Question

If A = \left[ {\begin{array}{*{20}{c}} 2&{ - 2} \\\ { - 2}&2 \end{array}} \right] , then ${A^n} = {2^k}A$ , where k= ?
A. ${2^{n - 1}}$
B. n+1
C. n-1
D. $2\left( {n - 1} \right)$

Answer 1

In this question to find the value of k we will find the value of ${A^2}$, ${A^3}$ and, ${A^4}$ with the help of the property of matrix i.e.
\left[ {\begin{array}{*{20}{c}} {ak}&{bk} \\\ {ck}&{dk} \end{array}} \right] = k\left[ {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right]
Now we will find the value of ${A^2}$ , ${A^3}$ and, ${A^4}$ in terms A so that we can compare these equations with the given equation i.e. ${A^n} = {2^k}A$ so that we can find the relation between n and k to get the required answer.

Complete step-by-step answer:
Given data: A = \left[ {\begin{array}{*{20}{c}} 2&{ - 2} \\\ { - 2}&2 \end{array}} \right]
We know that, \left[ {\begin{array}{*{20}{c}} {ak}&{bk} \\\ {ck}&{dk} \end{array}} \right] = k\left[ {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right]
Therefore using this property on matrix A
\Rightarrow A = 2\left[ {\begin{array}{*{20}{c}} 1&{ - 1} \\\ { - 1}&1 \end{array}} \right]...........(i)
Multiplying both sides by matrix A
\Rightarrow {A^2} = {2^2}\left[ {\begin{array}{*{20}{c}} 1&{ - 1} \\\ { - 1}&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&{ - 1} \\\ { - 1}&1 \end{array}} \right]
On multiplication of matrix and simplification we get,
\Rightarrow {A^2} = {2^2}\left[ {\begin{array}{*{20}{c}} {1 + 1}&{ - 1 - 1} \\\ { - 1 - 1}&{1 + 1} \end{array}} \right]
On simplifying the elements of the matrix we get,
\Rightarrow {A^2} = {2^2}\left[ {\begin{array}{*{20}{c}} 2&{ - 2} \\\ { - 2}&2 \end{array}} \right]..................(ii)
Now, we know that \left[ {\begin{array}{*{20}{c}} 2&{ - 2} \\\ { - 2}&2 \end{array}} \right] = A
$\Rightarrow {A^2} = {2^2}A.............(iii)$
Now using the property of a matrix i.e. \left[ {\begin{array}{*{20}{c}} {ak}&{bk} \\\ {ck}&{dk} \end{array}} \right] = k\left[ {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right] in equation(ii)
\Rightarrow {A^2} = {2^3}\left[ {\begin{array}{*{20}{c}} 1&{ - 1} \\\ { - 1}&1 \end{array}} \right]
On multiplying equation(i) and equation(ii)
\Rightarrow {A^3} = {2^4}\left[ {\begin{array}{*{20}{c}} 1&{ - 1} \\\ { - 1}&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&{ - 1} \\\ { - 1}&1 \end{array}} \right]
On multiplication of matrix and simplification
\Rightarrow {A^3} = {2^4}\left[ {\begin{array}{*{20}{c}} {1 + 1}&{ - 1 - 1} \\\ { - 1 - 1}&{1 + 1} \end{array}} \right]
On simplifying the elements of the matrix

2&{ - 2} \\\ { - 2}&2 \end{array}} \right].................(iv)$$ Now, we know that $\left[ {\begin{array}{*{20}{c}} 2&{ - 2} \\\ { - 2}&2 \end{array}} \right] = A$ $ \Rightarrow {A^3} = {2^4}A............(v)$ Now using the property of a matrix i.e. $\left[ {\begin{array}{*{20}{c}} {ak}&{bk} \\\ {ck}&{dk} \end{array}} \right] = k\left[ {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right]$ in equation(iv) $$ \Rightarrow {A^3} = {2^5}\left[ {\begin{array}{*{20}{c}} 1&{ - 1} \\\ { - 1}&1 \end{array}} \right]$$ Now multiplying equation(i) and equation(iv) $$ \Rightarrow {A^4} = {2^6}\left[ {\begin{array}{*{20}{c}} 1&{ - 1} \\\ { - 1}&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&{ - 1} \\\ { - 1}&1 \end{array}} \right]$$ On multiplication of matrix and simplification $ \Rightarrow {A^4} = {2^6}\left[ {\begin{array}{*{20}{c}} {1 + 1}&{ - 1 - 1} \\\ { - 1 - 1}&{1 + 1} \end{array}} \right]$ On simplifying the elements of the matrix $$ \Rightarrow {A^4} = {2^6}\left[ {\begin{array}{*{20}{c}} 2&{ - 2} \\\ { - 2}&2 \end{array}} \right]$$ Now, we know that $\left[ {\begin{array}{*{20}{c}} 2&{ - 2} \\\ { - 2}&2 \end{array}} \right] = A$ $ \Rightarrow {A^4} = {2^6}A...............(vi)$ Now on concluding from equation (iii), equation(v) and, equation(vi) We can say that, $k = 2\left( {n - 1} \right)$ **Hence, Option (D) is correct.** **Note:** While taking common any scalar elements from the square matrix some of the students apply the property of the determinant i.e. $\left| {\begin{array}{*{20}{c}} {ka}&{kb} \\\ {tc}&{td} \end{array}} \right| = kt\left| {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right|$ or $\left| {\begin{array}{*{20}{c}} {ka}&{kb} \\\ {kc}&{kd} \end{array}} \right| = {k^2}\left| {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right|$ which is not applicable for the matrix as the matrix follow the property i.e. $\left[ {\begin{array}{*{20}{c}} {ak}&{bk} \\\ {ck}&{dk} \end{array}} \right] = k\left[ {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right]$ so most of the students make mistakes while taking common like while finding the ${A^2}$ using the equation (ii) i.e. ${A^2} = {2^2}\left[ {\begin{array}{*{20}{c}} 2&{ - 2} \\\ { - 2}&2 \end{array}} \right]$ $ \Rightarrow {A^2} = {2^2} \times 2 \times 2\left[ {\begin{array}{*{20}{c}} 1&{ - 1} \\\ { - 1}&1 \end{array}} \right]$ , which is not correct and will take us to the wrong answer, so avoid making mistakes like this to get the correct answer.