Understanding Unique Solutions for Linear Systems

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

• Equation 1: \(2x + 3y + 5z = 9\)
• Equation 2: \(7x + 3y - 2z = 8\)
• Equation 3: \(2x + 3y + \lambda z = \mu\)

We want to find the condition under which this system of linear equations has a unique solution. A system of linear equations \(Ax = B\) has a unique solution if and only if the determinant of the coefficient matrix \(A\) is non-zero (\(|A| \ne 0\)).

Coefficient Matrix and Determinant Calculation

First, let's write down the coefficient matrix \(A\) for the given system of linear equations:

So, the coefficient matrix is:

$$ A = \begin{pmatrix} 2 & 3 & 5 \\ 7 & 3 & -2 \\ 2 & 3 & \lambda \end{pmatrix} $$

Next, we calculate the determinant of matrix \(A\), denoted by \(|A|\). We can expand along the first column or first row. Let's expand along the first row:

$$ |A| = 2 \times \det \begin{pmatrix} 3 & -2 \\ 3 & \lambda \end{pmatrix} - 3 \times \det \begin{pmatrix} 7 & -2 \\ 2 & \lambda \end{pmatrix} + 5 \times \det \begin{pmatrix} 7 & 3 \\ 2 & 3 \end{pmatrix} $$

Calculate the 2x2 determinants:

$$ \det \begin{pmatrix} 3 & -2 \\ 3 & \lambda \end{pmatrix} = (3)(\lambda) - (-2)(3) = 3\lambda + 6 $$

$$ \det \begin{pmatrix} 7 & -2 \\ 2 & \lambda \end{pmatrix} = (7)(\lambda) - (-2)(2) = 7\lambda + 4 $$

$$ \det \begin{pmatrix} 7 & 3 \\ 2 & 3 \end{pmatrix} = (7)(3) - (3)(2) = 21 - 6 = 15 $$

Substitute these values back into the determinant formula for \(|A|\):

$$ |A| = 2 \times (3\lambda + 6) - 3 \times (7\lambda + 4) + 5 \times (15) $$

$$ |A| = 6\lambda + 12 - 21\lambda - 12 + 75 $$

Combine like terms:

$$ |A| = (6\lambda - 21\lambda) + (12 - 12 + 75) $$

$$ |A| = -15\lambda + 75 $$

Condition for Unique Solutions

For the system of linear equations to have a unique solution, the determinant of the coefficient matrix must be non-zero.

$$ |A| \ne 0 $$

So, we set the determinant we calculated not equal to zero:

$$ -15\lambda + 75 \ne 0 $$

To solve for \(\lambda\), subtract 75 from both sides:

$$ -15\lambda \ne -75 $$

Now, divide by -15. Remember that when dividing or multiplying an inequality by a negative number, the inequality sign reverses. However, in this case, we have a 'not equal to' sign, so it remains 'not equal to'.

$$ \lambda \ne \frac{-75}{-15} $$

$$ \lambda \ne 5 $$

This condition, \(\lambda \ne 5\), ensures that the coefficient matrix is invertible, which directly leads to a unique solution for any right-hand side vector.

Role of μ in System Solutions

The value of \(\mu\) is a constant term on the right-hand side of the third equation. The condition for a unique solution for a system \(Ax=B\) depends solely on the properties of the coefficient matrix \(A\) (its determinant being non-zero or its rank being equal to the number of variables).

If \(|A| \ne 0\), the system \(Ax=B\) has a unique solution given by \(x = A^{-1}B\), regardless of the value of \(B\). In our case, the vector \(B\) contains the constants 9, 8, and \(\mu\). If \(\lambda \ne 5\) (which means \(|A| \ne 0\)), the system has a unique solution, and this is true for any real value of \(\mu\).

The value of \(\mu\) becomes important only when \(|A| = 0\) (i.e., when \(\lambda = 5\)). In that specific case (\(\lambda = 5\)), the system would need further analysis (e.g., checking the rank of the augmented matrix) to determine if it has no solution or infinitely many solutions, and \(\mu\) would play a role in that determination. But for the system to have a unique solution, the determinant condition \(|A| \ne 0\) is sufficient and necessary, and it only depends on \(\lambda\).

Conclusion for Unique Solution Condition

Based on the determinant calculation, the system of linear equations has a unique solution if and only if \(\lambda \ne 5\). The value of \(\mu\) can be any real number without affecting the existence of a unique solution when \(\lambda \ne 5\).

Therefore, the condition for the system to have unique solutions is \(\lambda \ne 5\) and \(\mu\) has any real value.

Revision Table: Types of Solutions for Linear Systems

Additional Information: Methods for Solving Linear Systems

There are several methods to solve a system of linear equations and understand the nature of its solutions:

• Gaussian Elimination: This method transforms the augmented matrix \([A|B]\) into row echelon form or reduced row echelon form. The number of pivots determines the rank of the matrix. If the rank of \(A\) equals the rank of \([A|B]\) and equals the number of variables, there is a unique solution.
• Matrix Inversion: As discussed, if the coefficient matrix \(A\) is square and non-singular (i.e., \(|A| \ne 0\)), its inverse \(A^{-1}\) exists, and the unique solution is given by \(x = A^{-1}B\).
• Cramer's Rule: This rule provides a direct formula for the variables using determinants, applicable only when \(|A| \ne 0\). Each variable \(x_i\) is the ratio of the determinant of a modified matrix \(A_i\) (where the \(i\)-th column is replaced by \(B\)) to the determinant of the original matrix \(A\). \(x_i = \frac{|A_i|}{|A|}\). This clearly shows that \(|A|\) must be non-zero for the solution to be defined and unique.

All these methods highlight the critical role of the coefficient matrix's properties in determining the existence and uniqueness of the solution for a system of linear equations.