જો $A = \begin{bmatrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$ હોય,તો $A(\operatorname{adj} A) = $

  • A
    $\begin{bmatrix} -1/3 & 0 & 0 \\ 0 & -1/3 & 0 \\ 0 & 0 & -1/3 \end{bmatrix}$
  • B
    $\begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}$
  • C
    $\begin{bmatrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$
  • D
    $\begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & 2 \\ 3 & 2 & 4 \end{bmatrix}$

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જો $A = \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}_{3 \times 3}$, અને $A^{-1} = \begin{bmatrix} \gamma & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2 \end{bmatrix}_{3 \times 3}$, તો $|\alpha \cdot \beta \cdot \gamma| = $ (જ્યાં $| \cdot |$ નિરપેક્ષ મૂલ્ય દર્શાવે છે)

જો $A = \begin{bmatrix} 3 & 4 \\ 5 & 7 \end{bmatrix}$ હોય,તો $A(adj A) = $

જો $A = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{bmatrix}$ હોય,તો $|\operatorname{adj} A|$ ની કિંમત શોધો.

જો $A=\left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right]$ હોય,તો ચકાસો કે $A^{3}-6 A^{2}+9 A-4 I=0$ અને તે પરથી $A^{-1}$ શોધો.

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જો $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ અને $X$ એ $2 \times 2$ શ્રેણિક છે કે જેથી $AX = I$ થાય,તો $X =$

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