3. Let A be a matrix of a linear transformation y : R3 → R³ with respect to the naturalbasis. Find a basis of the image and the kernel of p.1 2 3A =4 5 67 8 9||

Consider the given matrix as A=123456789 which represents the linear transformation between R3 and…

Answered: 3. Let A be a matrix of a linear…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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3. Let A be a matrix of a linear transformation y : R3 → R³ with respect to the natural
basis. Find a basis of the image and the kernel of p.
1 2 3
A =
4 5 6
7 8 9
||

Expert Solution

Step 1

Consider the given matrix as $A = (\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix})$ which represents the linear transformation between $R^{3}$ and $R^{3}$ .

For the kernel of the linear transformation set $A X = 0$ , where X is the column vector of the 3 cross 1 order.

Substitute the matrix in the equation $A X = 0$ .

$A X = 0 (\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix}) (\begin{matrix} x \\ y \\ z \end{matrix}) = 0 x + 2 y + 3 z = 0 4 x + 5 y + 6 z = 0 7 x + 8 y + 9 z = 0$

Step 2

Convert the matrix into row reduced echelon form by the set of row operations as below:

$A = (\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix}) ~ (\begin{matrix} 1 & 2 & 3 \\ 0 & - 3 & - 6 \\ 0 & - 6 & - 12 \end{matrix}) (\begin{matrix} R_{2} \to - 4 R_{1} + R_{2} \\ R_{3} \to - 7 R_{1} + R_{3} \end{matrix}) ~ (\begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{matrix}) (\begin{matrix} R_{2} \to - \frac{1}{3} 4 \\ R_{3} \to 6 R_{6} + R_{3} \end{matrix}) ~ (\begin{matrix} 1 & 0 & - 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{matrix}) (R_{1} \to - 2 R_{2} + R_{1})$

The above row reduced echelon form gives $x - z = 0, y + 2 z = 0$ or $x = z, y = - 2 z$ .

Thus, the basis for the kernel $φ$ is obtain below:

$(\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} z \\ - 2 z \\ z \end{matrix}) = z (\begin{matrix} 1 \\ - 2 \\ 1 \end{matrix}) B a s i s o f k e r φ = \{(\begin{matrix} 1 \\ - 2 \\ 1 \end{matrix})\}$

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