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 5E from Chapter 3.3 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 5E
Chapter:
Problem:
Use Cramer's rule to compute the solutions of the systems -5x1 + x2é=3—3x1 + 2x3 = 0x2 — 2x3 = 2
Step-by-Step Solution
Given Information
We are given with system of equations: \[ \begin{aligned} x _ { 1 } + x _ { 2 } & = 3 \\ - 3 x _ { 1 } + 2 x _ { 3 } & = 0 \\ x _ { 2 } - 2 x _ { 3 } & = 2 \end{aligned} \] We have to solve the system of equations using crammer's Rule
Step-1: The matrices
The matrices A and B for the equation $Ax=B$ are: \[ A = \left[ \begin{array} { c c c } { 1 } & { 1 } & { 0 } \\ { - 3 } & { 0 } & { 2 } \\ { 0 } & { 1 } & { - 2 } \end{array} \right] \text { and } \mathbf { b } = \left[ \begin{array} { l } { 3 } \\ { 0 } \\ { 2 } \end{array} \right] \] The matrices $ A _ { 1 } ( \mathbf { b } ) , A_ { 2 } ( \mathbf { b } ) \text { and } A _ { 3 } ( \mathbf { b } ) $ are: \[{A_1}({\bf{b}}) = \left[ {\begin{array}{*{20}{c}} 3&1&0\\ 0&0&2\\ 2&1&{ - 2} \end{array}} \right];\,\,{A_2}({\bf{b}}) = \left[ {\begin{array}{*{20}{c}} 1&3&0\\ { - 3}&0&2\\ 0&2&{ - 2} \end{array}} \right];\,\,{A_3}({\bf{b}}) = \left[ {\begin{array}{*{20}{c}} 1&1&3\\ { - 3}&0&0\\ 0&1&2 \end{array}} \right]\]
Step-2: The Determinants
Determinant of A: \[ \begin{aligned} \operatorname { det } A & = 1 \cdot \left| \begin{array} { c c } { 0 } & { 2 } \\ { 1 } & { - 2 } \end{array} \right| - 1 \cdot \left| \begin{array} { c c } { - 3 } & { 2 } \\ { 0 } & { - 2 } \end{array} \right| + 0 \cdot \left| \begin{array} { c c } { - 3 } & { 0 } \\ { 0 } & { 1 } \end{array} \right| \\ & = 1 ( - 2 \cdot 0 - 1 \cdot 2 ) - 1 ( 2 \cdot 3 - 0 \cdot 2 ) + 0 \\ & = - 2 - 6 \\ & = - 8 \end{aligned} \] Determinant of [$A_1b$]: \[ \begin{aligned} \operatorname { det } \left[ A _ { 1 } ( \mathbf { b } ) \right] & = 3 \cdot \left| \begin{array} { c c } { 0 } & { 2 } \\ { 1 } & { - 2 } \end{array} \right| - 1 \cdot \left| \begin{array} { c c } { 0 } & { 2 } \\ { 2 } & { - 2 } \end{array} \right| + 0 \cdot \left| \begin{array} { c c } { 0 } & { 0 } \\ { 2 } & { 1 } \end{array} \right| \\ & = 3 ( - 2 \cdot 0 - 1 \cdot 2 ) - 1 ( - 2 \cdot 0 - 2 \cdot 2 ) + 0 \\ & = - 6 + 4 + 0 \\ & = - 2 \end{aligned} \] Determinant of [$A_2b$]: \[ \begin{aligned} \operatorname { det } \left[ A _ { 2 } ( \mathbf { b } ) \right] & = 1 \cdot \left| \begin{array} { c c } { 0 } & { 2 } \\ { 2 } & { - 2 } \end{array} \right| - 3 \cdot \left| \begin{array} { c c } { - 3 } & { 2 } \\ { 0 } & { - 2 } \end{array} \right| + 0 \cdot \left| \begin{array} { c c } { - 3 } & { 0 } \\ { 0 } & { 2 } \end{array} \right| \\ & = 1 ( - 2 \cdot 