© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Write down the system of equations
We have three linear equations:
1) $x + 2y + 3z = 1$
2) $3x + 4y + 5z = \mu$
3) $4x + 4y + 4z = \delta$
We want to find the values of $\mu$ and $\delta$ for which this system is inconsistent.
Step 2: Recall condition for inconsistency
A system of linear equations is inconsistent if the determinant of the coefficient matrix is zero, and at least one of the determinants of the modified coefficient matrices (for $x, y,$ or $z$) is non-zero. Symbolically:
$\Delta = 0$
At least one of $\Delta_x, \Delta_y, \Delta_z \neq 0$
Step 3: Compute the determinant of the coefficient matrix $\Delta$
The coefficient matrix is:
$
\begin{pmatrix}
1 & 2 & 3\\
3 & 4 & 5\\
4 & 4 & 4
\end{pmatrix}
$
Thus,
$
\Delta =
\begin{vmatrix}
1 & 2 & 3\\
3 & 4 & 5\\
4 & 4 & 4
\end{vmatrix}.
$
We must have $\Delta = 0$ for potential inconsistency.
Let us quickly verify $\Delta$ is indeed $0$ (though it is given in the reference solution). Expanding along the first row, for instance:
$
\Delta
= 1 \times
\begin{vmatrix}
4 & 5\\
4 & 4
\end{vmatrix}
- 2 \times
\begin{vmatrix}
3 & 5\\
4 & 4
\end{vmatrix}
+ 3 \times
\begin{vmatrix}
3 & 4\\
4 & 4
\end{vmatrix}.
$
$
= 1 \times (4 \times 4 - 5 \times 4)
- 2 \times (3 \times 4 - 5 \times 4)
+ 3 \times (3 \times 4 - 4 \times 4).
$
$
= 1 \times (16 - 20)
- 2 \times (12 - 20)
+ 3 \times (12 - 16)
$
$
= 1 \times (-4)
- 2 \times (-8)
+ 3 \times (-4)
$
$
= -4 + 16 - 12 = 0.
$
Hence, $\Delta = 0$ is confirmed.
Step 4: Compute the determinant $\Delta_x$ and deduce condition
Let $\Delta_x$ be the determinant obtained by replacing the first column of the coefficient matrix with the constants on the right-hand side of the equations, that is:
$
\Delta_x
=
\begin{vmatrix}
1 & 2 & 3\\
\mu & 4 & 5\\
\delta & 4 & 4
\end{vmatrix}.
$
Expanding $\Delta_x$ (for instance along the first row again) yields:
$
\Delta_x
= 1 \times
\begin{vmatrix}
4 & 5\\
4 & 4
\end{vmatrix}
- 2 \times
\begin{vmatrix}
\mu & 5\\
\delta & 4
\end{vmatrix}
+ 3 \times
\begin{vmatrix}
\mu & 4\\
\delta & 4
\end{vmatrix}.
$
Let us simplify this step by step:
$
\begin{vmatrix}
4 & 5\\
4 & 4
\end{vmatrix}
= 4 \times 4 - 5 \times 4 = 16 - 20 = -4.
$
$
\begin{vmatrix}
\mu & 5\\
\delta & 4
\end{vmatrix}
= \mu \times 4 - 5 \times \delta = 4\mu - 5\delta.
$
$
\begin{vmatrix}
\mu & 4\\
\delta & 4
\end{vmatrix}
= \mu \times 4 - 4 \times \delta = 4\mu - 4\delta.
$
Hence,
$
\Delta_x
= 1 \times (-4)
- 2 \times (4\mu - 5\delta)
+ 3 \times (4\mu - 4\delta).
$
$
\Delta_x
= -4
- 2(4\mu - 5\delta)
+ 3(4\mu - 4\delta).
$
We seek a condition that makes $\Delta_x \neq 0$, because for inconsistency, at least one of $\Delta_x, \Delta_y, \Delta_z$ must be non-zero, given $\Delta = 0.$ A typical derived condition (from the reference above) is:
$2\mu \neq \delta + 2.$
Plugging in the options, only $(\mu,\delta) = (4,3)$ satisfies $2\mu \neq \delta + 2$. Indeed, for $\mu = 4$ and $\delta = 3$:
$2 \times 4 = 8 \quad \text{and} \quad 3 + 2 = 5 \quad \text{so} \quad 8 \neq 5.
$
Hence, $(4,3)$ is where $\Delta = 0$ but $\Delta_x \neq 0$, making the system inconsistent.
Final Answer
The ordered pair $(\mu, \delta) = (4, 3)$ makes the system of equations inconsistent.
