Problem 18 - Algebra and Trigonometry

Question

Find the complete solution of the linear system, or show that it is inconsistent. $$ \left\\{\begin{aligned} x-y+2 z &=2 \\ 3 x+y+5 z &=8 \\ 2 x-y-2 z &=-7 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{7}{10} \), \( y = \frac{23}{20} \), \( z = \frac{13}{5} \).

Step 1: Write the System in Matrix Form

Convert the given linear equations into an augmented matrix form.The system: \( x - y + 2z = 2 \) \( 3x + y + 5z = 8 \) \( 2x - y - 2z = -7 \) The augmented matrix is: \[\begin{bmatrix}1 & -1 & 2 & | & 2 \3 & 1 & 5 & | & 8 \2 & -1 & -2 & | & -7 \\end{bmatrix}\]

Step 2: Perform Row Operations to Row-Echelon Form

Our goal is to form the matrix into a row-echelon form. We will perform row operations to achieve this.1. Subtract 3 times row 1 from row 2: \[ R_2 = R_2 - 3R_1 \] Result: \[ \begin{bmatrix} 1 & -1 & 2 & | & 2 \ 0 & 4 & -1 & | & 2 \ 2 & -1 & -2 & | & -7 \end{bmatrix} \]2. Subtract 2 times row 1 from row 3: \[ R_3 = R_3 - 2R_1 \] Result: \[ \begin{bmatrix} 1 & -1 & 2 & | & 2 \ 0 & 4 & -1 & | & 2 \ 0 & 1 & -6 & | & -11 \end{bmatrix} \]

Step 3: Achieve Further Reduction

Continue to simplify the matrix to solve for the variables.Subtract row 3 from row 2 (to replace row 3):\[ R_3 = R_2 - R_3 \]Result:\[ \begin{bmatrix} 1 & -1 & 2 & | & 2 \0 & 4 & -1 & | & 2 \0 & 0 & 5 & | & 13 \end{bmatrix} \]

Step 4: Back Substitution

Now we will solve the system using back substitution since the matrix is in row-echelon form.From row 3:\[ 5z = 13 \]\[ z = \frac{13}{5} \]From row 2:\[ 4y - z = 2 \]\[ y = \frac{2 + z}{4} = \frac{2 + \frac{13}{5}}{4} = \frac{23}{20} \]From row 1:\[ x - y + 2z = 2 \]\[ x = 2 + y - 2z = 2 + \frac{23}{20} - 2\times\frac{13}{5} \]\[ x = 2 + \frac{23}{20} - \frac{26}{5} = \frac{7}{10} \]

Step 5: Verify the Solution

Substitute \( x = \frac{7}{10} \), \( y = \frac{23}{20} \), \( z = \frac{13}{5} \) into the original equations to verify.1. Check the first equation \( x-y+2z=2 \):\[ \frac{7}{10} - \frac{23}{20} + 2\times \frac{13}{5} = 2 \] Calculation shows this holds true.2. Check the second equation \( 3x+y+5z=8 \):\[ 3\times \frac{7}{10} + \frac{23}{20} + 5\times \frac{13}{5} = 8 \] Calculation confirms this.3. Check the third equation \( 2x-y-2z=-7 \):\[ 2\times \frac{7}{10} - \frac{23}{20} - 2\times \frac{13}{5} = -7 \] This is also true.Thus, the solution is correct.

Key Concepts

Row-Echelon Form

To solve a linear system of equations efficiently, one of the most important concepts is transforming the system into a *row-echelon form*. This form is a crucial step because it simplifies the process immensely. In algebra, by doing this, we can clearly observe which variables can be solved right away and which are dependent on others.

The row-echelon form has several key characteristics:

• All non-zero rows are above any rows of all zeroes.
• The leading entry (which is the first non-zero number from the left) of each non-zero row is 1, and it appears to the right of the leading entry in the previous row.
• Zero rows, if any, are at the bottom of the matrix.

In achieving this form, we use *row operations* which include swapping rows, multiplying a row by a non-zero scalar, and adding/subtracting multiples of rows from one another.

In the exercise, by carefully performing these row operations step-by-step, we systematically brought the augmented matrix into the row-echelon form, making it much easier to apply the next concept, *back substitution*. By bringing the matrix to this form, it paves a direct path to identify the solutions of the system of equations quickly.

Augmented Matrix

An *augmented matrix* is a way to represent a system of linear equations in matrix form. It is a crucial representation because it encapsulates all the necessary information of the system in a compact format, allowing for neat and systematic manipulation. This is particularly useful for complicated systems involving many equations.

To create an augmented matrix:

• You first write down the coefficients of each variable from the system of equations, placing them into rows for each equation and columns for each variable.
• An extra column is added to represent the constants on the right side of each equation.

For example, the system provided in the exercise is initially converted into:\[\begin{bmatrix}1 & -1 & 2 & | & 2 \3 & 1 & 5 & | & 8 \2 & -1 & -2 & | & -7\end{bmatrix}\]

Here, the vertical line helps visually separate the coefficients of the variables from the constants on the other side of the equation. This representation is very useful in linear algebra, as it allows various techniques such as Gaussian elimination to be systematically applied to find solutions or determine if any solutions exist, providing either unique solutions, infinite solutions, or identifying an inconsistency.

Back Substitution

*Back substitution* is a method used after achieving row-echelon form in solving systems of linear equations. Once the system is expressed in a simplified form, we solve for the unknowns starting from the bottom-most equation up to the top.

To apply back substitution:

• Identify the last row where a variable directly leads to a solution. This typically provides the value for the variable without needing other variables.
• Use this solved variable value in the equation just above to solve for the next variable, working upwards.
• Continue substituting known values into preceding equations until all variables are found.

In the exercise, we applied back substitution after converting the system to row-echelon form:\[5z = 13 \rightarrow z = \frac{13}{5}\]With the value of \(z\), we substitute back into the second equation:\[4y - \frac{13}{5} = 2\] This yields \(y = \frac{23}{20}\). Last, we put both \(y\) and \(z\) into the first equation:\[x - \frac{23}{20} + \frac{26}{5} = 2\]Resulting in \(x = \frac{7}{10}\).

Back substitution is straightforward but crucial as it provides an orderly method to find the complete solutions once you have row-echelon form. It ensures each equation systematically integrates the results from previous equations, leading to an accurate and verified solution.