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Step-by-Step Solution
Step 1: Write down the system of equations
The given system of linear equations is:
1) $x + y + z = 5$
2) $x + 2y + 2z = 6$
3) $x + 3y + \lambda z = \mu$
We need the system to have infinitely many solutions, which implies:
• The determinant of the coefficient matrix $= 0$.
• And the corresponding determinants for $x, y, z$ (if we replace the relevant columns) must also be $0$.
Step 2: Determine the condition for infinite solutions (determinant of coefficient matrix)
The coefficient matrix is:
$
\begin{vmatrix}
1 & 1 & 1 \\
1 & 2 & 2 \\
1 & 3 & \lambda
\end{vmatrix}
$
We set this determinant to zero:
$
\Delta
=
\begin{vmatrix}
1 & 1 & 1 \\
1 & 2 & 2 \\
1 & 3 & \lambda
\end{vmatrix}
= 0
$
Expanding along the first row:
$
\Delta = 1 \cdot (2 \cdot \lambda - 2 \cdot 3)
- 1 \cdot (1 \cdot \lambda - 2 \cdot 1)
+ 1 \cdot (1 \cdot 3 - 2 \cdot 1).
$
Simplify:
$
\Delta
= (2\lambda - 6) - (\lambda - 2) + (3 - 2)
= 2\lambda - 6 - \lambda + 2 + 1
= \lambda - 3
.
$
Since $\Delta = 0$, we get $\lambda - 3 = 0 \implies \lambda = 3.$
Step 3: Use conditions for infinite solutions (determinants for x, y, z) to find μ
For the system to have infinitely many solutions, when we replace the $x$ column or $y$ column or $z$ column in the coefficient matrix with the constants from the right-hand side, the determinant must be zero as well.
Determinant when replacing the x-column
$
\Delta_x
=
\begin{vmatrix}
5 & 1 & 1 \\
6 & 2 & 2 \\
\mu & 3 & 3
\end{vmatrix}
= 0.
$
We will not fully expand here if not needed, but this condition must be satisfied.
Determinant when replacing the y-column
$
\Delta_y
=
\begin{vmatrix}
1 & 5 & 1 \\
1 & 6 & 2 \\
1 & \mu & 3
\end{vmatrix}
= 0.
$
Perform operations for a simpler expansion (for instance, subtract the first row from the second and the third row):
Make the matrix:
$
\begin{vmatrix}
1 & 5 & 1 \\
0 & 1 & 1 \\
0 & \mu-5 & 2
\end{vmatrix}.
$
Expanding along the first column now is straightforward. The determinant becomes:
$
1 \times
\begin{vmatrix}
1 & 1 \\
\mu - 5 & 2
\end{vmatrix}
= 1 \times (1 \cdot 2 - (\mu - 5) \cdot 1).
$
$
= 2 - (\mu - 5)
= 2 - \mu + 5
= 7 - \mu.
$
For this to be $0$, we need $7 - \mu = 0 \implies \mu = 7.$
Determinant when replacing the z-column
$
\Delta_z
=
\begin{vmatrix}
1 & 1 & 5 \\
1 & 2 & 6 \\
1 & 3 & \mu
\end{vmatrix}.
$
Substituting the values of $\lambda$ and $\mu$ we found ($\lambda = 3, \mu = 7$) will also make $\Delta_z = 0$, satisfying the infinite solutions condition.
Step 4: Find the sum of λ and μ
With $\lambda = 3$ and $\mu = 7$, we get:
$
\lambda + \mu = 3 + 7 = 10.
$
Final Answer
The value of $\lambda + \mu$ is $10.$
