$A^2=\left[\begin{matrix}1 & 0 \\1 & 1 \\\end{matrix}\right]\left[\begin{matrix}1 & 0 \\1 & 1 \\\end{matrix} \right] =\left[\begin{matrix}1 & 0 \\2 & 1 \\\end{matrix}\right]$
$A^3=\left[\begin{matrix}1 & 0 \\2 & 1 \\\end{matrix}\right]\left[\begin{matrix}1 & 0 \\1 & 1 \\\end{matrix}\right] =\left[\begin{matrix}1 & 0 \\3 & 1 \\\end{matrix}\right]$
$A^n=\left[\begin{matrix}1 & 0 \\ n & 1 \\\end{matrix}\right] -------(1)$
$nA-(n-1)I$
$=n\left[\begin{matrix}1 & 0 \\1 & 1 \\\end{matrix}\right]-(n-1)\left[\begin{matrix}1 & 0 \\0 & 1 \\\end{matrix}\right]$
$=\left[\begin{matrix}n & 0 \\ n & n \\\end{matrix}\right]-\left[\begin{matrix}(n-1) & 0 \\0 & (n-1) \\\end{matrix}\right]$
$=\left[\begin{matrix}1 & 0 \\n & 1 \\\end{matrix}\right]=A^n$
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અહી શ્રેણિક $A=\left[a_{i j}\right]_{3 \times 3}$ કે જ્યાં
$a_{i j}=J_{6+i, 3}-J_{i+3,3}, \quad i \leq j$
$\quad\quad\quad\quad\quad\quad0 , \quad\quad\quad i>j$.
તો $\left|\operatorname{adj} A^{-1}\right|$ મેળવો.