Problem 18 - Algebra and Trigonometry

Question

\(17-36\) . Find the complete solution of the linear system, or show that it is inconsistent. $$ \left\\{\begin{aligned} x-y+z &=0 \\ y+2 z &=-2 \\ x+y-z &=2 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = 1 \), \( y = 0 \), \( z = -1 \). The system is consistent.

Step 1: Write Down the System of Equations

Identify and write the given system of equations in matrix form or as a list of equations.\[ \begin{align*} 1. & \quad x - y + z = 0 \ 2. & \quad y + 2z = -2 \ 3. & \quad x + y - z = 2 \end{align*}\]

Step 2: Use Substitution with Equation 2

From Equation 2, express \( y \) in terms of \( z \):\[ y = -2 - 2z \]

Step 3: Substitute into Equation 1

Replace \( y \) in Equation 1 with the expression from Step 2:\[ x - (-2 - 2z) + z = 0 \]Simplify the equation:\[ x + 2 + 2z + z = 0 \]\[ x + 3z = -2 \]

Step 4: Substitute into Equation 3

Replace \( y \) in Equation 3 with the expression from Step 2:\[ x + (-2 - 2z) - z = 2 \]Simplify the equation:\[ x - 2 - 2z - z = 2 \]\[ x - 3z = 4 \]

Step 5: Solve Linear Equations for x and z

Now solve the two equations obtained from the substitutions:\[ \begin{align*} x + 3z &= -2 \quad (Equation \; A) \ x - 3z &= 4 \quad (Equation \; B) \end{align*} \]Add the equations to eliminate \( z \):\[ (x + 3z) + (x - 3z) = -2 + 4 \]\[ 2x = 2 \]Divide by 2:\[ x = 1 \]

Step 6: Solve for z

Use \( Equation \; A\) to find \( z \):\[ 1 + 3z = -2 \]Subtract 1:\[ 3z = -3 \]Divide by 3:\[ z = -1 \]

Step 7: Solve for y

Substitute \( z = -1 \) back into the expression for \( y \):\[ y = -2 - 2(-1) \]\[ y = -2 + 2 \]\[ y = 0 \]

Step 8: Verify the Solution

Verify by substituting \( x = 1 \), \( y = 0 \), and \( z = -1 \) back into the original equations:1. \(1 - 0 - 1 = 0\) is true.2. \(0 + 2(-1) = -2\) is true.3. \(1 + 0 + 1 = 2\) is true.All original equations are satisfied.

Key Concepts

System of Equations

A system of equations is a set of equations with multiple variables that are related to each other. In our example, we have the following three equations:

• \( x - y + z = 0 \)
• \( y + 2z = -2 \)
• \( x + y - z = 2 \)

Each equation represents a plane in a 3-dimensional space where the variables \( x \), \( y \), and \( z \) intersect. Our goal is to find the values of these variables that simultaneously satisfy all these equations. Solving such a system means finding a point in 3D space where all planes intersect. If no such point exists, the system is inconsistent and has no solution. Otherwise, we derive values that satisfy all equations in the system.

Substitution Method

The substitution method is a powerful technique for solving a system of equations, especially when the system is not too large. The idea is to express one variable in terms of the others using one equation and then substitute this expression into the other equations.
For example, from our system, we expressed \( y \) from the second equation:

• \( y = -2 - 2z \)

This expression was then substituted into the other two equations, allowing us to eliminate \( y \) and form a new system with only \( x \) and \( z \). By doing so, we reduced the complexity of the system, making it easier to solve directly.
In this case, the substitution resulted in:

• \( x + 3z = -2 \)
• \( x - 3z = 4 \)

These simpler equations were then used to find the values of \( x \) and \( z \), and subsequently, \( y \) was found using the expression we derived initially.

Matrix Form

The matrix form of a system of equations is a compact and efficient way of representing the system.
Instead of writing each equation separately, we can use matrices to express the coefficients of the variables and the constants on the right-hand side.
Consider our system of equations:

• \( x - y + z = 0 \)
• \( y + 2z = -2 \)
• \( x + y - z = 2 \)

In matrix form, this system can be written as:\[\begin{bmatrix}1 & -1 & 1 \0 & 1 & 2 \1 & 1 & -1 \\end{bmatrix}\begin{bmatrix}x \y \z \\end{bmatrix}= \begin{bmatrix}0 \-2 \2 \\end{bmatrix}\]The left matrix represents the coefficients of the variables, the middle column matrix represents the variables themselves, and the right column matrix represents the constants.
This matrix form allows mathematicians and computer programs to utilize linear algebra techniques like Gaussian elimination or the inversion method to solve the system efficiently.