If $\begin{vmatrix} x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^2 \end{vmatrix} = \frac{9}{8}(103x+81)$,then $\lambda$ and $\frac{\lambda}{3}$ are the roots of the equation:

  • A
    $4x^2+24x-27=0$
  • B
    $4x^2-24x+27=0$
  • C
    $4x^2+24x+27=0$
  • D
    $4x^2-24x-27=0$

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For each real number $x$ such that $-1 < x < 1$,let $A(x)$ be the matrix $\frac{1}{1-x^2} \begin{bmatrix} 1 & -x \\ -x & 1 \end{bmatrix}$. If $z = \frac{x+y}{1+xy}$,then which of the following is true?

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$A = \frac{1}{\pi} \begin{bmatrix} \sin^{-1}(\pi x) & \tan^{-1}(\frac{x}{\pi}) \\ \sin^{-1}(\frac{x}{\pi}) & \cot^{-1}(\pi x) \end{bmatrix}$ and $B = \frac{1}{\pi} \begin{bmatrix} -\cos^{-1}(\pi x) & \tan^{-1}(\frac{x}{\pi}) \\ \sin^{-1}(\frac{x}{\pi}) & -\tan^{-1}(\pi x) \end{bmatrix}$. Then,$A - B = $ . . . . . . .

If $A = \begin{bmatrix} 1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1 \end{bmatrix}$,then show that $A^{3} - 23A - 40I = 0$.

Let $A$ be a $3 \times 3$ matrix such that $A \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 2 \\ 2 \end{bmatrix}$ and $A \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \\ 1 \end{bmatrix}$. If $\det(A) = 1$ and the matrix $A$ satisfies $A \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$,then $\det(\text{adj}(A^2 + A))$ is equal to:

Let $P$ be a square matrix such that $P^2 = I - P$. For $\alpha, \beta, \gamma, \delta \in N$,if $P^\alpha + P^\beta = \gamma I - 29 P$ and $P^\alpha - P^\beta = \delta I - 13 P$,then $\alpha + \beta + \gamma - \delta$ is equal to $........$.

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