \\ 2 \\ -3 \end{bmatrix} , \mathbf{v}_3 = \begin{bmatrix} 2 \\ 2 \\ 3 \end{bmatrix} .$ Suppose that $\mathbf{v}_4$ is another vector which is […]
• Inner Product, Norm, and Orthogonal Vectors Let $\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3$ are vectors in $\R^n$. Suppose that vectors $\mathbf{u}_1$, $\mathbf{u}_2$ are orthogonal and the norm of $\mathbf{u}_2$ is $4$ and $\mathbf{u}_2^{\trans}\mathbf{u}_3=7$. Find the value of the real number $a$ in […]
• Diagonalizable by an Orthogonal Matrix Implies a Symmetric Matrix Let $A$ be an $n\times n$ matrix with real number entries. Show that if $A$ is diagonalizable by an orthogonal matrix, then $A$ is a symmetric matrix.   Proof. Suppose that the matrix $A$ is diagonalizable by an orthogonal matrix $Q$. The orthogonality of the […]
• Unit Vectors and Idempotent Matrices A square matrix $A$ is called idempotent if $A^2=A$. (a) Let $\mathbf{u}$ be a vector in $\R^n$ with length $1$. Define the matrix $P$ to be $P=\mathbf{u}\mathbf{u}^{\trans}$. Prove that $P$ is an idempotent matrix. (b) Suppose that $\mathbf{u}$ and $\mathbf{v}$ be […]
• Sherman-Woodbery Formula for the Inverse Matrix Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in $\R^n$, and let $I$ be the $n \times n$ identity matrix. Suppose that the inner product of $\mathbf{u}$ and $\mathbf{v}$ satisfies $\mathbf{v}^{\trans}\mathbf{u}\neq -1.$ Define the matrix […]

#### You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

##### A Matrix Similar to a Diagonalizable Matrix is Also Diagonalizable

Let $A, B$ be matrices. Show that if $A$ is diagonalizable and if $B$ is similar to $A$, then $B$...

Close