Q65E Consider a function聽聽D聽from聽... [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 6: Q65E (page 277)

Consider a functionDfrom to that is linear in all three columns and alternating on the columns. Assume that $D (I_{3}) = 1$ . Using Exercises 62 through 64 as a guide, show that $D (A)$ for all $3 脳 3$ matrices.

Short Answer

Expert verified

$D = [\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}] = \det A$

Step by step solution

Matrix Definition.

Matrix is aset of numbers arranged in rows and columns so as to form a rectangulararray.

The numbers are called the elements, or entries, of the matrix.

If there are m rows and n columns, the matrix is said to be an 鈥� m by n 鈥� matrix, written 鈥� $m 脳 n$ .鈥�

To show that D(A)=detA

To show:

$D (A) = d e t A$

To solve,

Take L.H.S:

$D = [\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}]$

$D = [\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}] = a D = [\begin{matrix} 1 & b & c \\ 0 & e & f \\ 0 & h & i \end{matrix}] + d D [\begin{matrix} 0 & b & c \\ 1 & e & f \\ 0 & h & i \end{matrix}] + g D [\begin{matrix} 0 & b & c \\ 0 & e & f \\ 1 & h & i \end{matrix}]$

$D = [\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}] = a e D = [\begin{matrix} 1 & 0 & c \\ 0 & 1 & f \\ 0 & 0 & i \end{matrix}] + a h D [\begin{matrix} 1 & 0 & c \\ 0 & 0 & f \\ 0 & 1 & i \end{matrix}] + d b D [\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ g & h & i \end{matrix}] + d h D [\begin{matrix} 0 & 0 & c \\ 1 & 0 & f \\ 0 & 1 & i \end{matrix}] + g b D [\begin{matrix} 0 & 1 & c \\ 0 & 0 & f \\ 1 & 0 & i \end{matrix}] + g e D D = [\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}] = a e i D = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] + a h f D [\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}] + d b i D [\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}] + d h c D [\begin{matrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix}] + g b D [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}] + g e C D = [\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}] = a e i - a h f - d b i + d h c + g b f - g e c D = [\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}] = d e t A$

Therefore,

L.H.S = R.H.S

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91影视

Consider a functionDfrom to that is linear in all three columns and alternating on the columns. Assume that $D (I_{3}) = 1$ . Using Exercises 62 through 64 as a guide, show that $D (A)$ for all $3 脳 3$ matrices.

Short Answer

Step by step solution

Matrix Definition.

To show that D(A)=detA

One App. One Place for Learning.

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91影视

Consider a functionDfrom to that is linear in all three columns and alternating on the columns. Assume that D(I3)=1 . Using Exercises 62 through 64 as a guide, show thatD(A) for all3脳3 matrices.

Short Answer

Step by step solution

Matrix Definition.

To show that D(A)=detA

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Statistics

Logic and Functions

Discrete Mathematics

Decision Maths

Probability and Statistics

Study anywhere. Anytime. Across all devices.

Consider a functionDfrom to that is linear in all three columns and alternating on the columns. Assume that $D (I_{3}) = 1$ . Using Exercises 62 through 64 as a guide, show that $D (A)$ for all $3 脳 3$ matrices.