यदि $A = \begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}$ है,तो $(A^2 - 5A)^{-1}$ क्या होगा?

  • A
    $-\frac{1}{4} \begin{bmatrix} -3 & 1 \\ 7 & -1 \end{bmatrix}$
  • B
    $\frac{1}{4} \begin{bmatrix} -3 & 1 \\ 7 & -1 \end{bmatrix}$
  • C
    $\frac{1}{4} \begin{bmatrix} 3 & 1 \\ 7 & 1 \end{bmatrix}$
  • D
    $-\frac{1}{4} \begin{bmatrix} 3 & -1 \\ 7 & -1 \end{bmatrix}$

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आव्यूह $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}$ के लिए,दर्शाइए कि $A^{3} - 6A^{2} + 5A + 11I = 0$ है। अतः,$A^{-1}$ ज्ञात कीजिए।

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यदि $A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$ इस प्रकार है कि $A^2 - 4A + 3I = 0$,जहाँ $I$ कोटि $2$ का एक इकाई आव्यूह है,तो $A^{-1}$ है

${\left[ {\begin{array}{*{20}{c}}{ - 6}&5\\{ - 7}&6\end{array}} \right]^{ - 1}}$ =

यदि $\operatorname{det}(AB)=(\operatorname{det} A)(\operatorname{det} B)$ और $A$,$3 \times 3$ कोटि का एक व्युत्क्रमणीय (non-singular) आव्यूह है,तो $\operatorname{det}(\operatorname{adj} A)=$

यदि $A = \begin{bmatrix} 1 & 2 & i \\ 1 & 1 & 1 \\ 1 & 1 & 0 \end{bmatrix}$ है,तो $[\operatorname{adj}(\operatorname{adj} A)]^{-1} = $

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