Key Concepts and Formulas

For a system of three linear equations in three variables, represented as $AX=B$, to have infinitely many solutions, the following conditions based on Cramer's Rule must be met:

• Determinant of the coefficient matrix ($\Delta$) must be zero. $\Delta = \left|\begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right| = 0$
• All other major determinants ($\Delta_x, \Delta_y, \Delta_z$) must also be zero.
  • $\Delta_x$: Determinant formed by replacing the $x$-coefficients column in $\Delta$ with the constant terms.
  • $\Delta_y$: Determinant formed by replacing the $y$-coefficients column in $\Delta$ with the constant terms.
  • $\Delta_z$: Determinant formed by replacing the $z$-coefficients column in $\Delta$ with the constant terms. $\Delta_x = 0 \quad \text{AND} \quad \Delta_y = 0 \quad \text{AND} \quad \Delta_z = 0$
• If $\Delta = 0$ but at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero, the system has no solution. If $\Delta \neq 0$, the system has a unique solution.

Step-by-Step Solution

Step 1: Identify the coefficient matrix and constant terms. The given system of linear equations is: $\begin{aligned} & 2 x+\lambda y+3 z=5 \\ & 3 x+2 y-z=7 \\ & 4 x+5 y+\mu z=9 \end{aligned}$ From these equations, we can extract the coefficient matrix $A$ and the constant terms vector $B$: $A = \begin{pmatrix} 2 & \lambda & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \mu \end{pmatrix}, \quad B = \begin{pmatrix} 5 \\ 7 \\ 9 \end{pmatrix}$

Step 2: Set up the required determinants. Based on the coefficients and constant terms, we form the four determinants:

• $\Delta$ (coefficient determinant): $\Delta = \left|\begin{array}{ccc} 2 & \lambda & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \mu \end{array}\right|$
• $\Delta_x$ (replace $x$-coefficients with constants): $\Delta_x = \left|\begin{array}{ccc} 5 & \lambda & 3 \\ 7 & 2 & -1 \\ 9 & 5 & \mu \end{array}\right|$
• $\Delta_y$ (replace $y$-coefficients with constants): $\Delta_y = \left|\begin{array}{ccc} 2 & 5 & 3 \\ 3 & 7 & -1 \\ 4 & 9 & \mu \end{array}\right|$
• $\Delta_z$ (replace $z$-coefficients with constants): $\Delta_z = \left|\begin{array}{ccc} 2 & \lambda & 5 \\ 3 & 2 & 7 \\ 4 & 5 & 9 \end{array}\right|$

Step 3: Calculate $\Delta_z$ and solve for $\lambda$. We start with $\Delta_z$ because it contains only one unknown variable ($\lambda$), which will allow us to directly solve for $\lambda$. For infinitely many solutions, $\Delta_z$ must be zero. $\Delta_z = \left|\begin{array}{ccc} 2 & \lambda & 5 \\ 3 & 2 & 7 \\ 4 & 5 & 9 \end{array}\right|$ Expand the determinant along the first row: $\Delta_z = 2 \cdot (2 \cdot 9 - 7 \cdot 5) - \lambda \cdot (3 \cdot 9 - 7 \cdot 4) + 5 \cdot (3 \cdot 5 - 2 \cdot 4)$ $\Delta_z = 2 \cdot (18 - 35) - \lambda \cdot (27 - 28) + 5 \cdot (15 - 8)$ $\Delta_z = 2 \cdot (-17) - \lambda \cdot (-1) + 5 \cdot (7)$ $\Delta_z = -34 + \lambda + 35$ $\Delta_z = \lambda + 1$ Setting $\Delta_z = 0$ for infinitely many solutions: $\lambda + 1 = 0 \implies \lambda = -1$

