Analyzing the System of Linear Equations

We are given a system of three linear equations with three variables x, y, and z:

1. $x + y + z = 8$
2. $x - y + 2z = 6$
3. $3x - y + 5z = k$

To determine the nature of the solutions (unique, infinitely many, or no solution), we can analyze the determinant of the coefficient matrix and the consistency of the augmented matrix.

Representing the System with Matrices

The system can be written in the matrix form $AX = B$, where:

Coefficient matrix $A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ 3 & -1 & 5 \end{pmatrix}$

Variable matrix $X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$

Constant matrix $B = \begin{pmatrix} 8 \\ 6 \\ k \end{pmatrix}$

The augmented matrix is $[A|B] = \begin{pmatrix} 1 & 1 & 1 & | & 8 \\ 1 & -1 & 2 & | & 6 \\ 3 & -1 & 5 & | & k \end{pmatrix}$

Calculating the Determinant of the Coefficient Matrix

Let's calculate the determinant of matrix A, denoted as $|A|$:

$|A| = 1 \begin{vmatrix} -1 & 2 \\ -1 & 5 \end{vmatrix} - 1 \begin{vmatrix} 1 & 2 \\ 3 & 5 \end{vmatrix} + 1 \begin{vmatrix} 1 & -1 \\ 3 & -1 \end{vmatrix}$

$|A| = 1((-1)(5) - (2)(-1)) - 1((1)(5) - (2)(3)) + 1((1)(-1) - (-1)(3))$

$|A| = 1(-5 + 2) - 1(5 - 6) + 1(-1 + 3)$

$|A| = 1(-3) - 1(-1) + 1(2)$

$|A| = -3 + 1 + 2$

$|A| = 0$

Interpreting the Determinant Value

Since the determinant of the coefficient matrix $|A|$ is 0, the system of equations does not have a unique solution. A unique solution exists only if $|A| \neq 0$. This immediately tells us that statement 3 is incorrect.

Analyzing Consistency using Gaussian Elimination

To determine if the system has no solution or infinitely many solutions, we perform row operations on the augmented matrix to bring it to row-echelon form.

The last row of the row-echelon form of the augmented matrix corresponds to the equation $0x + 0y + 0z = k - 20$, which simplifies to $0 = k - 20$.

Evaluating the Statements

We can now evaluate each statement based on the resulting equation $0 = k - 20$:

Statement 1: They have no solution if k = 15

• If $k = 15$, the last row equation becomes $0 = 15 - 20$, which is $0 = -5$.
• This is a contradiction (0 cannot equal -5).
• A contradictory equation indicates that the system is inconsistent and has no solution.
• Therefore, statement 1 is correct.

Statement 2: They have infinitely many solutions, if k = 20

• If $k = 20$, the last row equation becomes $0 = 20 - 20$, which is $0 = 0$.
• This equation is always true and does not impose any restriction on the variables.
• In the row-echelon form, we have two non-zero rows corresponding to three variables. This means the rank of the coefficient matrix (2) is equal to the rank of the augmented matrix (2), but less than the number of variables (3).
• This condition leads to infinitely many solutions, where one variable can be chosen freely (a free variable).
• Therefore, statement 2 is correct.

Statement 3: They have a unique solution if k = 25

• For a unique solution, the determinant of the coefficient matrix must be non-zero ($|A| \neq 0$).
• We calculated $|A| = 0$.
• Since $|A| = 0$ for any value of k, the system can never have a unique solution.
• Therefore, statement 3 is incorrect for any value of k, including k = 25.

Based on our analysis, statements 1 and 2 are correct, while statement 3 is incorrect.

Conclusion on System Solutions

The system of equations has:

• No solution if $k \neq 20$ (e.g., when k = 15).
• Infinitely many solutions if $k = 20$.
• Never a unique solution.

Revision Table: System Analysis Summary

Additional Information on Linear Systems

For a system of linear equations $AX=B$ with n variables and n equations:

• Unique Solution: Exists if and only if the determinant of the coefficient matrix $|A| \neq 0$. In this case, the rank of A equals the rank of the augmented matrix [A|B], and both are equal to n (the number of variables).
• No Solution: Occurs if $|A| = 0$ and the system is inconsistent. This happens when the rank of A is less than the rank of the augmented matrix [A|B]. Gaussian elimination will result in a row like $[0 \ 0 \ ... \ 0 \ | \ \text{non-zero}]$.
• Infinitely Many Solutions: Occurs if $|A| = 0$ and the system is consistent. This happens when the rank of A equals the rank of the augmented matrix [A|B], and both are less than the number of variables n. Gaussian elimination will result in one or more rows of zeros in both the coefficient and augmented parts $[0 \ 0 \ ... \ 0 \ | \ 0]$. The number of free variables is n minus the rank.