High School Mathematics Extensions/Matrices/Problem Set/Solutions

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Matrices Problem Set

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{\begin{pmatrix}2&3\\3&5\end{pmatrix}}{\begin{pmatrix}?&?&?&?\\?&?&?&?\end{pmatrix}}={\begin{pmatrix}28&94&70&102\\44&153&112&163\end{pmatrix}}

{\begin{pmatrix}2&3\\3&5\end{pmatrix}}^{-1}{\begin{pmatrix}2&3\\3&5\end{pmatrix}}{\begin{pmatrix}?&?&?&?\\?&?&?&?\end{pmatrix}}={\begin{pmatrix}2&3\\3&5\end{pmatrix}}^{-1}{\begin{pmatrix}28&94&70&102\\44&153&112&163\end{pmatrix}}

{\begin{pmatrix}?&?&?&?\\?&?&?&?\end{pmatrix}}={\begin{pmatrix}2&3\\3&5\end{pmatrix}}^{-1}{\begin{pmatrix}28&94&70&102\\44&153&112&163\end{pmatrix}}

={\frac {1}{2\times 5-3\times 3}}{\begin{pmatrix}5&-3\\-3&2\end{pmatrix}}{\begin{pmatrix}28&94&70&102\\44&153&112&163\end{pmatrix}}

={\begin{pmatrix}(5\times 28+(-3)\times 44)&(5\times 94+(-3)\times 153)&(5\times 70+(-3)\times 112)&(5\times 102+(-3)\times 163)\\((-3)\times 28+2\times 44)&((-3)\times 94+2\times 153)&((-3)\times 70+2\times 112)&((-3)\times 102+2\times 163)\end{pmatrix}}

={\begin{pmatrix}8&11&14&21\\4&24&14&20\end{pmatrix}}

Therefore the message is "iloveyou"

Combine the two matrices together, we have

A{\begin{pmatrix}1&3\\2&4\end{pmatrix}}={\begin{pmatrix}1&0\\0&1\end{pmatrix}}

A{\begin{pmatrix}1&3\\2&4\end{pmatrix}}=I

Therefore the inverse of A is

{\begin{pmatrix}1&3\\2&4\end{pmatrix}}

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