Q57E In Exercises 55 through 65, show... [FREE SOLUTION]

Chapter 2: Q57E (page 100)

In Exercises 55 through 65, show that the given matrix $A$ is invertible, and find the inverse. Interpret the linear transformation $T (\overset{鈫�}{x}) = A \overset{鈫�}{x}$ and the inverse transformation $T^{- 1} (\overset{鈫�}{y}) = A^{- 1} \overset{鈫�}{y}$ geometrically. Interpret det $A$ geometrically. In your figure, show the angle $胃$ and the vectors $\overset{鈫�}{v}$ and $\overset{鈫�}{w}$ introduced in Theorem 2.4.10.
57 $[\begin{matrix} c o s 伪 & s i n 伪 \\ s i n 伪 & - c o s 伪 \end{matrix}]$ ..

Short Answer

Expert verified

The inverse of matrix is,

$A^{鈭� 1} = [\begin{matrix} \cos 伪 & \sin 伪 \\ \sin 伪 & 鈭� \cos 伪 \end{matrix}]$ . $| \det A | = 鈥� \overset{鈫�}{v} 鈥� 鈥夆赌� | \sin 胃 | 鈥夆赌� 鈥� \overset{鈫�}{w} 鈥�$ is the area of the parallelogram spanned by $\overset{鈫�}{v}$ and. $\overset{鈫�}{w}$

Step by step solution

To find the determinant

Let $A = [\begin{matrix} \cos 伪 & \sin 伪 \\ \sin 伪 & 鈭� \cos 伪 \end{matrix}]$ be the given matrix.

In order to find the inverse of matrix, we find the determinant of the matrix.

Therefore,

$\begin{matrix} | A | = | \begin{matrix} \cos 伪 & \sin 伪 \\ \sin 伪 & 鈭� \cos 伪 \end{matrix} | \\ = 鈭� \cos^{2} 伪鈭� \sin^{2} 伪 \\ = 鈭� (\cos^{2} 伪 + \sin^{2} 伪) \\ = 鈭� 1 \\ 鈮� 0 \end{matrix}$

$鈭� | A | 鈮� 0$ .

Thus, the inverse of matrix exists.

To find the inverse of matrix

Let us find the inverse of matrix.

Consider the system of matrices

$\begin{array}{l} [\begin{matrix} \cos 伪 & \sin 伪鈥夆赌夆€夆赌夆€夆赌夆€夆赌夆€� | & 1 & 0 \\ \sin 伪 & 鈭� \cos 伪鈥夆赌夆€� | & 0 & 1 \end{matrix}] \\ R_{1} 鈫� \frac{R_{1}}{\cos 伪} \end{array}$

$\begin{array}{l} ~ [\begin{matrix} \cos 伪 & \frac{\sin 伪}{\cos 伪} 鈥夆赌夆€夆赌夆€夆赌夆€夆赌夆€� | & \frac{1}{\cos 伪} & 0 \\ \sin 伪 & 鈭� \cos 伪鈥夆赌夆€� | & 0 & 1 \end{matrix}] \\ R_{2} 鈫� R_{2} 鈭� R_{1} \sin 伪 \\ ~ [\begin{matrix} \cos 伪 & \frac{\sin 伪}{\cos 伪} 鈥夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌� | & \frac{1}{\cos 伪} & 0 \\ \sin 伪 & 鈭� (\frac{\sin^{2} 伪 + \cos^{2} 伪}{\cos 伪}) 鈥夆赌夆€� | & 鈭� \frac{\sin 伪}{\cos 伪} & 1 \end{matrix}] \\ ~ [\begin{matrix} \cos 伪 & 鈥夆赌夆€� \frac{\sin 伪}{\cos 伪} 鈥夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌� | & \frac{1}{\cos 伪} & 0 \\ \sin 伪 & 鈭� \frac{1}{\cos 伪} 鈥夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€� | & 鈭� \frac{\sin 伪}{\cos 伪} & 1 \end{matrix}] \\ R_{1} 鈫� R_{1} + R_{2} \sin 伪 \\ ~ [\begin{matrix} 1 & 鈥夆赌夆€� 0 鈥夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌� | & \frac{1}{\cos 伪} 鈭� \frac{\sin^{2} 伪}{\cos 伪} & \sin 伪 \\ 0 & 鈭� \frac{1}{\cos 伪} 鈥夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€� | & 鈭� \frac{\sin 伪}{\cos 伪} & 1 \end{matrix}] \\ R_{2} 鈫� 鈭� R_{2} \cos 伪 \\ ~ [\begin{matrix} 1 & 鈥夆赌夆€� 0 鈥夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌� | & \cos 伪 & \sin 伪 \\ 0 & 1 鈥夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌夆€夆赌� | & \sin 伪 & 鈭� \cos 伪 \end{matrix}] \end{array}$

Hence, the inverse of the matrix is $[\begin{matrix} \cos 伪 & \sin 伪 \\ \sin 伪 & 鈭� \cos 伪 \end{matrix}]$ , .

To interpret the inverse transformation

Now, we have to interpret the linear transformation $T (\overset{鈫�}{x}) = A \overset{鈫�}{x}$ and the inverse transformation $T^{鈭� 1} (\overset{鈫�}{y}) = A^{鈭� 1} \overset{鈫�}{y}$ .

Then

$T (\overset{鈫�}{x}) = [\begin{matrix} \cos 伪 & \sin 伪 \\ \sin 伪 & 鈭� \cos 伪 \end{matrix}] \overset{鈫�}{x}$

And

$\begin{matrix} T^{鈭� 1} (\overset{鈫�}{y}) = A^{鈭� 1} \overset{鈫�}{y} \\ = [\begin{matrix} \cos 伪 & \sin 伪 \\ \sin 伪 & 鈭� \cos 伪 \end{matrix}] \overset{鈫�}{y} \end{matrix}$

Geometrically, $[\begin{matrix} \cos 伪 & \sin 伪 \\ \sin 伪 & 鈭� \cos 伪 \end{matrix}]$ and the inverse $[\begin{matrix} \cos 伪 & \sin 伪 \\ \sin 伪 & 鈭� \cos 伪 \end{matrix}]$ represents the reflection of the vector at an angle $伪$ .

To interpret the determinant geometrically

Now, let us interpret the determinant of $A$ geometrically.

Consider the matrix $A = [\begin{matrix} \cos 伪 & \sin 伪 \\ \sin 伪 & 鈭� \cos 伪 \end{matrix}]$ .

Let $\overset{鈫�}{v} = [\begin{matrix} \cos 伪 \\ \sin 伪 \end{matrix}], 鈥夆赌� \overset{鈫�}{w} = [\begin{matrix} \sin 伪 \\ 鈭� \cos 伪 \end{matrix}]$ .

Then , $\det A = 鈥� \overset{鈫�}{v} 鈥� \sin 胃鈥� \overset{鈫�}{w} 鈥�$ whereis the oriented angle from $\overset{鈫�}{v}$ to $\overset{鈫�}{w}$ , $鈭� 蟺 < 胃 < 蟺$ .

Thus, $| \det A | = 鈥� \overset{鈫�}{v} 鈥� 鈥夆赌� | \sin 胃 | 鈥夆赌� 鈥� \overset{鈫�}{w} 鈥�$ is the area of the parallelogram spanned by $\overset{鈫�}{v}$ and. $\overset{鈫�}{w}$

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Short Answer

Step by step solution

To find the determinant

To find the inverse of matrix

To interpret the inverse transformation

To interpret the determinant geometrically

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