Consider the matrix $f(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$. Given below are two statements:
Statement $I$: $f(-x)$ is the inverse of the matrix $f(x)$.
Statement $II$: $f(x) f(y) = f(x+y)$.
In the light of the above statements,choose the correct answer from the options given below:

  • A
    Statement $I$ is false but Statement $II$ is true.
  • B
    Both Statement $I$ and Statement $II$ are false.
  • C
    Statement $I$ is true but Statement $II$ is false.
  • D
    Both Statement $I$ and Statement $II$ are true.

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Let $A$ and $B$ be two $3 \times 3$ real matrices such that $(A^{2}-B^{2})$ is an invertible matrix. If $A^{5}=B^{5}$ and $A^{3} B^{2}=A^{2} B^{3}$,then the value of the determinant of the matrix $A^{3}+B^{3}$ is equal to:

Let $A = \begin{bmatrix} 2 & b & 1 \\ b & b^2+1 & b \\ 1 & b & 2 \end{bmatrix}$ where $b > 0$. Then the minimum value of $\frac{\det(A)}{b}$ is

If $\Delta _1 = \left| \begin{array}{ccc} b^5c^6(c^3 - b^3) & a^4c^6(a^3 - c^3) & a^4b^5(b^3 - a^3) \\ b^2c^3(b^6 - c^6) & ac^3(c^6 - a^6) & ab^2(a^6 - b^6) \\ b^2c^3(c^3 - b^3) & ac^3(a^3 - c^3) & ab^2(b^3 - a^3) \end{array} \right|$ and $\Delta _2 = \left| \begin{array}{ccc} a & b^2 & c^3 \\ a^4 & b^5 & c^6 \\ a^7 & b^8 & c^9 \end{array} \right|$,then $\Delta _1 \Delta _2$ is equal to

$\left|\begin{array}{ll}2 & 1 \\ 3 & 1\end{array}\right|+\left|\begin{array}{cc}1 & 1/3 \\ 3 & 1\end{array}\right|+\left|\begin{array}{cc}1/2 & 1/9 \\ 3 & 1\end{array}\right|+\left|\begin{array}{cc}1/4 & 1/27 \\ 3 & 1\end{array}\right|+\ldots \infty=$

For a square matrix $B$ of order $3$, if $B^T=B^{-1}$ and $|B|=1$, then $|B-I|=$

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