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Question: Matrix A such that \({{A}^{2}}=2A-I\), where I is the identity matrix. Then for \(n\ge 2\),\({{A}^{n...

Matrix A such that A2=2A−I{{A}^{2}}=2A-I, where I is the identity matrix. Then for n≥2n\ge 2,An{{A}^{n}} is equal to
(a) nA−(n−1)InA - (n - 1)I
(b) nA–InA – I
(c) 2n−1A−(n−1)I{{2}^{n-1}}A-\left( n-1 \right)I
(d) 2n−1A−I{{2}^{n-1}}A-I

Explanation

Solution

Hint:Find A3{{A}^{3}} by multiplying the given equation A2=2A−I→A{{A}^{2}}=2A-I\to A, further calculate A4{{A}^{4}}by multiplying the equation of A3{{A}^{3}}, similarly, get A5,A6,A7......{{A}^{5}},{{A}^{6}},{{A}^{7}}...... by multiplying the previous equation formed by A. and use the equation A2=2A−I{{A}^{2}}=2A-I to make all the equations of A3,A4,A5......{{A}^{3}},{{A}^{4}},{{A}^{5}}...... only in terms of A and I. Now, observe all the equations and get the values of An{{A}^{n}}by observing the equations only.

Complete step-by-step answer:
Equation with matrix is given as
A2=2A−I{{A}^{2}}=2A-I……………..(i)
Where, I is the identity matrix.
And with the help of equation (i), we need to calculate the value of An{{A}^{n}}, for n≥2n\ge 2.
As, we know multiplication of any matrix M to M will give resultant as M2{{M}^{2}} and similarly, if we multiply M2{{M}^{2}}by M, we will get M3{{M}^{3}} and so on.
So, let us multiply the equation (i) by ‘A’ to both sides. Hence, we get
A.A2=(2A−I)A A3=2A.A−IA A3=2A2−A \begin{aligned} & A.{{A}^{2}}=\left( 2A-I \right)A \\\ & {{A}^{3}}=2A.A-IA \\\ & {{A}^{3}}=2{{A}^{2}}-A \\\ \end{aligned}
Where, we know multiplication of an identity matrix with any other matrix will give the same matrix.
So, IA = A.
Hence, we get equation after multiplying the equation (i) by ‘A’ as
A3=2A2−A{{A}^{3}}=2{{A}^{2}}-A …………..(ii)
Now, replace A2→{{A}^{2}}\to 2A – I from the equation (i) to the equation (ii).
Hence, we get
A3=2(2A−I)−A A3=4A−A−2I \begin{aligned} & {{A}^{3}}=2\left( 2A-I \right)-A \\\ & {{A}^{3}}=4A-A-2I \\\ \end{aligned}
A3=3A−2I{{A}^{3}}=3A-2I……………..(iii)
Now, let us multiply the above equation by A to both sides. So, we get
A.A3=(3A−2I)A A4=3A2−2AI,A4=3A2−2A \begin{aligned} & A.{{A}^{3}}=\left( 3A-2I \right)A \\\ & {{A}^{4}}=3{{A}^{2}}-2AI,{{A}^{4}}=3{{A}^{2}}-2A \\\ \end{aligned}
Now, we can replace A2{{A}^{2}}by 2A – I from the equation (i) to the above equation. So, we get
A4=3(2A−I)−2A A4=6A−3I−2A A4=6A−2A−3I \begin{aligned} & {{A}^{4}}=3\left( 2A-I \right)-2A \\\ & {{A}^{4}}=6A-3I-2A \\\ & {{A}^{4}}=6A-2A-3I \\\ \end{aligned}
A4=4A−3I{{A}^{4}}=4A-3I……………….(iv)
Now, similarly, multiply the equation (iv) by A to both sides as well, so, we get’

& A.{{A}^{4}}=\left( 4A-3I \right)A \\\ & {{A}^{5}}=4{{A}^{2}}-3IA,{{A}^{5}}=4{{A}^{2}}-3A \\\ \end{aligned}$$ Now, replace$${{A}^{2}}\to 2A-I$$in the above equation from the equation (i). so, we get $\begin{aligned} & {{A}^{5}}=4\left( 2A-I \right)-3A \\\ & {{A}^{5}}=8A-4I-3A \\\ \end{aligned}$ ${{A}^{5}}=5A-4I$……………(v) Now, we can observe the equation (i), (ii), (iv) and (v) and hence get that the coefficient of A is following the sequence 2, 3, 4, 5…… and coefficient of I in all the equations is following the sequence (-1, -2, -3, -4……..) It means, coefficient of A for ${{A}^{2}}=2$ Coefficient of A for ${{A}^{3}}=3$ Coefficient of A for $${{A}^{4}}=4$$ Coefficient of A for ${{A}^{5}}=5$ Similarly, the coefficient of A for ${{A}^{n}}$ should be = n. And coefficient of I for ${{A}^{2}}=-1=-\left( 2-1 \right)$ Coefficient of I for${{A}^{3}}=-2=-\left( 3-1 \right)$ Coefficient of I for${{A}^{4}}=-3=\left( 4-1 \right)$ Coefficient of I for${{A}^{5}}=-4=-\left( 5-1 \right)$ Hence, coefficient of I for ${{A}^{n}}$ id given as – (n - 1). Hence, we get the value of ${{A}^{n}}$, with the help of above observations as ${{A}^{n}}=nA-\left( n-1 \right)I$ So, option (a) is the correct answer. Note: One may go wrong if he/she applies the following approach as $\begin{aligned} & {{A}^{2}}=2A-I \\\ & {{A}^{2}}-2A+I=0 \\\ & {{\left( A-I \right)}^{2}}=0 \\\ \end{aligned}$ Now, one may take A = I and hence, will get ${{A}^{n}}\to I$, which is wrong. So, take care with this approach. Observing the pattern of ${{A}^{2}},{{A}^{3}},{{A}^{4}}.........{{A}^{n}}$ was the key point of the solution. One may always use this approach for these kind of questions.