Solve The image displays a mathematical expression involving vectors and matrices, commonly used in linear algebra. The expression is as follows:\[ \mathbf{x'} = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} \mathbf{x} \]### Explanation:- $\mathbf{x'}$ represents a transformed vector resulting from the matrix multiplication.- The matrix $\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}$ is a 2x2 matrix that transforms the vector $\mathbf{x}$.- This particular matrix appears to perform a shearing transformation in a two-dimensional space, where the x-component is adjusted by subtracting the y-component, but the y-component remains unchanged.This content might be used in a section discussing linear transformations or matrices in a linear algebra course on an educational website.

Given: X'=1-10 1X Let A=1-10 1 Since A is a upper triangular matrix . So, its eigenvalues are just…

Answered: 1 -1 X%3D X. 0 1

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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The image displays a mathematical expression involving vectors and matrices, commonly used in linear algebra. The expression is as follows:

\[ \mathbf{x'} = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} \mathbf{x} \]

### Explanation:
- $\mathbf{x'}$ represents a transformed vector resulting from the matrix multiplication.
- The matrix $\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}$ is a 2x2 matrix that transforms the vector $\mathbf{x}$.
- This particular matrix appears to perform a shearing transformation in a two-dimensional space, where the x-component is adjusted by subtracting the y-component, but the y-component remains unchanged.

This content might be used in a section discussing linear transformations or matrices in a linear algebra course on an educational website.

Expert Solution

Step 1

Given: $X' = [\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix}] X$

Let $A = [\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix}]$

Since $A$ is a upper triangular matrix . So, its eigenvalues are just the diagonal entries .

So, eigenvalue of $A$ is 1 with multiplicity 2.

Now we find eigen vector corresponding to this eigenvalue.

Consider $(A - I) X = 0$

$[\begin{matrix} 1 - 1 & - 1 - 0 \\ 0 & 1 - 1 \end{matrix}] X = 0 [\begin{matrix} 0 & - 1 \\ 0 & 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = 0 \Rightarrow - x_{2} = 0 x_{2} = 0$

So, $X = [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = [\begin{matrix} 1 \\ 0 \end{matrix}] x_{1}$ where $x_{1}$ can take any value.

So, $[\begin{matrix} 1 \\ 0 \end{matrix}]$ is the eigen vector

Step 2

Now, to find another linearly independent vector , consider

$(A - I) Y = X [\begin{matrix} 0 & - 1 \\ 0 & 0 \end{matrix}] [\begin{matrix} y_{1} \\ y_{2} \end{matrix}] = [\begin{matrix} 1 \\ 0 \end{matrix}] \Rightarrow y_{2} = - 1$

So, $[\begin{matrix} y_{1} \\ y_{2} \end{matrix}] = [\begin{matrix} 0 \\ - 1 \end{matrix}] y_{2}$ where $y_{2}$ can take any value.

So, another linearly independent vector is $[\begin{matrix} 0 \\ - 1 \end{matrix}]$

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