Key Concept: Conditions for Infinitely Many Solutions in a System of Linear Equations

For a system of linear equations in three variables, represented in matrix form as $AX=B$:

$$\begin{aligned} a_1x + b_1y + c_1z &= d_1 \\ a_2x + b_2y + c_2z &= d_2 \\ a_3x + b_3y + c_3z &= d_3 \end{aligned}$$

to have infinitely many solutions, the following conditions, derived from Cramer's Rule, must be satisfied:

1. The determinant of the coefficient matrix, denoted as $\Delta$ (or $D$), must be equal to zero. $\Delta = \left| \begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array} \right| = 0$
2. All the determinants formed by replacing a column of the coefficient matrix with the constant terms (i.e., $\Delta_x$, $\Delta_y$, and $\Delta_z$) must also be equal to zero. $\Delta_x = \left| \begin{array}{ccc} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{array} \right| = 0$ $\Delta_y = \left| \begin{array}{ccc} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{array} \right| = 0$ $\Delta_z = \left| \begin{array}{ccc} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{array} \right| = 0$ If $\Delta = 0$ but at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero, the system has no solution.

Given System of Equations: $x+2y+3z=3$ $4x+3y-4z=4$ $8x+4y-\lambda z=9+\mu$

From these equations, we identify the coefficients and constant terms: $a_1=1, b_1=2, c_1=3, d_1=3$ $a_2=4, b_2=3, c_2=-4, d_2=4$ $a_3=8, b_3=4, c_3=-\lambda, d_3=9+\mu$

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Step 1: Calculate $\Delta$ and set it to zero to find $\lambda$.

First, we write the coefficient matrix and calculate its determinant $\Delta$. This step is crucial because $\Delta=0$ is the primary condition for a system to have either no solution or infinitely many solutions.

