Find $A^{2}-5 A+6 I$ if $A=\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]$

Let $A=\left[a_{i j}\right]$ be a $3 \times 3$ matrix, where

$a_{i j}= 1 , \quad\quad\text { if } i=j$

$\quad\quad-x ,\quad \text { if }|i-j|=1$

$\quad\quad2 x+1, \text { otherwise }$

Let a function f: $\mathrm{R} \rightarrow \mathrm{R}$ be defined as $\mathrm{f}(\mathrm{x})=\operatorname{det}(\mathrm{A})$. Then the sum of maximum and minimum values of $f$ on $R$ is equal to:

The matrix $A^2 + 4A – 5I$, where $I$ is identity matrix and $A = \left[ {\begin{array}{*{20}{c}}
1&2\\
4&{ – 3}
\end{array}} \right]$ , equals

Two farmers Ramkishan and Gurcharan Singh cultivates only three varieties of rice namely Basmati, Permal and Naura. The sale (in Rupees) of these varieties of rice by both the farmers in the month of September and October are given by the following matrices $A$ and $B$.

September Sales (in Rupees)

$A=$  $\left[ {\begin{array}{*{20}{c}}
  {{\text{ Basmati }}}&{{\text{ Permal }}}&{{\text{ Naura }}} \\ 
  {10,000}&{20,000}&{30,000} \\ 
  {50,000}&{30,000}&{10,000} 
\end{array}} \right]\,$ $\begin{matrix}
   {}  \\
    \mathrm {Ramkrishan} \,\,\,\,\,\,\,\,\,\,\,\,\,  \\
    \mathrm {Gurcharan}\,\, \mathrm {Singh}  \\
\end{matrix}$

October Sales (in Rupees)

$B=\left[\begin{array}{ccc}\text { Basmati } & \text { Permal } & \text { Naura } \\ 5000 & 10,000 & 6000 \\ 20,000 & 10,000 & 10,000\end{array}\right]$ $\begin{matrix}
   {}  \\
    \mathrm {Ramkrishan} \,\,\,\,\,\,\,\,\,\,\,\,\,  \\
   \mathrm {Gurcharan}\,\,\mathrm {Singh}  \\
\end{matrix}$

$(i)$ Find the combined sales in September and October for each farmer in each variety.

$(ii)$ Find the decrease in sales from September to October.

$(iii)$ If both farmers receive $2\%$ profit on gross sales, compute the profit for each farmer and for each variety sold in October.

If $A = \left[ {\begin{array}{*{20}{c}}1&2&{ – 1}\\3&0&{\,\,2}\\4&5&{\,\,0}\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{c}}1&0&0\\2&1&0\\0&1&3\end{array}} \right],$then $AB$ is