Q 64E Question: In Exercise 55 through... [FREE SOLUTION]

91影视

Linear Algebra With Applications

Otto Bretscher

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 442 pages

ISBN: 9780321796974

Chapter 2: Q 64E (page 100)

Question: In Exercise 55 through 65, show that the given matrix A is invertible, and find the inverse. Interpret the linear transformation $T^{鈭� 1} (\overset{鈫�}{x}) = A \overset{鈫�}{x}$ and the inverse transformation $T^{鈭� 1} (\overset{鈫�}{y}) = A^{鈭� 1} \overset{鈫�}{y}$ geometrically. Interpret det A geometrically. In your figure, show the angle 胃 and the vectors $\overset{鈫�}{v} a n d \overset{鈫�}{w}$ introduced in Theorem 2.4.10.
64. $[\begin{matrix} 3 & 4 \\ 鈭� 4 & 3 \end{matrix}]$

Short Answer

Expert verified

Thus, the inverse is $\frac{1}{25} [\begin{matrix} 3 & 鈭� 4 \\ 4 & 3 \end{matrix}]$ and the value of $|\det A|$ is 25.

The required sketch for angel and the vectors are obtained.

Step by step solution

Inverse of the given matrices.

The inverse of a matrix of the form $[\begin{matrix} a & b \\ c & d \end{matrix}]$ is equal to $\frac{1}{a d - b c} [\begin{matrix} d & - b \\ - c & a \end{matrix}]$ .

Thus, the inverse of $[\begin{matrix} 3 & 4 \\ 鈭� 4 & 3 \end{matrix}]$ is equal to;

$\frac{1}{9 + 16} [\begin{matrix} 3 & 鈭� 4 \\ 4 & 3 \end{matrix}] = \frac{1}{25} [\begin{matrix} 3 & 鈭� 4 \\ 4 & 3 \end{matrix}]$

Geometrically $[\begin{matrix} 3 & 4 \\ 鈭� 4 & 3 \end{matrix}]$ represents a scaling of $\sqrt{25} = 5$ and a clockwise rotation while $\frac{1}{25} [\begin{matrix} 3 & 鈭� 4 \\ 4 & 3 \end{matrix}]$ represents a scaling of $\frac{1}{5}$ and a counterclockwise rotation.

Hence, the inverse is $\frac{1}{25} [\begin{matrix} 3 & 鈭� 4 \\ 4 & 3 \end{matrix}]$ .

Evaluate det A geometrically.

We know that the determinant of 2 脳 2 matrix of the form $[\begin{matrix} a & b \\ c & d \end{matrix}]$ is ad - bc.

Thus, the determinant of the given matrix is:

$\begin{matrix} \det [\begin{matrix} 3 & 4 \\ 鈭� 4 & 3 \end{matrix}] = 3 鈰� 3 鈭� (鈭� 4) 鈰� 4 \\ = 9 + 16 \\ = 25 \end{matrix}$

Geometrically, it is written as follows;

$\begin{matrix} |\det A| = |[\begin{matrix} 3 \\ 鈭� 4 \end{matrix}] \sin 胃 [\begin{matrix} 4 \\ 3 \end{matrix}]| \\ = |[\begin{matrix} 3 \\ 鈭� 4 \end{matrix}]| |\sin \frac{蟺}{2}| |[\begin{matrix} 4 \\ 3 \end{matrix}]| \\ = 25 \end{matrix}$