Problem 29 - Advanced Engineering Mathematics

Question

$$ \left(\begin{array}{rrr} 1 & -2 & 1 \\ 0 & 1 & 2 \\ 2 & 6 & 1 \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right)=\left(\begin{array}{r} 14 \\ -42 \\ 7 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \( x_1 = 25 \), \( x_2 = -4 \), and \( x_3 = -19 \).

Step 1: 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: \[ \begin{align*} x_1 - 2x_2 + x_3 &= 14, \ x_2 + 2x_3 &= -42, \ 2x_1 + 6x_2 + x_3 &= 7. \end{align*} \]

Step 2: 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 \( x_2 \).

Step 3: 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: \[ x_1 = -70 - 5x_3. \]

Step 4: 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 \[ 2x_1 - 11x_3 = 259. \]

Step 5: 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 \( x_3 \).

Step 6: 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 \( x_3 \): \( x_3 = -19 \).

Step 7: 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 \( x_2 \): \[ x_2 = -42 - 2(-19) = -42 + 38 = -4. \]

Step 8: Final Solution

The solutions for the variables are \( x_1 = 25 \), \( x_2 = -4 \), \( x_3 = -19 \).

Key Concepts

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 \( 7 \).

To solve a matrix equation, you'll extract each row from \( A \) and \( \mathbf{b} \), forming individual equations. This method simplifies visualizing complex systems, making the handling and computation more systematic. It's as if you are summarizing the teaching of a course into a single sentence! As students, focus on converting these matrix forms into familiar linear equations if you're working by hand, then solve using algebraic methods.

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.

In our example, once you have \( x_1 \) and \( x_3 \), back-substitute these to find \( x_2 \). This transformative process helps in isolating variables step-by-step, making it easier to solve the whole 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.

For example, if you manage to line up terms such that the \( x_2 \) terms cancel, you'll focus on resolving the remaining variables, such as \( x_1 \) and \( x_3 \). Often combined with scaling equations (multiplying by constants) to facilitate clear cancellations, this method is especially useful when substitution might seem lengthy. It's like methodically peeling an onion layer by layer.