0 - 2 \cdot 2 ) - 3 ( 2 \cdot 3 - 0 \cdot 2 ) + 0 \\ & = - 4 - 18 \\ & = - 22 \end{aligned} \] Determinant of [$A_3b$]: \[ \begin{aligned} \operatorname { det } \left[ A _ { 3 } ( \mathbf { b } ) \right] & = 1 \cdot \left| \begin{array} { c c } { 0 } & { 0 } \\ { 1 } & { 2 } \end{array} \right| - 1 \cdot \left| \begin{array} { c c } { - 3 } & { 0 } \\ { 0 } & { 2 } \end{array} \right| + 3 \cdot \left| \begin{array} { c c } { - 3 } & { 0 } \\ { 0 } & { 1 } \end{array} \right| \\ & = 1 ( 2 \cdot 0 - 0 \cdot 1 ) - 1 ( - 3 \cdot 2 - 0 \cdot 0 ) + 3 ( - 3 \cdot 1 - 0 \cdot 0 ) \\ & = 6 - 9 \\ & = - 3 \end{aligned} \]
Step-3: The Solution
Solve for $x_1$ \[ \begin{aligned} x _ { 1 } & = \dfrac { \operatorname { det } \left[ A _ { 1 } ( \mathbf { b } ) \right] } { \operatorname { det } A } \\ & = \dfrac { - 2 } { - 8 } \\ & = \dfrac { 1 } { 4 } \end{aligned} \] Solve for $x_2$ \[ \begin{aligned} x _ { 2 } & = \dfrac { \operatorname { det } \left[ A _ { 2 } ( \mathbf { b } ) \right] } { \operatorname { det } A } \\ & = \dfrac { - 22 } { - 8 } \\ & = \dfrac { 11 } { 4 } \end{aligned} \] Solve for $x_3$ \[ \begin{aligned} x _ { 3 } & = \dfrac { \operatorname { det } \left[ A _ { 3 } ( \mathbf { b } ) \right] } { \operatorname { det } A } \\ & = \dfrac { - 3 } { - 8 } \\ & = \dfrac { 3 } { 8 } \end{aligned} \] Therefore, the solution is:
\[{\bf{x}} = \left[ {\begin{array}{*{20}{c}} {\dfrac{1}{4}}\\ {\dfrac{{11}}{4}}\\ {\dfrac{3}{8}} \end{array}} \right]\]
We are given with system of equations: \[ \begin{aligned} x _ { 1 } + x _ { 2 } & = 3 \\ - 3 x _ { 1 } + 2 x _ { 3 } & = 0 \\ x _ { 2 } - 2 x _ { 3 } & = 2 \end{aligned} \] We have to solve the system of equations using crammer's Rule
Step-1: The matrices
The matrices A and B for the equation $Ax=B$ are: \[ A = \left[ \begin{array} { c c c } { 1 } & { 1 } & { 0 } \\ { - 3 } & { 0 } & { 2 } \\ { 0 } & { 1 } & { - 2 } \end{array} \right] \text { and } \mathbf { b } = \left[ \begin{array} { l } { 3 } \\ { 0 } \\ { 2 } \end{array} \right] \] The matrices $ A _ { 1 } ( \mathbf { b } ) , A_ { 2 } ( \mathbf { b } ) \text { and } A _ { 3 } ( \mathbf { b } ) $ are: \[{A_1}({\bf{b}}) = \left[ {\begin{array}{*{20}{c}} 3&1&0\\ 0&0&2\\ 2&1&{ - 2} \end{array}} \right];\,\,{A_2}({\bf{b}}) = \left[ {\begin{array}{*{20}{c}} 1&3&0\\ { - 3}&0&2\\ 0&2&{ - 2} \end{array}} \right];\,\,{A_3}({\bf{b}}) = \left[ {\begin{array}{*{20}{c}} 1&1&3\\ { - 3}&0&0\\ 0&1&2 \end{array}} \right]\]
Step-2: The Determinants
