Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 38E from Chapter 3.1 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 38E
Chapter:
Problem:
Let and let k be a scalar. Find a formula that relates det kA to k and det A.
Step-by-Step Solution
Given Information
We are given with a matrix: \[ A = \left[ \begin{array} { l l } { a } & { b } \\ { c } & { d } \end{array} \right] \] Objective is to find a formula that relates det(kA) to k and det(A)
Step-1:
The product of scalar k and matrix A is: \[ \begin{aligned} k A & = k \left[ \begin{array} { l l } { a } & { b } \\ { c } & { d } \end{array} \right] \\ & = \left[ \begin{array} { l l } { k a } & { k b } \\ { k c } & { k d } \end{array} \right] \end{aligned} \]
Step-2: Determinant of matrix A
\[ \begin{aligned} \operatorname { det } A & = \left| \begin{array} { l l } { a } & { b } \\ { c } & { d } \end{array} \right| \\ & = a d - b c \end{aligned} \]
Step-3: Determinant of matrix kA
\[ \begin{aligned} \operatorname { det } k A & = \left| \begin{array} { c c } { | k a } & { k b } \\ { k c } & { k d } \end{array} \right| \\ & = ( k a ) ( k d ) - ( k b ) ( k c ) \\ & = k ^ { 2 } a d - k ^ { 2 } b c \\ & = k ^ { 2 } ( a d - b c ) \end{aligned} \] Therefore, the formula relating det(kA) to det(A) is:
\[ det (kA) = {k^2}( det A)\]
We are given with a matrix: \[ A = \left[ \begin{array} { l l } { a } & { b } \\ { c } & { d } \end{array} \right] \] Objective is to find a formula that relates det(kA) to k and det(A)
Step-1:
The product of scalar k and matrix A is: \[ \begin{aligned} k A & = k \left[ \begin{array} { l l } { a } & { b } \\ { c } & { d } \end{array} \right] \\ & = \left[ \begin{array} { l l } { k a } & { k b } \\ { k c } & { k d } \end{array} \right] \end{aligned} \]
Step-2: Determinant of matrix A
\[ \begin{aligned} \operatorname { det } A & = \left| \begin{array} { l l } { a } & { b } \\ { c } & { d } \end{array} \right| \\ & = a d - b c \end{aligned} \]
Step-3: Determinant of matrix kA
\[ \begin{aligned} \operatorname { det } k A & = \left| \begin{array} { c c } { | k a } & { k b } \\ { k c } & { k d } \end{array} \right| \\ & = ( k a ) ( k d ) - ( k b ) ( k c ) \\ & = k ^ { 2 } a d - k ^ { 2 } b c \\ & = k ^ { 2 } ( a d - b c ) \end{aligned} \] Therefore, the formula relating det(kA) to det(A) is:
\[ det (kA) = {k^2}( det A)\]