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Step-by-step Solution
Step 1: Write down the system of equations
We have the system of three linear equations:
1) $x + y + z = 2$
2) $x + 2y + 3z = 5$
3) $x + 3y + \lambda z = \mu$
For this system to have infinitely many solutions, the determinant of the coefficient matrix must be zero, and the system must satisfy certain additional conditions with the augmented matrices.
Step 2: Form the coefficient matrix and set its determinant to zero
The coefficient matrix is
$
\begin{pmatrix}
1 & 1 & 1 \\
1 & 2 & 3 \\
1 & 3 & \lambda
\end{pmatrix}.
$
Let $D$ be the determinant of this matrix. Then,
$
D
= \begin{vmatrix}
1 & 1 & 1 \\
1 & 2 & 3 \\
1 & 3 & \lambda
\end{vmatrix}.
$
Expanding along the first row (using cofactor expansion):
$
D
= 1 \bigl(2 \cdot \lambda - 3 \cdot 3\bigr)
- 1 \bigl(1 \cdot \lambda - 1 \cdot 3\bigr)
+ 1 \bigl(1 \cdot 3 - 2 \cdot 1\bigr).
$
Simplify each term:
First term: $1 \times (2\lambda - 9) = 2\lambda - 9$.
Second term: $-1 \times (\lambda - 3) = -\lambda + 3$.
Third term: $1 \times (3 - 2) = 1$.
Hence,
$
D = (2\lambda - 9) - (\lambda - 3) + (3 - 2)
= (2\lambda - 9) - \lambda + 3 + 1.
$
This simplifies to:
$
D = 2\lambda - \lambda - 9 + 3 + 1 = \lambda - 5.
$
For $D = 0$, we must have:
$
\lambda - 5 = 0 \quad\Longrightarrow\quad \lambda = 5.
$
Step 3: Use the augmented matrix condition for infinite solutions
We substitute $\lambda = 5$ back into the system and require the determinant of the appropriately augmented matrix also be zero. One standard approach is to check the determinant $D_1$, $D_2$, or $D_3$ (depending on which columns we replace by the constants) to ensure consistency for infinite solutions. Let us replace the first column (the $x$-column) by the constants on the right-hand side and call that determinant $D_1$.
The augmented matrix for $D_1$ becomes:
$
\begin{pmatrix}
2 & 1 & 1 \\
5 & 2 & 3 \\
\mu & 3 & 5
\end{pmatrix}.
$
Step 4: Set $D_1 = 0$ and solve for $\mu$
Let
$
D_1
= \begin{vmatrix}
2 & 1 & 1 \\
5 & 2 & 3 \\
\mu & 3 & 5
\end{vmatrix}.
$
Expand along the first row again:
$
D_1
= 2 \begin{vmatrix} 2 & 3 \\ 3 & 5 \end{vmatrix}
- 1 \begin{vmatrix} 5 & 3 \\ \mu & 5 \end{vmatrix}
+ 1 \begin{vmatrix} 5 & 2 \\ \mu & 3 \end{vmatrix}.
$
Compute each minor:
$
\begin{vmatrix} 2 & 3 \\ 3 & 5 \end{vmatrix}
= 2 \times 5 - 3 \times 3 = 10 - 9 = 1.
$
$
\begin{vmatrix} 5 & 3 \\ \mu & 5 \end{vmatrix}
= 5 \times 5 - 3 \times \mu = 25 - 3\mu.
$
$
\begin{vmatrix} 5 & 2 \\ \mu & 3 \end{vmatrix}
= 5 \times 3 - 2 \times \mu = 15 - 2\mu.
$
So,
$
D_1
= 2(1) - 1(25 - 3\mu) + 1(15 - 2\mu).
$
Simplifying:
$
D_1
= 2 - (25 - 3\mu) + (15 - 2\mu)
= 2 - 25 + 3\mu + 15 - 2\mu
= (3\mu - 2\mu) + (2 - 25 + 15)
= \mu - 8.
$
For $D_1 = 0$, we get:
$
\mu - 8 = 0 \quad\Longrightarrow\quad \mu = 8.
$
Step 5: State the final values of $\lambda$ and $\mu$
From our calculations,
$
\lambda = 5
$
and
$
\mu = 8.
$
These values ensure the system has infinitely many solutions.
Answer
The values of $\lambda$ and $\mu$ are $5$ and $8$, respectively.
