$If\ A = \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix}\ and\ I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix},$ then which one of the following holds for all $n \geq 1$, by the principle of mathematical induction?

Question

Dr. Abubacker · Accepted Answer

(d)
$A^{2} = \begin{bmatrix}
1 & 0 \
1 & 1
\end{bmatrix}\begin{bmatrix}
1 & 0 \
1 & 1
\end{bmatrix} = \begin{bmatrix}
1 & 0 \
2 & 1
\end{bmatrix}$
$A^{3} = \begin{bmatrix}
1 & 0 \
2 & 1
\end{bmatrix}\begin{bmatrix}
1 & 0 \
1 & 1
\end{bmatrix} = \begin{bmatrix}
1 & 0 \
3 & 1
\end{bmatrix}$
$A^{n} = \begin{bmatrix}
1 & 0 \
1 & 1
\end{bmatrix}$ can be verified by induction. Now, taking option
$(b)\begin{bmatrix}
1 & 0 \
n & 1
\end{bmatrix} = \begin{bmatrix}
n & 0 \
n & n
\end{bmatrix} + \begin{bmatrix}
n - 1 & 0 \
0 & n - 1
\end{bmatrix}$
$\Longrightarrow \begin{bmatrix}
1 & 0 \
n & 1
\end{bmatrix} 
eq \begin{bmatrix}
2n - 1 & 0 \
1 & 2n - 1
\end{bmatrix}$
$(d)nA - (n - 1)I = \begin{bmatrix}
n & 0 \
n & n
\end{bmatrix} - \begin{bmatrix}
n - 1 & 0 \
1 & n - 1
\end{bmatrix}$
$= \begin{bmatrix}
1 & 0 \
n & 1
\end{bmatrix}{= A}^{n}$

Q. $If\ A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\ and\ I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},$ then which one of the following holds for all $n \geq 1$, by the principle of mathematical induction?

Explanation

Reviewed by Dr. Abubacker

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