Q86E Consider the matricesA=[10204001... [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 3: Q86E (page 146)

Consider the matrices
$A = [\begin{matrix} 1 & 0 & 2 & 0 & 4 & 0 \\ 0 & 1 & 3 & 0 & 5 & 0 \\ 0 & 0 & 0 & 1 & 6 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix}] and B = [\begin{matrix} 1 & 0 & 2 & 0 & 4 & 0 \\ 0 & 1 & 3 & 0 & 5 & 0 \\ 0 & 0 & 0 & 1 & 7 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix}]$
Show that the kernels of the matrices A and B are different

Short Answer

Expert verified

$k e r (A) = s p a n \{[\begin{matrix} - 2 \\ - 3 \\ 1 \\ 0 \\ 0 \\ 0 \end{matrix}], [\begin{matrix} - 4 \\ - 5 \\ 0 \\ - 6 \\ 1 \\ 0 \end{matrix}]\} a n d k e r (B) = s p a n \{[\begin{matrix} - 2 \\ - 3 \\ 1 \\ 0 \\ 0 \\ 0 \end{matrix}], [\begin{matrix} - 4 \\ - 5 \\ 0 \\ - 7 \\ 1 \\ 0 \end{matrix}]\} 鈬� k e r (A) 鈮� k e r (B)$

Step by step solution

Finding the ker(A)

Consider $A x 鈨� = 0$

$鈬� [\begin{matrix} 1 & 0 & 2 & 0 & 4 & 0 \\ 0 & 1 & 3 & 0 & 5 & 0 \\ 0 & 0 & 0 & 1 & 6 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \\ x_{6} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix}]$

$鈬� x_{1} + 2 x_{3} + 4 x_{5} = 0 x_{2} + 3 x_{3} + 5 x_{5} = 0 x_{4} + 6 x_{5} = 0 x_{6} = 0$

Since the pivot elements are in first, second, fourth, and sixth columns, therefore $x_{3} and x_{5}$ are free variables.

Let $x_{3} = k_{1} a n d x_{5} = k_{2}$

$鈬� x_{1} = - 2 k_{1} - 4 k_{2} x_{2} = - 3 k_{1} - 5 k_{2} x_{4} = - 6 k_{2} x_{6} = 0$

$鈬� [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \\ x_{6} \end{matrix}] = k_{1} [\begin{matrix} - 2 \\ - 3 \\ 1 \\ 0 \\ 0 \\ 0 \end{matrix}] + k_{2} [\begin{matrix} - 4 \\ - 5 \\ 0 \\ - 6 \\ 1 \\ 0 \end{matrix}] 鈬� k e r (A) = s p a n \{[\begin{matrix} - 2 \\ - 3 \\ 1 \\ 0 \\ 0 \\ 0 \end{matrix}], [\begin{matrix} - 4 \\ - 5 \\ 0 \\ - 6 \\ 1 \\ 0 \end{matrix}]\}$

Finding the ker(B)

Let $B x 鈨� = 0$

$鈬� [\begin{matrix} 1 & 0 & 2 & 0 & 4 & 0 \\ 0 & 1 & 3 & 0 & 5 & 0 \\ 0 & 0 & 0 & 1 & 7 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \\ x_{6} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix}]$

$鈬� x_{1} + 2 x_{3} + 4 x_{5} = 0 x_{2} + 3 x_{3} + 5 x_{5} = 0 x_{4} + 7 x_{5} = 0 x_{6} = 0$

Since the pivot elements are in first, second, fourth, and sixth columns, therefore $x_{3}$ are $x_{5}$ free variables.

Let $x_{3} = k_{1} a n d x_{5} = k_{2}$

$鈬� x_{1} = - 2 k_{1} - 4 k_{2} x_{2} = - 3 k_{1} - 5 k_{2} x_{4} = - 7 k_{2} x_{6} = 0$

鈬� [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \\ x_{6} \end{matrix}] = k_{1} [\begin{matrix} - 2 \\ - 3 \\ 1 \\ 0 \\ 0 \\ 0 \end{matrix}] + k_{2} [\begin{matrix} - 4 \\ - 5 \\ 0 \\ - 7 \\ 1 \\ 0 \end{matrix}]

$鈬� k e r (B) = s p a n \{[\begin{matrix} - 2 \\ - 3 \\ 1 \\ 0 \\ 0 \\ 0 \end{matrix}], [\begin{matrix} - 4 \\ - 5 \\ 0 \\ - 7 \\ 1 \\ 0 \end{matrix}]\}$

Final Answer

Since, for the matrix A and B,

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91影视

Short Answer

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Finding the ker(A)

Finding the ker(B)

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91影视

Consider the matricesA=[102040013050000160000001]andB=[102040013050000170000001]Show that the kernels of the matrices A and B are different

Short Answer

Step by step solution

Finding the ker(A)

Finding the ker(B)

Final Answer

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Applied Mathematics

Theoretical and Mathematical Physics

Geometry

Pure Maths

Statistics

Study anywhere. Anytime. Across all devices.