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10.18 Matrices

Tutorials and Workshops.pdf

Workshop

  1. Calculate \((1 + i)^{-5}\).

    First, \(1+i=\sqrt2cos(\frac{\pi}{4})+\sqrt2i\sin\left(\frac{\pi}{4}\right)=\left(\sqrt2,\frac{\pi}{4}\right)\)

    Then \(\left(1+i\right)^{-5}=\left(\left(\sqrt2^{-5},\frac{\pi}{4}\times\left(-5\right)\right)\right)=\left(\frac{1}{4\sqrt2},-\frac{5\pi}{4}\right)=\frac{1}{4\sqrt2}\cos\left(\frac{3\pi}{4}\right)+\frac{1}{4\sqrt2}i\sin\left(\frac{3\pi}{4}\right)=\frac{1}{4\sqrt2}\left(-\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\right)=-\frac18+\frac18i\)

  2. Let \(A = \begin{pmatrix} 3 & -6 \\ -1 & 2 \end{pmatrix}\). Find a \(2 \times 2\) matrix \(B\) such that \(AB = 0\).

    Suppose \(B=\begin{pmatrix}a & b\\ c & d\end{pmatrix}\), then \(A\cdot B=\begin{pmatrix}3 & -6\\ -1 & 2\end{pmatrix}\cdot\begin{pmatrix}a & b\\ c & d\end{pmatrix}=\begin{pmatrix}3a-6c & 3b-6d\\ -a+2c & -b+2d\end{pmatrix}=\begin{pmatrix}0 & 0\\ 0 & 0\end{pmatrix}\)

    Then we have \(a=2c,b=2d\)

    Therefore, one of \(B\) can be \(\begin{pmatrix}2 & 2\\ 1 & 1\end{pmatrix}\)

  3. Find the inverse of the matrix \(A = \begin{pmatrix} 2 & 6 \\ 3 & 5 \end{pmatrix} \in M_{2 \times 2}(\mathbb{Z}_{11})\).

    Easily, \(B=\begin{pmatrix}-\frac58 & \frac34\\ \frac38 & -\frac14\end{pmatrix}\)

    Since \(8\cdot7\equiv1\) (mod \(11\)) and \(-\frac58\equiv-5\cdot7\equiv-35\equiv9\)

    Since \(8\cdot7\equiv1\) (mod \(11\)) and \(\frac68\equiv6\cdot7\equiv42\equiv9\)

    Since \(8\cdot7\equiv1\) (mod \(11\)) and \(\frac38\equiv3\cdot7\equiv21\equiv10\)

    Since \(8\cdot7\equiv1\) (mod \(11\)) and \(-\frac28\equiv-2\cdot7\equiv-14\equiv8\)

    Thus \(A^{-1}=\begin{pmatrix}9 & 9\\ 10 & 8\end{pmatrix}\)

  4. Find all the roots of \(p(x) = 5x^4 - 15x^2 - 20\) and their multiplicities.

    Since \(5x^4-15x^2-20=5(x^4-3x^2-4)=5(x^2-4)(x^2+1)=5(x+2)(x-2)(x-i)(x+i)\)

    Thus the roots are \(2,-2,i,-i\)

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