Question

Mod challenge

Prove : $\begin{pmatrix}1&2\\ 3&4\end{pmatrix}^3=\begin{pmatrix}37&54\\ 81&118\end{pmatrix}$

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Answer 1

Question :

☯ Prove $\large \begin{pmatrix}1&2\\ 3&4\end{pmatrix}^3=\begin{pmatrix}37&54\\ 81&118\end{pmatrix}$

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Given :

✭ Matrix given : $\begin{pmatrix}1&2\\ 3&4\end{pmatrix}^3$

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Solution :

We know a³ = a² × a

$\begin{pmatrix}1&2\\ 3&4\end{pmatrix}^3 = \begin{pmatrix}1&2\\ 3&4\end{pmatrix}^2+\begin{pmatrix}1&2\\ 3&4\end{pmatrix}$

$\longrightarrow \bf{ Solve\:\:\begin{pmatrix}1&2\\ 3&4\end{pmatrix}^2}$

$\bf{Multiply\:the\:rows\:of\:the\:first\:matrix\:by\:the\:columns\:of\:the\:second\:matrix}$

$\begin{pmatrix}1&2\end{pmatrix}\begin{pmatrix}1\\ 3\end{pmatrix}=1\cdot \:1+2\cdot \:3$

$\begin{pmatrix}1&2\end{pmatrix}\begin{pmatrix}2\\ 4\end{pmatrix}=1\cdot \:2+2\cdot \:4$

$\begin{pmatrix}3&4\end{pmatrix}\begin{pmatrix}1\\ 3\end{pmatrix}=3\cdot \:1+4\cdot \:3$

$\begin{pmatrix}3&4\end{pmatrix}\begin{pmatrix}2\\ 4\end{pmatrix}=3\cdot \:2+4\cdot \:4$

$=\large \begin{pmatrix}1\cdot \:1+2\cdot \:3&1\cdot \:2+2\cdot \:4\\ 3\cdot \:1+4\cdot \:3&3\cdot \:2+4\cdot \:4\end{pmatrix}$

$\bf{Simplify\:each\:element}$

$=\begin{pmatrix}7&10\\ 15&22\end{pmatrix}$

$\longrightarrow \bf{ Now\:\:Solve\:\:\begin{pmatrix}1&2\\ 3&4\end{pmatrix}^2+\begin{pmatrix}1&2\\ 3&4\end{pmatrix}}$

Substitute value of : $\begin{pmatrix}1&2\\ 3&4\end{pmatrix}^2$

$\longrightarrow \bf{\begin{pmatrix}1&2\\ 3&4\end{pmatrix}^2+\begin{pmatrix}1&2\\ 3&4\end{pmatrix}} = \begin{pmatrix}7&10\\ 15&22\end{pmatrix}\begin{pmatrix}1&2\\ 3&4\end{pmatrix}$

$\mathrm{Multiply\:the\:rows\:of\:the\:first\:matrix\:by\:the\:columns\:of\:the\:second\:matrix}$

$\begin{pmatrix}7&10\end{pmatrix}\begin{pmatrix}1\\ 3\end{pmatrix}=7\cdot \:1+10\cdot \:3$

$\begin{pmatrix}7&10\end{pmatrix}\begin{pmatrix}2\\ 4\end{pmatrix}=7\cdot \:2+10\cdot \:4$

$\begin{pmatrix}15&22\end{pmatrix}\begin{pmatrix}1\\ 3\end{pmatrix}=15\cdot \:1+22\cdot \:3$

$\begin{pmatrix}15&22\end{pmatrix}\begin{pmatrix}2\\ 4\end{pmatrix}=15\cdot \:2+22\cdot \:4$

$=\begin{pmatrix}7\cdot \:1+10\cdot \:3&7\cdot \:2+10\cdot \:4\\ 15\cdot \:1+22\cdot \:3&15\cdot \:2+22\cdot \:4\end{pmatrix}$

$\mathrm{Simplify\:each\:element}$

$=\begin{pmatrix}37&54\\ 81&118\end{pmatrix}$

✭ LHS = RHS

Hence proved !!

Answer 2

Answer:

$\begin{pmatrix}1&2\\ 3&4\end{pmatrix}^3 = \begin{pmatrix}1&2\\ 3&4\end{pmatrix}^2+\begin{pmatrix}1&2\\ 3&4\end{pmatrix}$

$=\large \begin{pmatrix}1\cdot \:1+2\cdot \:3&1\cdot \:2+2\cdot \:4\\ 3\cdot \:1+4\cdot \:3&3\cdot \:2+4\cdot \:4\end{pmatrix}$

$\longrightarrow \bf{ Now\:\:Solve\:\:\begin{pmatrix}1&2\\ 3&4\end{pmatrix}^2+\begin{pmatrix}1&2\\ 3&4\end{pmatrix}}$

$\longrightarrow \bf{\begin{pmatrix}1&2\\ 3&4\end{pmatrix}^2+\begin{pmatrix}1&2\\ 3&4\end{pmatrix}} = \begin{pmatrix}7&10\\ 15&22\end{pmatrix}\begin{pmatrix}1&2$

$\begin{pmatrix}7&10\end{pmatrix}\begin{pmatrix}1\\ 3\end{pmatrix}=7\cdot \:1+10\cdot \:3$

$=\begin{pmatrix}7\cdot \:1+10\cdot \:3&7\cdot \:2+10\cdot \:4\\ 15\cdot \:1+22\cdot \:3&15\cdot \:2+22\cdot \:4\end{pmatrix}$

$\mathrm{Simplify\:each\:element}$

$=\begin{pmatrix}37&54\\ 81&118\end{pmatrix}$