Determinant of A: \[ \begin{aligned} \operatorname { det } A & = 1 \cdot \left| \begin{array} { c c } { 0 } & { 2 } \\ { 1 } & { - 2 } \end{array} \right| - 1 \cdot \left| \begin{array} { c c } { - 3 } & { 2 } \\ { 0 } & { - 2 } \end{array} \right| + 0 \cdot \left| \begin{array} { c c } { - 3 } & { 0 } \\ { 0 } & { 1 } \end{array} \right| \\ & = 1 ( - 2 \cdot 0 - 1 \cdot 2 ) - 1 ( 2 \cdot 3 - 0 \cdot 2 ) + 0 \\ & = - 2 - 6 \\ & = - 8 \end{aligned} \] Determinant of [$A_1b$]: \[ \begin{aligned} \operatorname { det } \left[ A _ { 1 } ( \mathbf { b } ) \right] & = 3 \cdot \left| \begin{array} { c c } { 0 } & { 2 } \\ { 1 } & { - 2 } \end{array} \right| - 1 \cdot \left| \begin{array} { c c } { 0 } & { 2 } \\ { 2 } & { - 2 } \end{array} \right| + 0 \cdot \left| \begin{array} { c c } { 0 } & { 0 } \\ { 2 } & { 1 } \end{array} \right| \\ & = 3 ( - 2 \cdot 0 - 1 \cdot 2 ) - 1 ( - 2 \cdot 0 - 2 \cdot 2 ) + 0 \\ & = - 6 + 4 + 0 \\ & = - 2 \end{aligned} \] Determinant of [$A_2b$]: \[ \begin{aligned} \operatorname { det } \left[ A _ { 2 } ( \mathbf { b } ) \right] & = 1 \cdot \left| \begin{array} { c c } { 0 } & { 2 } \\ { 2 } & { - 2 } \end{array} \right| - 3 \cdot \left| \begin{array} { c c } { - 3 } & { 2 } \\ { 0 } & { - 2 } \end{array} \right| + 0 \cdot \left| \begin{array} { c c } { - 3 } & { 0 } \\ { 0 } & { 2 } \end{array} \right| \\ & = 1 ( - 2 \cdot 0 - 2 \cdot 2 ) - 3 ( 2 \cdot 3 - 0 \cdot 2 ) + 0 \\ & = - 4 - 18 \\ & = - 22 \end{aligned} \] Determinant of [$A_3b$]: \[ \begin{aligned} \operatorname { det } \left[ A _ { 3 } ( \mathbf { b } ) \right] & = 1 \cdot \left| \begin{array} { c c } { 0 } & { 0 } \\ { 1 } & { 2 } \end{array} \right| - 1 \cdot \left| \begin{array} { c c } { - 3 } & { 0 } \\ { 0 } & { 2 } \end{array} \right| + 3 \cdot \left| \begin{array} { c c } { - 3 } & { 0 } \\ { 0 } & { 1 } \end{array} \right| \\ & = 1 ( 2 \cdot 0 - 0 \cdot 1 ) - 1 ( - 3 \cdot 2 - 0 \cdot 0 ) + 3 ( - 3 \cdot 1 - 0 \cdot 0 ) \\ & = 6 - 9 \\ & = - 3 \end{aligned} \]
Step-3: The Solution
Solve for $x_1$ \[ \begin{aligned} x _ { 1 } & = \dfrac { \operatorname { det } \left[ A _ { 1 } ( \mathbf { b } ) \right] } { \operatorname { det } A } \\ & = \dfrac { - 2 } { - 8 } \\ & = \dfrac { 1 } { 4 } \end{aligned} \] Solve for $x_2$ \[ \begin{aligned} x _ { 2 } & = \dfrac { \operatorname { det } \left[ A _ { 2 } ( \mathbf { b } ) \right] } { \operatorname { det } A } \\ & = \dfrac { - 22 } { - 8 } \\ & = \dfrac { 11 } { 4 } \end{aligned} \] Solve for $x_3$ \[ \begin{aligned} x _ { 3 } & = \dfrac { \operatorname { det } \left[ A _ { 3 } ( \mathbf { b } ) \right] } { \operatorname { det } A } \\ & = \dfrac { - 3 } { - 8 } \\ & = \dfrac { 3 } { 8 } \end{aligned} \] Therefore, the solution is:
\[{\bf{x}} = \left[ {\begin{array}{*{20}{c}} {\dfrac{1}{4}}\\ {\dfrac{{11}}{4}}\\ {\dfrac{3}{8}} \end{array}} \right]\]