Question

If \[A = \left| {\begin{array}{*{20}{c}}
  1&0 \\
  1&1
\end{array}} \right|\]and\[I = \left| {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right|\], then which one of the following holds for all\[n \geqslant 1\], by the principle of mathematical induction
A) \[{A^n} = nA - \left( {n - 1} \right)I\]
B) \[{A^n} = {2^{n - 1}}A - \left( {n - 1} \right)I\]
C) \[{A^n} = nA + \left( {n - 1} \right)I\]
D) \[{A^n} = {2^{n - 1}}A + \left( {n - 1} \right)I\]

Answer 1

Hint: Use mathematical induction theorem, which is a mathematical technique that is used to prove a statement, a formula, or a theorem is true for natural value.
In this question, start from checking the options whether the given equation satisfies the given matrix, by finding the value of\[{A^n}\]and then checking for R.H.S of the equation.
A matrix is a rectangular array of tables, symbols, or expressions, arranged in rows and columns.

Complete step by step answer:
 \[A = \left| {\begin{array}{*{20}{c}}
  1&0 \\
  1&1
\end{array}} \right|\]
\[I = \left| {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right|\]
In the given options, we can see \[{A^n}\] common for every option; hence find \[{A^n}\]
\[
  A = \left| {\begin{array}{*{20}{c}}
  1&0 \\
  1&1
\end{array}} \right| \\
  {A^2} = \left| {\begin{array}{*{20}{c}}
  1&0 \\
  1&1
\end{array}} \right|\left| {\begin{array}{*{20}{c}}
  1&0 \\
  1&1
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  {1 + 0}&0 \\
  {1 + 1}&{0 + 1}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  1&0 \\
  2&1
\end{array}} \right| \\
  {A^3} = A{A^2} = \left| {\begin{array}{*{20}{c}}
  1&0 \\
  2&1
\end{array}} \right|\left| {\begin{array}{*{20}{c}}
  1&0 \\
  1&1
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  {1 + 0}&{0 + 0} \\
  {2 + 1}&{0 + 1}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  1&0 \\
  3&1
\end{array}} \right| \\
  . \\
  . \\
  . \\
 \]
Hence by observing the value of exponents of matrix A, we can write,
\[{A^n} = \left| {\begin{array}{*{20}{c}}
  1&0 \\
  n&1
\end{array}} \right|\]
Since we have got the value of \[{A^n}\], now check the options for R.H.S
\[{A^n} = nA - \left( {n - 1} \right)I\]
Given R.H.S\[ = nA - \left( {n - 1} \right)I\]
Where
\[nA = n\left| {\begin{array}{*{20}{c}}
  1&0 \\
  1&1
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  n&0 \\
  n&n
\end{array}} \right|\]
\[\left( {n - 1} \right)I = \left( {n - 1} \right)\left| {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  {n - 1}&0 \\
  0&{n - 1}
\end{array}} \right|\]
Therefore
\[ = nA - \left( {n - 1} \right)I = \left| {\begin{array}{*{20}{c}}
  n&0 \\
  n&n
\end{array}} \right| - \left| {\begin{array}{*{20}{c}}
  {n - 1}&0 \\
  0&{n - 1}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  1&0 \\
  n&1
\end{array}} \right|\]
Hence L.H.S=R.H.S
\[{A^n} = nA - \left( {n - 1} \right)I\]
Condition satisfies
\[{A^n} = {2^{n - 1}}A - \left( {n - 1} \right)I\]
\[{2^{n - 1}}A = {2^{n - 1}}\left| {\begin{array}{*{20}{c}}
  1&0 \\
  1&1
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  {{2^{n - 1}}}&0 \\
  {{2^{n - 1}}}&{{2^{n - 1}}}
\end{array}} \right|\]
Hence R.H.S
\[{2^{n - 1}}A - \left( {n - 1} \right)I = \left| {\begin{array}{*{20}{c}}
  {{2^{n - 1}}}&0 \\
  {{2^{n - 1}}}&{{2^{n - 1}}}
\end{array}} \right| - \left| {\begin{array}{*{20}{c}}
  {n - 1}&0 \\
  0&{n - 1}
\end{array}} \right|\]
\[L.H.S \ne R.H.S\]
\[{A^n} = nA + \left( {n - 1} \right)I\]
\[nA = n\left| {\begin{array}{*{20}{c}}
  1&0 \\
  1&1
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  n&0 \\
  n&n
\end{array}} \right|\]
\[\left( {n - 1} \right)I = \left( {n - 1} \right)\left| {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  {n - 1}&0 \\
  0&{n - 1}
\end{array}} \right|\]
Therefore
\[ = nA + \left( {n - 1} \right)I = \left| {\begin{array}{*{20}{c}}
  n&0 \\
  n&n
\end{array}} \right| + \left| {\begin{array}{*{20}{c}}
  {n - 1}&0 \\
  0&{n - 1}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  n&{ - 1}
\end{array}} \right|\]
\[L.H.S \ne R.H.S\]
\[{A^n} = {2^{n - 1}}A + \left( {n - 1} \right)I\]
\[{2^{n - 1}}A = {2^{n - 1}}\left| {\begin{array}{*{20}{c}}
  1&0 \\
  1&1
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  {{2^{n - 1}}}&0 \\
  {{2^{n - 1}}}&{{2^{n - 1}}}
\end{array}} \right|\]
Hence R.H.S
\[{2^{n - 1}}A + \left( {n - 1} \right)I = \left| {\begin{array}{*{20}{c}}
  {{2^{n - 1}}}&0 \\
  {{2^{n - 1}}}&{{2^{n - 1}}}
\end{array}} \right| + \left| {\begin{array}{*{20}{c}}
  {n - 1}&0 \\
  0&{n - 1}
\end{array}} \right|\]
\[L.H.S \ne R.H.S\]
Hence option (A) is correct.

Note: Students must note to do the multiplication of matrix, and this is possible only if the number of rows of the first matrix must be equal to the number of columns of the second matrix and the result will have the same number of rows as the first matrix and the same number of columns as the second matrix.