Question

If $A = \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 1}&3
\end{array}} \right]$ and ${A^6} = KA - 205I$ then
(A) $K = $$11$
(B) $K = $$22$
(C) $K = $$33$
(D) $K = $$44$

Answer 1

Hint:
Find ${A^6}$ in multiple steps, i.e., ${A^2} = A \times A$, ${A^3} = {A^2} \times A$ and ${A^6} = {A^3} \times {A^3}$. Then use the given relation ${A^6} = KA - 205I$ to find the value of $K$.

Complete step by step solution:
 Given, $A = \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 1}&3
\end{array}} \right]$
We have to calculate ${A^6}$. For this, we find firstly ${A^2}$ and ${A^3}$.
\[{A^2} = A \times A = \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 1}&3
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 1}&3
\end{array}} \right]\]
$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
  {1 \times 1 + 2 \times - 1}&{1 \times 2 + 2 \times 3} \\
  { - 1 \times 1 + 3 \times - 1}&{ - 1 \times 2 + 3 \times 3}
\end{array}} \right]$
$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
  {1 - 2}&{2 + 6} \\
  { - 1 - 3}&{ - 2 + 9}
\end{array}} \right]$
$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
  { - 1}&8 \\
  { - 4}&7
\end{array}} \right]$
And \[{A^3} = {A^2} \times A = \left[ {\begin{array}{*{20}{c}}
  { - 1}&8 \\
  { - 4}&7
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 1}&3
\end{array}} \right]\]
$ \Rightarrow {A^3} = \left[ {\begin{array}{*{20}{c}}
  { - 1 \times 1 + 8 \times - 1}&{ - 1 \times 2 + 8 \times 3} \\
  { - 4 \times 1 + 7 \times - 1}&{ - 4 \times 2 + 7 \times 3}
\end{array}} \right]$
$ \Rightarrow {A^3} = \left[ {\begin{array}{*{20}{c}}
  { - 1 - 8}&{ - 2 + 24} \\
  { - 4 - 7}&{ - 8 + 21}
\end{array}} \right]$
$ \Rightarrow {A^3} = \left[ {\begin{array}{*{20}{c}}
  { - 9}&{22} \\
  { - 11}&{13}
\end{array}} \right]$
Now, \[{A^6} = {A^3} \times {A^3} = \left[ {\begin{array}{*{20}{c}}
  { - 9}&{22} \\
  { - 11}&{13}
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
  { - 9}&{22} \\
  { - 11}&{13}
\end{array}} \right]\]
$ \Rightarrow {A^6} = \left[ {\begin{array}{*{20}{c}}
  { - 9 \times - 9 + 22 \times - 11}&{ - 9 \times 22 + 22 \times 13} \\
  { - 11 \times - 9 + 13 \times - 11}&{ - 11 \times 22 + 13 \times 13}
\end{array}} \right]$
$ \Rightarrow {A^6} = \left[ {\begin{array}{*{20}{c}}
  {81 - 242}&{ - 198 + 286} \\
  {99 - 143}&{ - 242 + 169}
\end{array}} \right]$
$ \Rightarrow {A^6} = \left[ {\begin{array}{*{20}{c}}
  { - 161}&{88} \\
  { - 44}&{ - 73}
\end{array}} \right]$
We have ${A^6} = KA - 205I$
$ \Rightarrow $${A^6} + 205I = KA$ ….. (1)
Substitute the value of ${A^6}$, $I$ and $A$ in (1);
$\left[ {\begin{array}{*{20}{c}}
  { - 161}&{88} \\
  { - 44}&{ - 73}
\end{array}} \right] + 205\left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right] = K\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 1}&3
\end{array}} \right]$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  { - 161}&{88} \\
  { - 44}&{ - 73}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  {205}&0 \\
  0&{205}
\end{array}} \right] = K\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 1}&3
\end{array}} \right]$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  { - 161 + 205}&{88 + 0} \\
  { - 44 + 0}&{ - 73 + 205}
\end{array}} \right] = K\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 1}&3
\end{array}} \right]$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  {44}&{88} \\
  { - 44}&{132}
\end{array}} \right] = K\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 1}&3
\end{array}} \right]$
$ \Rightarrow 44\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 1}&3
\end{array}} \right] = K\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  { - 1}&3
\end{array}} \right]$
On comparing both sides, we get-
\[ \Rightarrow K = 44\]

Hence, option (D) is the correct answer.

Note:
Here, the symbol $I$ is used for identity matrices that have an order $n \times n$. The entries on the diagonal from the upper left to the bottom right are all $1's$ and all other entries are $0$. For ex-
${I_2} = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]$ , ${I_3} = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]$
The identity matrix plays a similar role in operations with matrices as the number $1$ plays in operations with real numbers.