Question

If the system of equations \[2x + 7y + \lambda z = 3,\]\[3x + 2y + 5z = 4,\]\[x + \mu y + 32z = -1\]has infinitely many solutions, then $(\lambda - \mu)$ is equal to ________.

Answer 1

Correct Answer: 38

The problem provides a system of three linear equations in variables \(x, y, z\) with two parameters, \(\lambda\) and \(\mu\). We are told that the system has infinitely many solutions, and we need to find the value of \((\lambda - \mu)\).

Concept Used:

For a system of linear equations of the form \(AX = B\), where \(A\) is the coefficient matrix and \(B\) is the constant vector, the condition for having infinitely many solutions is that the determinant of the coefficient matrix is zero, and the determinants of the matrices formed by replacing a column of \(A\) with \(B\) are also zero. In terms of Cramer's rule notation, this means:

\[ \Delta = \det(A) = 0 \]

and

\[ \Delta_x = \Delta_y = \Delta_z = 0 \]

where \(\Delta_x, \Delta_y, \Delta_z\) are the determinants of the matrices obtained by replacing the x, y, and z columns of \(A\) with the constant vector, respectively.

Step-by-Step Solution:

Step 1: Write down the determinants \(\Delta\) and \(\Delta_y\).

The given system of equations is:

\[ 2x + 7y + \lambda z = 3 \] \[ 3x + 2y + 5z = 4 \] \[ x + \mu y + 32z = -1 \]

The determinant of the coefficient matrix is:

\[ \Delta = \begin{vmatrix} 2 & 7 & \lambda \\ 3 & 2 & 5 \\ 1 & \mu & 32 \end{vmatrix} \]

The determinant \(\Delta_y\) is obtained by replacing the second column (coefficients of y) with the constant terms:

\[ \Delta_y = \begin{vmatrix} 2 & 3 & \lambda \\ 3 & 4 & 5 \\ 1 & -1 & 32 \end{vmatrix} \]

Since \(\Delta_y\) does not contain the parameter \(\mu\), it is advantageous to evaluate it first to find \(\lambda\).

Step 2: Set \(\Delta_y = 0\) and solve for \(\lambda\).

For the system to have infinitely many solutions, we must have \(\Delta_y = 0\).

\[ \begin{vmatrix} 2 & 3 & \lambda \\ 3 & 4 & 5 \\ 1 & -1 & 32 \end{vmatrix} = 0 \]

Expanding along the first row:

\[ 2(4 \cdot 32 - 5 \cdot (-1)) - 3(3 \cdot 32 - 5 \cdot 1) + \lambda(3 \cdot (-1) - 4 \cdot 1) = 0 \] \[ 2(128 + 5) - 3(96 - 5) + \lambda(-3 - 4) = 0 \] \[ 2(133) - 3(91) - 7\lambda = 0 \] \[ 266 - 273 - 7\lambda = 0 \] \[ -7 - 7\lambda = 0 \] \[ 7\lambda = -7 \implies \lambda = -1 \]

Step 3: Set \(\Delta = 0\) and substitute the value of \(\lambda\) to solve for \(\mu\).

For infinitely many solutions, we must also have \(\Delta = 0\).

\[ \Delta = \begin{vmatrix} 2 & 7 & \lambda \\ 3 & 2 & 5 \\ 1 & \mu & 32 \end{vmatrix} = 0 \]

Expanding along the first row:

\[ 2(2 \cdot 32 - 5 \cdot \mu) - 7(3 \cdot 32 - 5 \cdot 1) + \lambda(3 \cdot \mu - 2 \cdot 1) = 0 \] \[ 2(64 - 5\mu) - 7(96 - 5) + \lambda(3\mu - 2) = 0 \] \[ 128 - 10\mu - 7(91) + \lambda(3\mu - 2) = 0 \] \[ 128 - 10\mu - 637 + \lambda(3\mu - 2) = 0 \] \[ -509 - 10\mu + \lambda(3\mu - 2) = 0 \]

Now, substitute the value \(\lambda = -1\) into this equation:

\[ -509 - 10\mu + (-1)(3\mu - 2) = 0 \] \[ -509 - 10\mu - 3\mu + 2 = 0 \] \[ -507 - 13\mu = 0 \] \[ 13\mu = -507 \] \[ \mu = -\frac{507}{13} = -39 \]

Step 4: Calculate the value of \((\lambda - \mu)\).

We have found \(\lambda = -1\) and \(\mu = -39\).

\[ \lambda - \mu = (-1) - (-39) = -1 + 39 = 38 \]

The value of \((\lambda - \mu)\) is 38.

Answer 2

For the system to have infinitely many solutions, the determinant of the coefficient matrix must be zero. We set:
\[D = D_1 = D_2 = D_3 = 0\]
Calculating \( D_3 \):
\[D_3 = \begin{vmatrix}2 & 7 & 3 \\3 & 2 & 4 \\1 & \mu & -1\end{vmatrix}= 0\]

Expanding, we get:
\[2 \begin{vmatrix}2 & 4 \\\mu & -1\end{vmatrix}- 7 \begin{vmatrix}3 & 4 \\1 & -1\end{vmatrix}+ 3 \begin{vmatrix}3 & 2 \\1 & \mu\end{vmatrix}= 0\]
Solving for \( \mu \), we find:
\[\mu = -39\]
Now, calculating \( D \) with \( \lambda \) in place:
\[D = \begin{vmatrix}2 & 7 & \lambda \\3 & 2 & 5 \\1 & -39 & 32\end{vmatrix}= 0\]
Solving this determinant, we get:
\[\lambda = -1\]

Thus, \( \lambda - \mu = -1 - (-39) = 38 \).