Adding a Multiple of One Row to Another

We are given the system of linear equations:

$$x + y + z = 6 \quad \cdots (1)$$

$$2x + 5y + az = 36 \quad \cdots (2)$$

$$x + 2y + 3z = b \quad \cdots (3)$$

The coefficient matrix is:

$$A = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & a \\ 1 & 2 & 3 \end{vmatrix}$$

Its determinant is

$$\det(A) = 1(15 - 2a) - 1(6 - a) + 1(4 - 5) = 15 - 2a - 6 + a - 1 = 8 - a.$$

Since $$\det(A) = 8 - a\,, $$ for $$a = 8$$ we have $$\det(A) = 0$$, so the system does not have a unique solution and Options 1 and 3 are eliminated.

Substituting $$a = 8$$ into the augmented matrix gives

$$\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 2 & 5 & 8 & | & 36 \\ 1 & 2 & 3 & | & b \end{pmatrix}.$$

Applying $$R_2 \to R_2 - 2R_1$$ yields $$(0, 3, 6 \mid 24)$$ and then $$R_3 \to R_3 - R_1$$ yields $$(0, 1, 2 \mid b - 6).$$ Finally, $$R_3 \to R_3 - \tfrac{1}{3}R_2$$ gives $$(0, 0, 0 \mid b - 14).$$

For consistency (infinitely many solutions) we require $$b - 14 = 0\,, $$ so $$b = 14.$$ If instead $$b = 16\,, $$ then $$b - 14 = 2 \neq 0$$ and the system is inconsistent (no solution). Therefore the correct answer is Option 4: infinitely many solutions for $$a = 8$$ and $$b = 14$$.

Let $$t=\frac{\sin x}{x}\,,$$ so that $$\displaystyle\lim_{x\to 0}t=1$$ because $$\frac{\sin x}{x}\to 1$$ as $$x\to 0$$.

With this notation the determinant becomes
$$f(x)=\left|\begin{matrix}a+t & 1 & b\\ a & 1+t & b\\ a & 1 & b+t\end{matrix}\right|.$$

Perform the row operation $$R_1\rightarrow R_1-R_2$$ (subtract the second row from the first):
$$f(x)=\left|\begin{matrix}t & -t & 0\\ a & 1+t & b\\ a & 1 & b+t\end{matrix}\right|.$$

Factor out $$t$$ from the first row:
$$f(x)=t\;\left|\begin{matrix}1 & -1 & 0\\ a & 1+t & b\\ a & 1 & b+t\end{matrix}\right|.$$

Call the remaining determinant $$D(t)$$, i.e.
$$D(t)=\left|\begin{matrix}1 & -1 & 0\\ a & 1+t & b\\ a & 1 & b+t\end{matrix}\right|.$$

Expand $$D(t)$$ along the first row:
$$\begin{aligned} D(t)=&\;1\;\left|\begin{matrix}1+t & b\\ 1 & b+t\end{matrix}\right| \;-\;(-1)\;\left|\begin{matrix}a & b\\ a & b+t\end{matrix}\right| \;+\;0\\[4pt] =&\;(1+t)(b+t)-b + a\,t. \end{aligned}$$

Simplify the expression:
$$(1+t)(b+t)-b = (1+t)(b+t)-b = (1+t)(b+t)-b.$$
Therefore
$$D(t)=(1+t)(b+t)-b+a\,t.$$

Now take the limit as $$x\to 0$$, i.e. $$t\to 1$$:
$$\begin{aligned} \lim_{x\to 0}f(x) &=\lim_{t\to 1}\;t\,D(t)\\ &=1\;\bigl[(1+1)(b+1)-b+a\cdot 1\bigr]\\ &=(2)(b+1)-b+a\\ &=2b+2-b+a\\ &=a+b+2. \end{aligned}$$

Hence
$$\lambda+\mu a+\nu b=a+b+2,$$
so $$\lambda=2,\;\mu=1,\;\nu=1.$$

The required value is
$$(\lambda+\mu+\nu)^2=(2+1+1)^2=4^2=16.$$

Therefore, the correct option is Option A (16).

A homogeneous system $$AX = 0$$ has a non-trivial solution if and only if the determinant of the coefficient matrix is zero ($$|A| = 0$$).

Set up the determinant:

$$\begin{vmatrix} 1 & \sqrt{2}\sin\alpha & \sqrt{2}\cos\alpha \\ 1 & \cos\alpha & \sin\alpha \\ 1 & \sin\alpha & -\cos\alpha \end{vmatrix} = 0$$

Perform row operations ($$R_2 \to R_2 - R_1$$ and $$R_3 \to R_3 - R_1$$) or expand directly. Expanding along the first column:

$$1(-\cos^2\alpha - \sin^2\alpha) - 1(-\sqrt{2}\sin\alpha\cos\alpha - \sqrt{2}\sin\alpha\cos\alpha) + 1(\sqrt{2}\sin^2\alpha - \sqrt{2}\cos^2\alpha) = 0$$

Simplify:

$$-1 + 2\sqrt{2}\sin\alpha\cos\alpha - \sqrt{2}(\cos^2\alpha - \sin^2\alpha) = 0$$

$$-1 + \sqrt{2}\sin 2\alpha - \sqrt{2}\cos 2\alpha = 0 \implies \sin 2\alpha - \cos 2\alpha = \frac{1}{\sqrt{2}}$$

Divide by $$\sqrt{2}$$: $$\frac{1}{\sqrt{2}}\sin 2\alpha - \frac{1}{\sqrt{2}}\cos 2\alpha = \frac{1}{2} \implies \sin(2\alpha - \frac{\pi}{4}) = \frac{1}{2}$$

$$2\alpha - \frac{\pi}{4} = \frac{\pi}{6} \implies 2\alpha = \frac{\pi}{6} + \frac{\pi}{4} = \frac{5\pi}{12} \implies \alpha = \frac{5\pi}{24}$$