Question

If there are three square matrices A, B and C of the same order satisfying the equation \[{{A}^{2}}={{A}^{-1}}\] and let \[B={{A}^{{{2}^{n}}}}\] and \[C={{A}^{2\left( n-2 \right)}}\] then (B – C)?

Answer 1

Hint: We have \[{{A}^{2}}={{A}^{-1}}\] and we also have \[{{A}^{2}}={{A}^{-1}}\] and we also have \[B={{A}^{{{2}^{n}}}}\] and \[C={{A}^{2\left( n-2 \right)}}\] and we start with \[B={{A}^{{{2}^{n}}}}\] then we use n = n – 1 + 1 to expand B, then we will use \[{{a}^{x}}.{{a}^{y}}={{a}^{x+y}}.\] After expanding the value of B, we will use \[{{\left( {{a}^{x}} \right)}^{y}}={{\left( {{a}^{y}} \right)}^{x}}\] to simplify more. Lastly, we will get the relation between B and C and we use that to subtract C from B.

Complete step by step answer:
We are given that A, B and C are three square matrices of the same order. We know that the square matrix is the one in which the number of columns = number of rows. We are given that A is such a matrix that its square is the same as its inverse means
\[{{A}^{2}}={{A}^{-1}}......\left( i \right)\]
We are given our matrix B is defined as \[B={{A}^{{{2}^{n}}}}\]
We will rearrange our matrix B. We know that, \[{{a}^{x}}.{{a}^{y}}={{a}^{x+y}}.\]
So, we know that n can be written as n = n – 1 + 1
So, \[{{2}^{n}}={{2}^{n-1+1}}\]
Now, as \[{{2}^{x+y}}={{2}^{x}}{{.2}^{y}},\] so,
\[{{2}^{n}}={{2}^{n-1+1}}\]
\[\Rightarrow {{2}^{n}}={{2}^{n-1}}{{.2}^{1}}\]
So, we get,
\[\Rightarrow {{2}^{n}}={{2.2}^{n-1}}......\left( ii \right)\]
We mark it as equation (ii)
Now, we have,
\[B={{A}^{{{2}^{n}}}}\]
Now using (ii), we get,
\[\Rightarrow B={{A}^{{{2.2}^{n-1}}}}\]
On simplifying, we get,
\[\Rightarrow B={{\left( {{A}^{2}} \right)}^{{{2}^{n-1}}}}\]
As, \[{{A}^{2}}={{A}^{-1}},\] using (i), we get,
\[\Rightarrow B=\left( {{\left( {{A}^{-1}} \right)}^{{{2}^{n-1}}}} \right)\]
As, we know \[{{\left( {{a}^{x}} \right)}^{y}}={{\left( {{a}^{y}} \right)}^{x}},\]
So, we have,
\[B={{\left( {{A}^{{{2}^{n-1}}}} \right)}^{-1}}\]
Now, again
\[{{2}^{n-1}}={{2}^{n-2+1}}\]
\[\Rightarrow {{2}^{n-1}}={{2.2}^{n-2}}\]
So, we get,
\[\Rightarrow B={{\left( {{A}^{{{2.2}^{n-2}}}} \right)}^{-1}}\]
On simplifying, we get,
\[\Rightarrow B={{\left[ {{\left( {{A}^{2}} \right)}^{{{2}^{n-2}}}} \right]}^{-1}}\]
As \[{{A}^{2}}={{A}^{-1}},\] so we get,
\[\Rightarrow B={{\left[ {{\left( {{A}^{-1}} \right)}^{{{2}^{n-2}}}} \right]}^{-1}}\]
Again, as \[\left[ {{\left( {{a}^{x}} \right)}^{y}} \right]=\left[ {{\left( {{a}^{y}} \right)}^{x}} \right],\] so we get,
\[\Rightarrow B={{\left[ {{\left( {{A}^{{{2}^{n-2}}}} \right)}^{-1}} \right]}^{-1}}\]
On simplifying, we get,
\[\Rightarrow B={{A}^{{{2}^{\left( n-2 \right)}}}}\]
But we are given that \[C={{A}^{{{2}^{\left( n-2 \right)}}}}\] so we get,
\[B=C\]
Now,
\[B-C=B-B\left[ \text{As }B=C \right]\]
\[\Rightarrow B-C=0\]
So, we get,
\[\Rightarrow B-C=0\]

Note:
 While simplifying we should be careful about the powers of matrix A. Always remember that the product of two negative is positive.
\[{{\left( {{2}^{-1}} \right)}^{-1}}={{2}^{-1\times -1}}\]
\[\Rightarrow {{\left( {{2}^{-1}} \right)}^{-1}}=2\]
Also, when we add and subtract the same term, then it won’t affect the original term that is why we write it as n = n – 1 + 1.