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 12E from Chapter 2.1 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 12E
Chapter:
Problem:
Construct a 2 × 2 matrix B such that AB is the zero matrix. Use two different nonzero columns for B:
Step-by-Step Solution
Given Information
Given Matrix:\[A = \left[ {\begin{array}{*{20}{c}}3&{ - 6}\\{ - 1}&2\end{array}} \right]\]We have to find a $2 \times 2$ matrix such that AB is the zero matrixes by using two different nonzero columns for B.
Step 1:
Let:\[B = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\] Then the Product is given by: \[\begin{array}{l}AB = \left[ {\begin{array}{*{20}{c}}3&{ - 6}\\{ - 1}&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{3a - 6c}&{3b - 6d}\\{ - a + 2c}&{ - b + 2d}\end{array}} \right]\end{array}\]
Step 2
Equate the product with zero Matrix\[\begin{array}{l}AB = \left[ {\begin{array}{*{20}{c}}0&0\\0&0\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{3a - 6c}&{3b - 6d}\\{ - a + 2c}&{ - b + 2d}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0&0\\0&0\end{array}} \right]\end{array}\]
Step 3
Compare the equations:\[\begin{array}{l}3a - 6c = 0\\3b - 6d = 0\\ - a + 2c = 0\\ - b + 2d = 0\end{array}\]Let,\[a = 2 \Rightarrow c = 1\]and \[b = 4 \Rightarrow d = 2\]Thus the Matrix is
\[B = \left[ {\begin{array}{*{20}{c}}2&4\\1&2\end{array}} \right]\]
Given Matrix:\[A = \left[ {\begin{array}{*{20}{c}}3&{ - 6}\\{ - 1}&2\end{array}} \right]\]We have to find a $2 \times 2$ matrix such that AB is the zero matrixes by using two different nonzero columns for B.
Step 1:
Let:\[B = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\] Then the Product is given by: \[\begin{array}{l}AB = \left[ {\begin{array}{*{20}{c}}3&{ - 6}\\{ - 1}&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{3a - 6c}&{3b - 6d}\\{ - a + 2c}&{ - b + 2d}\end{array}} \right]\end{array}\]
Step 2
Equate the product with zero Matrix\[\begin{array}{l}AB = \left[ {\begin{array}{*{20}{c}}0&0\\0&0\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{3a - 6c}&{3b - 6d}\\{ - a + 2c}&{ - b + 2d}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0&0\\0&0\end{array}} \right]\end{array}\]
Step 3
Compare the equations:\[\begin{array}{l}3a - 6c = 0\\3b - 6d = 0\\ - a + 2c = 0\\ - b + 2d = 0\end{array}\]Let,\[a = 2 \Rightarrow c = 1\]and \[b = 4 \Rightarrow d = 2\]Thus the Matrix is
\[B = \left[ {\begin{array}{*{20}{c}}2&4\\1&2\end{array}} \right]\]