1 & 2 & 3 \\ 4 & 3 & -4 \\ 8 & 4 & -\lambda \end{array} \right| $$ Expand the determinant along the first row: $$ \Delta = 1 \cdot (3(-\lambda) - (-4)(4)) - 2 \cdot (4(-\lambda) - (-4)(8)) + 3 \cdot (4(4) - 3(8)) $$ $$ \Delta = 1 \cdot (-3\lambda + 16) - 2 \cdot (-4\lambda + 32) + 3 \cdot (16 - 24) $$ $$ \Delta = -3\lambda + 16 + 8\lambda - 64 + 3 \cdot (-8) $$ $$ \Delta = -3\lambda + 16 + 8\lambda - 64 - 24 $$ $$ \Delta = 5\lambda - 72 $$ For infinitely many solutions, we must have $\Delta = 0$: $$ 5\lambda - 72 = 0 $$ $$ 5\lambda = 72 $$ $$ \lambda = \frac{72}{5} $$ --- **Step 2: Calculate $\Delta_z$ and set it to zero to find $\mu$.** Next, we need to ensure that the system is consistent, meaning there are actual solutions (not just no solutions). For infinitely many solutions, $\Delta_x$, $\Delta_y$, and $\Delta_z$ must all be zero. We can choose any one of these that involves $\mu$ to solve for it. Let's choose $\Delta_z$ as it often leads to simpler calculations. We substitute the constant terms ($d_1, d_2, d_3$) into the third column of the coefficient matrix. $$ \Delta_z = \left| \begin{array}{ccc} 1 & 2 & 3 \\ 4 & 3 & 4 \\ 8 & 4 & 9+\mu \end{array} \right| $$ Expand the determinant along the first row: $$ \Delta_z = 1 \cdot (3(9+\mu) - 4(4)) - 2 \cdot (4(9+\mu) - 4(8)) + 3 \cdot (4(4) - 3(8)) $$ $$ \Delta_z = 1 \cdot (27 + 3\mu - 16) - 2 \cdot (36 + 4\mu - 32) + 3 \cdot (16 - 24) $$ $$ \Delta_z = (11 + 3\mu) - 2 \cdot (4 + 4\mu) + 3 \cdot (-8) $$ $$ \Delta_z = 11 + 3\mu - 8 - 8\mu - 24 $$ Combine constant terms and $\mu$ terms: $$ \Delta_z = (11 - 8 - 24) + (3\mu - 8\mu) $$ $$ \Delta_z = (3 - 24) - 5\mu $$ $$ \Delta_z = -21 - 5\mu $$ For infinitely many solutions, we must have $\Delta_z = 0$: $$ -21 - 5\mu = 0 $$ $$ -5\mu = 21 $$ $$ \mu = -\frac{21}{5} $$ **Verification (Optional but Recommended): Using $\Delta_x$ or $\Delta_y$** To be absolutely sure, we can verify the value of $\mu$ using another determinant, say $\Delta_x$. This helps catch any calculation errors. Substitute the constant terms ($d_1, d_2, d_3$) into the first column of the coefficient matrix and the calculated $\lambda = \frac{72}{5}$ into the third column. $$ \Delta_x = \left| \begin{array}{ccc} 3 & 2 & 3 \\ 4 & 3 & -4 \\ 9+\mu & 4 & -\frac{72}{5} \end{array} \right| $$ Expand along the first row: $$ \Delta_x = 3 \left( 3(-\frac{72}{5}) - (-4)(4) \right) - 2 \left( 4(-\frac{72}{5}) - (-4)(9+\mu) \right) + 3 \left( 4(4) - 3(9+\mu) \right) $$ $$ \Delta_x = 3 \left( -\frac{216}{5} + 16 \right) - 2 \left( -\frac{288}{5} + 36 + 4\mu \right) + 3 \left( 16 - 27 - 3\mu \right) $$ $$ \Delta_x = 3 \left( \frac{-216 + 80}{5} \right) - 2 \left( \frac{-288 + 180}{5} + 4\mu \right) + 3 \left( -11 - 3\mu \right) $$ $$ \Delta_x = 3 \left( -\frac{136}{5} \right) - 2 \left( -\frac{108}{5} + 4\mu \right) - 33 - 9\mu $$ $$ \Delta_x = -\frac{408}{5} + \frac{216}{5} - 8\mu - 33 - 9\mu $$ $$ \Delta_x = -\frac{192}{5} - 17\mu - 33 $$ $$ \Delta_x = -\frac{192}{5} - \frac{165}{5} - 17\mu $$ $$ \Delta_x = -\frac{357}{5} - 17\mu $$ For infinitely many solutions, $\Delta_x = 0$: $$ -\frac{357}{5} - 17\mu = 0 $$ $$ -17\mu = \frac{357}{5} $$ $$ \mu = -\frac{357}{5 \times 17} = -\frac{21}{5} $$ Both $\Delta_z$ and $\Delta_x$ yield the same value for $\mu$, confirming our result. --- **Summary and Final Answer** For the given system of linear equations to have infinitely many solutions, we found: * $\lambda = \frac{72}{5}$ * $\mu = -\frac{21}{5}$ Therefore, the ordered pair $(\lambda, \mu)$ is $\left( \frac{72}{5}, -\frac{21}{5} \right)$. Comparing this with the given options: (A) $\left( {{{72} \over 5},{{21} \over 5}} \right)$ (B) $\left( { - {{72} \over 5}, - {{21} \over 5}} \right)$ (C) $\left( { - {{72} \over 5},{{21} \over 5}} \right)$ (D) $\left( {{{72} \over 5}, - {{21} \over 5}} \right)$ Our calculated pair matches option (D). **Tip for JEE Aspirants:** * Always be meticulous with determinant calculations, especially with signs. A single sign error can lead to an incorrect answer. * When finding conditions for infinitely many solutions, remember that *all* determinants ($\Delta$, $\Delta_x$, $\Delta_y$, $\Delta_z$) must be zero. This distinguishes it from "no solution" where $\Delta=0$ but at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero. * If possible, use row/column operations to simplify determinants before expanding, especially for larger matrices. For $3 \times 3$ matrices, direct expansion is often fastest if done carefully. * If you find yourself stuck or unsure about a calculation, try using a different determinant (e.g., $\Delta_y$ instead of $\Delta_x$) or a different method (like Gaussian elimination/row reduction) to cross-verify your results. As shown in the verification step, both $\Delta_x$ and $\Delta_z$ yield the same $\mu$, increasing confidence in the result.