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 1E from Chapter 3.1 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 1E
Chapter:
Problem:
Compute the determinants in Exercises 1-8 using a cofactor expansion across the first row. In Exercises 1-4, also compute the determinant by a cofactor expansion down the second column...
Step-by-Step Solution
Step 1
Given Determinant:\[D = \left| {\begin{array}{*{20}{l}}3&0&4\\2&3&2\\0&5&{ - 1}\end{array}} \right|\]We have to Compute the determinant given above using a cofactorexpansion across the first row. Also, we have to find the determinant by a cofactor expansion down the second row.
Step 2: Determinant using a cofactor expansion across the first row
\[\begin{array}{l}D = \left| {\begin{array}{*{20}{c}}3&0&4\\2&3&2\\0&5&{ - 1}\end{array}} \right|\\\\ = 3\left| {\begin{array}{*{20}{c}}3&2\\5&{ - 1}\end{array}} \right| - 0\left| {\begin{array}{*{20}{c}}2&2\\0&{ - 1}\end{array}} \right| + 4\left| {\begin{array}{*{20}{c}}2&3\\0&5\end{array}} \right|\\\\ = 0 + 3 \times \left( { - 3 - 0} \right) - 5 \times \left( {6 - 8} \right)\\\\ = {\rm{ }} - 9 + 10\\\\ = 1\end{array}\]
Step 3: Determinant using a cofactor expansion across the second row
\[\begin{array}{l}D = \left| {\begin{array}{*{20}{c}}3&0&4\\2&3&2\\0&5&{ - 1}\end{array}} \right|\\\\ = {( - 1)^{1 + 2}} \times 0\left| {\begin{array}{*{20}{c}}2&2\\0&{ - 1}\end{array}} \right| + {( - 1)^{2 + 2}} \times 3\left| {\begin{array}{*{20}{c}}3&4\\0&{ - 1}\end{array}} \right| + {( - 1)^{3 + 2}}\cdot \times 5\left| {\begin{array}{*{20}{c}}3&4\\2&2\end{array}} \right|\\\\ = 0 + 3\left( { - 3 - 0} \right) - 5\left( {6 - 8} \right)\\\\ = - 9 + 10\\\\ = - 1\end{array}\]
ANSWER
D=1
Given Determinant:\[D = \left| {\begin{array}{*{20}{l}}3&0&4\\2&3&2\\0&5&{ - 1}\end{array}} \right|\]We have to Compute the determinant given above using a cofactorexpansion across the first row. Also, we have to find the determinant by a cofactor expansion down the second row.
Step 2: Determinant using a cofactor expansion across the first row
\[\begin{array}{l}D = \left| {\begin{array}{*{20}{c}}3&0&4\\2&3&2\\0&5&{ - 1}\end{array}} \right|\\\\ = 3\left| {\begin{array}{*{20}{c}}3&2\\5&{ - 1}\end{array}} \right| - 0\left| {\begin{array}{*{20}{c}}2&2\\0&{ - 1}\end{array}} \right| + 4\left| {\begin{array}{*{20}{c}}2&3\\0&5\end{array}} \right|\\\\ = 0 + 3 \times \left( { - 3 - 0} \right) - 5 \times \left( {6 - 8} \right)\\\\ = {\rm{ }} - 9 + 10\\\\ = 1\end{array}\]
Step 3: Determinant using a cofactor expansion across the second row
\[\begin{array}{l}D = \left| {\begin{array}{*{20}{c}}3&0&4\\2&3&2\\0&5&{ - 1}\end{array}} \right|\\\\ = {( - 1)^{1 + 2}} \times 0\left| {\begin{array}{*{20}{c}}2&2\\0&{ - 1}\end{array}} \right| + {( - 1)^{2 + 2}} \times 3\left| {\begin{array}{*{20}{c}}3&4\\0&{ - 1}\end{array}} \right| + {( - 1)^{3 + 2}}\cdot \times 5\left| {\begin{array}{*{20}{c}}3&4\\2&2\end{array}} \right|\\\\ = 0 + 3\left( { - 3 - 0} \right) - 5\left( {6 - 8} \right)\\\\ = - 9 + 10\\\\ = - 1\end{array}\]
ANSWER
D=1