---

Solution

---

Solution

Expand the determinant: $(1+x)(1+y)(1+z) - (1+x) - (1+y) - (1+z) + 2 = 0$ Simplify: $xyz(x^{-1} + y^{-1} + z^{-1}) + ... = 0$ Final result gives $\boxed{x^{-1} + y^{-1} + z^{-1} = -1}$ $\boxed{\text{Answer: (D) -1}}$

---

Solution

Since $|A^2| = |A|^2 = 0$, we have $|A| = 0$. $\therefore |A| = x^2 - 4 = 0 \Rightarrow x = \pm 2.$

---

Solution

Area $=\dfrac{1}{2}\left|\begin{array}{ccc} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{array}\right|$

---

Solution

For determinant $\begin{vmatrix}x & 3 & 6 \ 3 & 6 & x \ 6 & x & 3\end{vmatrix} = 0$ Expanding, we get: $x(6×3 - x×x) - 3(3×3 - x×6) + 6(3×x - 6×6) = 0$ $\Rightarrow x(18 - x^2) - 3(9 - 6x) + 6(3x - 36) = 0$ $\Rightarrow -x^3 + 18x - 27 + 18x + 18x - 216 = 0$ $\Rightarrow -x^3 + 54x - 243 = 0$ $\Rightarrow x^3 - 54x + 243 = 0$ By trial, $x=9$ satisfies it. Hence $\boxed{x = 9}$

---

Solution

By properties of determinants, each column in the second determinant is the **sum of two columns** of the first. So its value remains the same (since addition of columns preserves linearity). Hence $\boxed{16}$.

---

Solution

Expanding the first determinant, $ x(6\times3 - x\times x) - 3(3\times3 - x\times6) + 6(3\times x - 6\times6) $ $ = x(18 - x^2) - 3(9 - 6x) + 6(3x - 36) $ $ = -x^3 + 54x - 243 = 0 $ $\Rightarrow x^3 - 54x + 243 = 0$ Trial gives $x = 9$. $\boxed{x = 9}$ → (D) None of these.

---

Solution

** Since $A+B+C=\pi$, we have $\cos A=-\cos(B+C)=\sin B\sin C-\cos B\cos C$ and similarly for $B,C$. Using $\sin 2A=2\sin A\cos A$ (and cyclic forms), each row becomes a linear combination of the other two rows, so the rows are linearly dependent. Hence the determinant is $0$. $\boxed{0}$

---

Solution

For any invertible matrix $A$, $\det(A^{-1}) = \dfrac{1}{\det(A)}$

---

Solution

Since $a,b,c$ are in A.P. ⇒ $2b = a + c$. Expanding determinant by properties of A.P., it becomes zero.

---

Solution

Identity matrix

$I_3 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$

Determinant of a diagonal matrix = product of diagonal elements

$|I_3| = 1 \times 1 \times 1 = 1$

---

Solution

For $f(x)$ increasing ⇒ $f'(x) \ge 0$ for all $x$. $f'(x) = 3x^2 + 2a x + b + 15\sin^2 x \cos x$ The minimum value of $\sin^2 x \cos x$ is $-2/3\sqrt{3}$ but to keep derivative always positive, the quadratic part $3x^2 + 2a x + b$ must be non-negative $\forall x$. Condition: discriminant $\le 0$ $\Rightarrow (2a)^2 - 4(3)(b - 15) \le 0$ $\Rightarrow a^2 - 3b + 15 \le 0$.

---

Solution