Question
If there is a statement involving the natural number n such that:
  1. The statement is true for $n = 1$
  2. When the statement is true for $n = k ($where $k$ is some positive integer$),$ then the statement is also true for $n = k + 1.$
Then, the statement is true for all natural numbers n.
Also, if $A$ is a square matrix of order n, then $A^2$ is defined as $AA$. In general, $A^m = AA .... A (m$ times$)$. where m is any positive integer.
Based on the above information, answer the following questions.
  1. If $\text{A}=\begin{bmatrix}3&-4\\1&-1\end{bmatrix},$ then for any positive integer n,
  1. $\text{A}^\text{n}=\begin{bmatrix}3\text{n}&-4\text{n}\\\text{n}&-\text{n}\end{bmatrix}$
  2. $\text{A}^\text{n}=\begin{bmatrix}1+2\text{n}&-4\text{n}\\\text{n}&1-2\text{n}\end{bmatrix}$
  3. $\text{A}^\text{n}=\begin{bmatrix}3\text{n}&-8\text{n}\\1&-\text{n}\end{bmatrix}$
  4. $\text{A}^\text{n}=\begin{bmatrix}1+3\text{n}&-4\text{n}\\\text{n}&1-3\text{n}\end{bmatrix}$
  1. If $\text{A}=\begin{bmatrix}1&2\\0&1\end{bmatrix},$ then $|A^n|$, where $\text{n}\in\text{ N},$ is equal to:
  1. $2^n$
  2. $3^n$
  3. $n$
  4. $1$
  1. If $\text{A}=\begin{bmatrix}1&0\\1&1\end{bmatrix}$ and $\text{I}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ then which of the following holds for all natural numbers $\text{n}\geq1?$
  1. $A^n= nA - (n - 1)I$
  2. $A^n = 2^{n-1} A - (n - 1)I$
  3. $A^n= nA + (n - 1)I$
  4. $A^n = 2^{n-1} A + (n - 1)I$
  1. Let $\text{A}=\begin{bmatrix}\text{a}&0&0\\0&\text{a}&0\\0&0&\text{a}\end{bmatrix}$ and $\text{A}^\text{n}=[\text{a}_{\text{ij}}]_{3\times3}$ for some positive integer n, then the cofactor of $a_{13}$ is:
  1. $a^n$
  2. $-a^n$
  3. $2a^n$
  4. $0$
  1. If $A$ is a square matrix such that $|A| = 2,$ then for any positive integer n, $|A^n|$ is equal to:
  1. $0$
  2. $2n$
  3. $2^n$
  4. $n^2$

Answer

  1. (b) $\text{A}^\text{n}=\begin{bmatrix}1+2\text{n}&-4\text{n}\\\text{n}&1-2\text{n}\end{bmatrix}$
Solution:
We have, $\text{A}=\begin{bmatrix}3&-4\\1&-1\end{bmatrix}$
$\therefore\text{A}^2=\begin{bmatrix}3&-4\\1&-1\end{bmatrix}\begin{bmatrix}3&-4\\1&-1\end{bmatrix}=\begin{bmatrix}5&-8\\2&-3\end{bmatrix},$ which can be obtained from $\text{A}^\text{n}=\begin{bmatrix}1+2\text{n}&-4\text{n}\\\text{n}&1-2\text{n}\end{bmatrix}$ for $n = 2.$
  1. (d) $1$
Solution:
We have, $\text{A}=\begin{bmatrix}1&2\\0&1\end{bmatrix}$
$\therefore|\text{A}|=\begin{vmatrix}1&2\\0&1\end{vmatrix}=1-0=1$
Also, $|A^n| = |A· A ...... A(n$ times$)| = |A|^n = 1^n = 1$
  1. (a) $A^n= nA - (n - 1)I$
Solution:
For $n = 1,$ all options are true.
$\text{A}^2=\text{A}\cdot\text{A}=\begin{bmatrix}1&0\\1&1\end{bmatrix}\begin{bmatrix}1&0\\1&1\end{bmatrix}=\begin{bmatrix}1&0\\2&1\end{bmatrix}$
and $\text{A}^3=\text{A}^2\cdot\text{A}=\begin{bmatrix}1&0\\2&1\end{bmatrix}\begin{bmatrix}1&0\\1&1\end{bmatrix}=\begin{bmatrix}1&0\\3&1\end{bmatrix}$
Putting $n = 3$, in $(a),$ we get $A^3 = 3A - 2I$
$=3\begin{bmatrix}1&0\\1&1\end{bmatrix}-\begin{bmatrix}2&0\\0&2\end{bmatrix}$
$=\begin{bmatrix}3&0\\3&3\end{bmatrix}-\begin{bmatrix}2&0\\0&2\end{bmatrix}=\begin{bmatrix}1&0\\3&1\end{bmatrix},$ which is true.
All other options are different from $A^3 = 3A -2I$ for $n = 3.$
  1. (d) $0$
Solution:
We have, $\text{A}=\begin{bmatrix}\text{a}&0&0\\0&\text{a}&0\\0&0&\text{a}\end{bmatrix}$
$\therefore\text{A}^2=\text{A}\cdot\text{A}=\begin{bmatrix}\text{a}&0&0\\0&\text{a}&0\\0&0&\text{a}\end{bmatrix}\begin{bmatrix}\text{a}&0&0\\0&\text{a}&0\\0&0&\text{a}\end{bmatrix}$
$=\begin{bmatrix}\text{a}^2&0&0\\0&\text{a}^2&0\\0&0&\text{a}^2\end{bmatrix}$
Similarly, $\text{A}^\text{n}=\begin{bmatrix}\text{a}^\text{n}&0&0\\0&\text{a}^\text{n}&0\\0&0&\text{a}^\text{n}\end{bmatrix}$
Now, cofactor of $\text{a}_{13}=(-1)^{1+3}\begin{vmatrix}0&\text{a}^\text{n}\\0&0\end{vmatrix}=0$
  1. (c) $2^n$​​​​​​​
Solution:
We have, $|A| = 2$ and $|A^n| = |A·A ...... A(n- $ times$)|$
$= |A| |A| ...... |A|(n -$ times$) = |A|^n = 2^n$​​​​​​​

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