ધારો કે A = begin{bmatrix} 0 & 1 & 2 a & 0 & | vedclass

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(B) આપેલ છે $A = \begin{bmatrix} 0 & 1 & 2 \ a & 0 & 3 \ 1 & c & 0 \end{bmatrix}$.$A^2 = \begin{bmatrix} 0 & 1 & 2 \ a & 0 & 3 \ 1 & c & 0 \end{bm

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$2$

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(B) આપેલ છે $A = \begin{bmatrix} 0 & 1 & 2 \ a & 0 & 3 \ 1 & c & 0 \end{bmatrix}$.$A^2 = \begin{bmatrix} 0 & 1 & 2 \ a & 0 & 3 \ 1 & c & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \ a & 0 & 3 \ 1 & c & 0 \end{bmatrix} = \begin{bmatrix} a+2 & 2c & 3 \ 3 & a+3c & 2a \ ac & 1 & 2+3c \end{bmatrix}$.$A^3 = A^2 \cdot A = \begin{bmatrix} a+2 & 2c & 3 \ 3 & a+3c & 2a \ ac & 1 & 2+3c \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \ a & 0 & 3 \ 1 & c & 0 \end{bmatrix}$.$A^3 = A$ સરખાવતા,$(1,1)$ ઘટક માટે: $2ac + 3 = 0 \implies ac = -\frac{3}{2}$.$(1,2)$ ઘટક માટે: $a + 2 + 3c = 1 \implies a + 3c = -1$.$c = -\frac{3}{2a}$ ને $a + 3c = -1$ માં મુકતા:$a + 3(-\frac{3}{2a}) = -1 \implies a - \frac{9}{2a} = -1 \implies 2a^2 + 2a - 9 = 0$.દ્વિઘાત સૂત્રનો ઉપયોગ કરતા: $a = \frac{-2 \pm \sqrt{4 - 4(2)(-9)}}{2(2)} = \frac{-2 \pm \sqrt{76}}{4} = \frac{-1 \pm \sqrt{19}}{2}$.$a > 0$ હોવાથી,$a = \frac{\sqrt{19} - 1}{2}$.$4 આમ,$a \in (1, 2]$,જેનો અર્થ છે કે $n = 2$.

યાદી-$I$	યાદી-$II$
$A$. $\|k^{-1} A^{-1}\|$	$I$. $BA^k + A^kB$
$B$. $\|\text{Adj}(A^{-1})\|$	$II$. $\frac{B\text{Adj}(B)}{\|B\|}$
$C$. $BAB^{-1} = I \Rightarrow BA^kB^{-1} =$	$III$. $\frac{1}{\|B\|^3\|A\|}$
$D$. $\text{Adj}(\text{Adj}(A^{-1})) =$	$IV$. $\frac{1}{\|A\|}(A^{-1})$
	$V$. $\frac{1}{\|A\|^2}$

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