Define the linear transformation I by T(x) = Ax. Find ker(T), nullity(T), range(T), and rank(T). 8 8 4 3 3 8 8 4 A = 3 3 4 4 2 3 (a) ker(T) STEP 1: The kernel of T is given by the solution to the equation T(x) = 0. Let x = (x,, Xa, x,) and find x such that T(x) = 0. (If there are an infinite number of solutions use t and s as your parameters.) X = STEP 2: Use your result from Step 1 to find the kernel of T. (If there are an infinite number of solutions use t and s as your parameters.) ker(T) = 3. (b) nullity(T) STEP 3: Use the fact that nullity(T) = dim(ker(T)) to compute nullity(T). (c) range(T) STEP 4: Transpose A and find its equivalent reduced row-echelon form. 8 8 4 4 STEP 5: Use your result from Step 4 to find the range of T. OR
Define the linear transformation I by T(x) = Ax. Find ker(T), nullity(T), range(T), and rank(T). 8 8 4 3 3 8 8 4 A = 3 3 4 4 2 3 (a) ker(T) STEP 1: The kernel of T is given by the solution to the equation T(x) = 0. Let x = (x,, Xa, x,) and find x such that T(x) = 0. (If there are an infinite number of solutions use t and s as your parameters.) X = STEP 2: Use your result from Step 1 to find the kernel of T. (If there are an infinite number of solutions use t and s as your parameters.) ker(T) = 3. (b) nullity(T) STEP 3: Use the fact that nullity(T) = dim(ker(T)) to compute nullity(T). (c) range(T) STEP 4: Transpose A and find its equivalent reduced row-echelon form. 8 8 4 4 STEP 5: Use your result from Step 4 to find the range of T. OR
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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![Define the linear transformation \( T \) by \( T(x) = Ax \). Find ker(\( T \)), nullity(\( T \)), range(\( T \)), and rank(\( T \)).
\[
A = \begin{bmatrix}
8 & -8 & 4 \\
3 & 3 & 3 \\
8 & -8 & -4 \\
3 & 3 & 3 \\
4 & -4 & 2 \\
3 & 3 & 3
\end{bmatrix}
\]
**(a) ker(\( T \))**
**STEP 1:** The kernel of \( T \) is given by the solution to the equation \( T(x) = 0 \). Let \( x = (x_1, x_2, x_3) \) and find \( x \) such that \( T(x) = 0 \). (If there are an infinite number of solutions use \( t \) and \( s \) as your parameters.)
\[ x = \_\_\_ \]
**STEP 2:** Use your result from Step 1 to find the kernel of \( T \). (If there are an infinite number of solutions use \( t \) and \( s \) as your parameters.)
\[ \text{ker}(T) = \{ \_\_\_ : s, t \in \mathbb{R} \} \]
**(b) nullity(\( T \))**
**STEP 3:** Use the fact that nullity(\( T \)) = \(\dim (\text{ker}(T))\) to compute nullity(\( T \)).
\[ \_\_\_ \]
**(c) range(\( T \))**
**STEP 4:** Transpose \( A \) and find its equivalent reduced row-echelon form.
\[
A^T = \begin{bmatrix}
8 & 3 & 8 & 3 & 4 & 3 \\
-8 & 3 & -8 & 3 & -4 & 3 \\
4 & 3 & -4 & 3 & 2 & 3
\end{bmatrix}
\quad \Rightarrow \quad
\begin{bmatrix}
\_\_ & \_\_ & \_\_ \\
\_\_ & \_\_ & \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9a7c37e1-d117-46e5-8cc2-cb36983cacd6%2F5ee4c515-9378-44a0-ae08-e5687d9e5246%2Fhb5kt3k_processed.png&w=3840&q=75)
Transcribed Image Text:Define the linear transformation \( T \) by \( T(x) = Ax \). Find ker(\( T \)), nullity(\( T \)), range(\( T \)), and rank(\( T \)).
\[
A = \begin{bmatrix}
8 & -8 & 4 \\
3 & 3 & 3 \\
8 & -8 & -4 \\
3 & 3 & 3 \\
4 & -4 & 2 \\
3 & 3 & 3
\end{bmatrix}
\]
**(a) ker(\( T \))**
**STEP 1:** The kernel of \( T \) is given by the solution to the equation \( T(x) = 0 \). Let \( x = (x_1, x_2, x_3) \) and find \( x \) such that \( T(x) = 0 \). (If there are an infinite number of solutions use \( t \) and \( s \) as your parameters.)
\[ x = \_\_\_ \]
**STEP 2:** Use your result from Step 1 to find the kernel of \( T \). (If there are an infinite number of solutions use \( t \) and \( s \) as your parameters.)
\[ \text{ker}(T) = \{ \_\_\_ : s, t \in \mathbb{R} \} \]
**(b) nullity(\( T \))**
**STEP 3:** Use the fact that nullity(\( T \)) = \(\dim (\text{ker}(T))\) to compute nullity(\( T \)).
\[ \_\_\_ \]
**(c) range(\( T \))**
**STEP 4:** Transpose \( A \) and find its equivalent reduced row-echelon form.
\[
A^T = \begin{bmatrix}
8 & 3 & 8 & 3 & 4 & 3 \\
-8 & 3 & -8 & 3 & -4 & 3 \\
4 & 3 & -4 & 3 & 2 & 3
\end{bmatrix}
\quad \Rightarrow \quad
\begin{bmatrix}
\_\_ & \_\_ & \_\_ \\
\_\_ & \_\_ & \
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