Chapter 8: Problem 29
$$ \left(\begin{array}{rrr} 1 & -2 & 1 \\ 0 & 1 & 2 \\ 2 & 6 & 1 \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right)=\left(\begin{array}{r} 14 \\ -42 \\ 7 \end{array}\right) $$
Short Answer
Expert verified
The solution is \( x_1 = 25 \), \( x_2 = -4 \), and \( x_3 = -19 \).
Step by step solution
01
Write Down the System of Equations
The given matrix equation can be translated into a system of linear equations. By multiplying each row of the first matrix by the vector \( \begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} \), we get three equations.\[ \begin{align*} 1 \cdot x_1 - 2 \cdot x_2 + 1 \cdot x_3 &= 14, \ 0 \cdot x_1 + 1 \cdot x_2 + 2 \cdot x_3 &= -42, \ 2 \cdot x_1 + 6 \cdot x_2 + 1 \cdot x_3 &= 7. \end{align*} \] This results in the system: \[ \begin{align*} x_1 - 2x_2 + x_3 &= 14, \ x_2 + 2x_3 &= -42, \ 2x_1 + 6x_2 + x_3 &= 7. \end{align*} \]
02
Simplify the System by Substitution or Elimination
We start by solving the second equation for \( x_2 \): \[ x_2 = -42 - 2x_3. \] Substitute \( x_2 \) from this equation into the first and third equations to eliminate \( x_2 \).
03
Substitute \( x_2 \) into the First Equation
Substitute \( x_2 = -42 - 2x_3 \) into the first equation: \[ x_1 - 2(-42 - 2x_3) + x_3 = 14. \] Simplify to find: \[ x_1 + 84 + 4x_3 + x_3 = 14, \] which simplifies to \[ x_1 + 5x_3 = -70. \] So we have: \[ x_1 = -70 - 5x_3. \]
04
Substitute \( x_2 \) into the Third Equation
Now, substitute \( x_2 = -42 - 2x_3 \) into the third equation: \[ 2x_1 + 6(-42 - 2x_3) + x_3 = 7. \] Simplify: \[ 2x_1 - 252 - 12x_3 + x_3 = 7. \] This gives \[ 2x_1 - 11x_3 = 259. \]
05
Solve for \( x_1 \) using \( 2x_1 - 11x_3 = 259 \)
From Step 4, we have the equation \( 2x_1 - 11x_3 = 259 \). Substitute \( x_1 = -70 - 5x_3 \) (from Step 3) into this equation: \[ 2(-70 - 5x_3) - 11x_3 = 259. \] Simplify \(-140 - 10x_3 - 11x_3 = 259\), which becomes \(-140 - 21x_3 = 259\). Solve for \( x_3 \).
06
Solve for \( x_3 \)
From Step 5: \(-140 - 21x_3 = 259\). Simplify by adding 140 to both sides: \( -21x_3 = 399 \). Divide by -21 to solve for \( x_3 \): \( x_3 = -19 \).
07
Substitute \( x_3 \) Back to Find \( x_1 \) and \( x_2 \)
With \( x_3 = -19 \), substitute back into \( x_1 = -70 - 5x_3 \) to find \( x_1 \): \[ x_1 = -70 - 5(-19) = -70 + 95 = 25. \] Use \( x_3 = -19 \) in \( x_2 = -42 - 2x_3 \) to find \( x_2 \): \[ x_2 = -42 - 2(-19) = -42 + 38 = -4. \]
08
Final Solution
The solutions for the variables are \( x_1 = 25 \), \( x_2 = -4 \), \( x_3 = -19 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Matrix Equation
A matrix equation is a concise way to represent a system of linear equations using matrices. In this form, you express the system with a single equation, typically written as \[ A\mathbf{x} = \mathbf{b} \]where:
- \( A \) is a matrix composed of coefficients for the system.
- \( \mathbf{x} \) is a column matrix or vector, containing the variables \( x_1, x_2, \) and \( x_3 \).
- \( \mathbf{b} \) is a column matrix containing constants \( 14, -42, \) and \( 7 \).
Substitution Method
The substitution method involves solving for one variable and substituting it into other equations. It's a very systematic approach! Start with the simplest equation to make solving easy. For instance, if you have:\[ x_2 + 2x_3 = -42\]You can express \( x_2 \) in terms of \( x_3 \):\[ x_2 = -42 - 2x_3 \]. Substitute this expression for \( x_2 \) into other equations to eliminate \( x_2 \). This reduces the number of variables and equations you need to tackle. Breakdown:
- Solve one equation for one variable.
- Substitute this expression into the remaining equations.
- Simplify and solve the reduced system.
Elimination Method
The elimination method focuses on cancelling out variables in a system of linear equations. This method usually involves multiple steps:The primary goal in elimination is to align coefficients of selected variables across equations so that adding or subtracting the equations cancels out one of the variables:
- Add or subtract equations to eliminate one variable.
- Once a variable is eliminated, the new equation will have fewer variables.
- Solve this simplified system and backtrack to find other variables.