Step 4: Calculate $\Delta$ and solve for $\mu$. Next, we use the determinant $\Delta$, which involves both $\lambda$ and $\mu$. Since we have found $\lambda = -1$, we can substitute this value into $\Delta$ and then solve for $\mu$. For infinitely many solutions, $\Delta$ must also be zero. $\Delta = \left|\begin{array}{ccc} 2 & \lambda & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \mu \end{array}\right|$ Expand the determinant along the first row: $\Delta = 2 \cdot (2 \cdot \mu - (-1) \cdot 5) - \lambda \cdot (3 \cdot \mu - (-1) \cdot 4) + 3 \cdot (3 \cdot 5 - 2 \cdot 4)$ $\Delta = 2 \cdot (2\mu + 5) - \lambda \cdot (3\mu + 4) + 3 \cdot (15 - 8)$ $\Delta = 4\mu + 10 - 3\lambda\mu - 4\lambda + 3 \cdot (7)$ $\Delta = 4\mu + 10 - 3\lambda\mu - 4\lambda + 21$ $\Delta = 31 - 4\lambda + 4\mu - 3\lambda\mu$ Now, substitute $\lambda = -1$ into the equation $\Delta = 0$: $31 - 4(-1) + 4\mu - 3(-1)\mu = 0$ $31 + 4 + 4\mu + 3\mu = 0$ $35 + 7\mu = 0$ $7\mu = -35$ $\mu = -5$

Step 5: Verify with $\Delta_x$ and $\Delta_y$ (Recommended for completeness). To ensure our values of $\lambda = -1$ and $\mu = -5$ are correct, we should verify that $\Delta_x = 0$ and $\Delta_y = 0$ with these values.

• Check $\Delta_x$: Substitute $\lambda = -1$ and $\mu = -5$: $\Delta_x = \left|\begin{array}{ccc} 5 & -1 & 3 \\ 7 & 2 & -1 \\ 9 & 5 & -5 \end{array}\right|$ Expand along the first row: $\Delta_x = 5(2(-5) - (-1)5) - (-1)(7(-5) - (-1)9) + 3(7 \cdot 5 - 2 \cdot 9)$ $\Delta_x = 5(-10 + 5) + 1(-35 + 9) + 3(35 - 18)$ $\Delta_x = 5(-5) + 1(-26) + 3(17)$ $\Delta_x = -25 - 26 + 51$ $\Delta_x = -51 + 51 = 0$ This condition is satisfied.

• Check $\Delta_y$: Substitute $\lambda = -1$ and $\mu = -5$: $\Delta_y = \left|\begin{array}{ccc} 2 & 5 & 3 \\ 3 & 7 & -1 \\ 4 & 9 & -5 \end{array}\right|$ Expand along the first row: $\Delta_y = 2(7(-5) - (-1)9) - 5(3(-5) - (-1)4) + 3(3 \cdot 9 - 7 \cdot 4)$ $\Delta_y = 2(-35 + 9) - 5(-15 + 4) + 3(27 - 28)$ $\Delta_y = 2(-26) - 5(-11) + 3(-1)$ $\Delta_y = -52 + 55 - 3$ $\Delta_y = 3 - 3 = 0$ This condition is also satisfied. Since all four determinants are zero for $\lambda = -1$ and $\mu = -5$, these values are correct.

Step 6: Calculate the required expression $\left(\lambda^2+\mu^2\right)$. We need to find the value of $\lambda^2 + \mu^2$: $\lambda^2 + \mu^2 = (-1)^2 + (-5)^2$ $\lambda^2 + \mu^2 = 1 + 25$ $\lambda^2 + \mu^2 = 26$

Common Mistakes & Tips

• Arithmetic and Sign Errors: Determinant calculations can be prone to small errors, especially with negative numbers. Always double-check each step.
• Incomplete Conditions: A common mistake is only setting $\Delta = 0$. Remember that for infinitely many solutions, all four determinants ($\Delta, \Delta_x, \Delta_y, \Delta_z$) must be zero. If $\Delta = 0$ but any of $\Delta_x, \Delta_y, \Delta_z$ is non-zero, there are no solutions.
• Strategic Calculation: Choose the determinants to calculate first based on how many unknowns they contain. Solving determinants with fewer unknowns first can simplify the overall process.

Summary

To find the values of $\lambda$ and $\mu$ for which the system has infinitely many solutions, we applied the conditions from Cramer's Rule, which state that all four determinants ($\Delta, \Delta_x, \Delta_y, \Delta_z$) must be zero. We strategically calculated $\Delta_z$ first, as it contained only $\lambda$, yielding $\lambda = -1$. Substituting this value into $\Delta = 0$ allowed us to solve for $\mu = -5$. Finally, we verified these values by ensuring $\Delta_x = 0$ and $\Delta_y = 0$. With $\lambda = -1$ and $\mu = -5$, the expression $\lambda^2 + \mu^2$ was calculated as $26$.

Final Answer

The final answer is $\boxed{\text{26}}$, which corresponds to option (B).