High School Mathematics Extensions/Matrices/Problem Set/Solutions

High School Mathematics Extensions

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Supplementary Chapters — Primes and Modular Arithmetic — Logic

Mathematical Proofs — Set Theory and Infinite Processes — Counting and Generating Functions — Discrete Probability

Matrices — Further Modular Arithmetic — Mathematical Programming — Markov Chains

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Matrices Problem Set

At the moment, the main focus is on authoring the main content of each chapter. Therefore this exercise solutions section may be out of date and appear disorganised.

If you have a question please leave a comment in the "discussion section" or contact the author or any of the major contributors.

1.

$\left(\begin{array}{cc}2& 3\\ 3& 5\end{array}\right)\left(\begin{array}{cccc}?& ?& ?& ?\\ ?& ?& ?& ?\end{array}\right)=\left(\begin{array}{cccc}28& 94& 70& 102\\ 44& 153& 112& 163\end{array}\right)$

${\left(\begin{array}{cc}2& 3\\ 3& 5\end{array}\right)}^{-1}\left(\begin{array}{cc}2& 3\\ 3& 5\end{array}\right)\left(\begin{array}{cccc}?& ?& ?& ?\\ ?& ?& ?& ?\end{array}\right)={\left(\begin{array}{cc}2& 3\\ 3& 5\end{array}\right)}^{-1}\left(\begin{array}{cccc}28& 94& 70& 102\\ 44& 153& 112& 163\end{array}\right)$

$\left(\begin{array}{cccc}?& ?& ?& ?\\ ?& ?& ?& ?\end{array}\right)={\left(\begin{array}{cc}2& 3\\ 3& 5\end{array}\right)}^{-1}\left(\begin{array}{cccc}28& 94& 70& 102\\ 44& 153& 112& 163\end{array}\right)$

$=\frac{1}{2\times 5-3\times 3}\left(\begin{array}{cc}5& -3\\ -3& 2\end{array}\right)\left(\begin{array}{cccc}28& 94& 70& 102\\ 44& 153& 112& 163\end{array}\right)$

$=\left(\begin{array}{cccc}(5\times 28+(-3)\times 44)& (5\times 94+(-3)\times 153)& (5\times 70+(-3)\times 112)& (5\times 102+(-3)\times 163)\\ ((-3)\times 28+2\times 44)& ((-3)\times 94+2\times 153)& ((-3)\times 70+2\times 112)& ((-3)\times 102+2\times 163)\end{array}\right)$

$=\left(\begin{array}{cccc}8& 11& 14& 21\\ 4& 24& 14& 20\end{array}\right)$

Therefore the message is "iloveyou"

2.

Combine the two matrices together, we have

$A\left(\begin{array}{cc}1& 3\\ 2& 4\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$

$A\left(\begin{array}{cc}1& 3\\ 2& 4\end{array}\right)=I$

Therefore the inverse of A is

$\left(\begin{array}{cc}1& 3\\ 2& 4\end{array}\right)$