Question

$
{\text{Find }}{A^2}{\text{ if }}A = \left[ {\begin{array}{*{20}{c}}
  1&2&5 \\
  3&4&1 \\
  1&{ - 1}&2
\end{array}} \right]
$

Answer 1

\[
   \Rightarrow {\text{Let }}A = \left[ {\begin{array}{*{20}{c}}
  1&2&5 \\
  3&4&1 \\
  1&{ - 1}&2
\end{array}} \right]{\text{ (1)}} \\
  {\text{So, now we have to find the value of }}{A^2}.{\text{ So,}} \\
   \Rightarrow {A^2} = A*A{\text{ (2)}} \\
  {\text{Putting value of }}A{\text{ in the RHS}}{\text{ of equation 2 we get,}} \\
   \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
  1&2&5 \\
  3&4&1 \\
  1&{ - 1}&2
\end{array}} \right]*\left[ {\begin{array}{*{20}{c}}
  1&2&5 \\
  3&4&1 \\
  1&{ - 1}&2
\end{array}} \right] \\
  {\text{As, we know that for a matrix }}X{\text{ if we have to find }}{X^2}{\text{ then we find it as given below}} \\
   \Rightarrow {\text{Let }}X = \left[ {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right]{\text{ }} \\
  {\text{So, }}{X^2} = X*X,{\text{ is calculated as,}} \\
   \Rightarrow {X^2} = \left[ {\begin{array}{*{20}{c}}
  {(a*a) + (b*d) + (c*g)}&{(a*b) + (b*e) + (c*h)}&{(a*c) + (b*f) + (c*i)} \\
  {(d*a) + (e*d) + (f*g)}&{(d*b) + (e*e) + (f*h)}&{(d*c) + (e*f) + (f*i)} \\
  {(g*a) + (h*d) + (i*g)}&{(g*b) + (h*e) + (i*h)}&{(g*c) + (h*f) + (i*i)}
\end{array}} \right] \\
  {\text{So, like }}{X^2}{\text{ we can also find the value of }}{A^2}. \\
  {\text{So, for calculating the value of }}{A^2}. \\
   \Rightarrow {{\text{A}}^2} = \left[ {\begin{array}{*{20}{c}}
  {(1*1) + (2*3) + (5*1)}&{(1*2) + (2*4) + (5*( - 1))}&{(1*5) + (2*1) + (5*2)} \\
  {(3*1) + (4*3) + (1*1)}&{(3*2) + (4*4) + (1*( - 1))}&{(3*5) + (4*1) + (1*2)} \\
  {(1*1) + ( - 1*3) + (2*1)}&{(1*2) + ( - 1*4) + (2*( - 1))}&{(1*5) + (( - 1)*1) + (2*2)}
\end{array}} \right]{\text{ So,}} \\
   \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
  {1 + 6 + 5}&{2 + 8 - 5}&{5 + 2 + 10} \\
  {3 + 12 + 1}&{6 + 16 - 1}&{15 + 4 + 2} \\
  {1 - 3 + 2}&{2 - 4 - 2}&{5 - 1 + 4}
\end{array}} \right]{\text{ So on solving this it becomes}} \\
  {\text{Hence , }}{A^2} = \left[ {\begin{array}{*{20}{c}}
  {12}&5&{17} \\
  {16}&{21}&{21} \\
  0&{ - 4}&8
\end{array}} \right] \\
  {\text{NOTE: - Whenever you came up with this type of problem then make calculations proper}} \\
  {\text{As, in this type of problem there are many chances of getting calculation mistakes}}{\text{.}} \\
    \\
 \]