Linear Algebra I: Homework 3

Due: Friday, September 8, 2017
  1. Find the inverse \(A^{-1}\) of the matrix \(A\):

    \[A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8 \end{pmatrix}\]

    We can solve for \(A^{-1}\) using Gaussian elimination on the augmented matrix \((A \mid I)\) to get \((I \mid A^{-1})\):

    \[\left( \begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 2 & 5 & 3 & 0 & 1 & 0 \\ 1 & 0 & 8 & 0 & 0 & 1 \\ \end{array} \right) \\ \stackrel{(2)=(2)-2(1) \\ (3)=(3)-(1)}{\sim} \left( \begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & -3 & -2 & 1 & 0 \\ 0 & -2 & 5 & -1 & 0 & 1 \\ \end{array} \right) \\ \stackrel{(3)=(3)+2(2)}{\sim} \left( \begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & -3 & -2 & 1 & 0 \\ 0 & 0 & -1 & -5 & 2 & 1 \\ \end{array} \right) \\ \stackrel{(3)=-(3)}{\sim} \left( \begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & -3 & -2 & 1 & 0 \\ 0 & 0 & 1 & 5 & -2 & -1 \\ \end{array} \right) \\ \stackrel{(2)=(2)+3(3) \\ (1)=(1)-3(3)}{\sim} \left( \begin{array}{ccc|ccc} 1 & 2 & 0 & -14 & 6 & 3 \\ 0 & 1 & 0 & 13 & -5 & -3 \\ 0 & 0 & 1 & 5 & -2 & -1 \\ \end{array} \right) \\ \stackrel{(1)=(1)-2(2)}{\sim} \left( \begin{array}{ccc|ccc} 1 & 0 & 0 & -40 & 16 & 9 \\ 0 & 1 & 0 & 13 & -5 & -3 \\ 0 & 0 & 1 & 5 & -2 & -1 \\ \end{array} \right)\]

    So,

    \[A^{-1} = \begin{pmatrix} -40&16&9\\13&-5&-3\\5&-2&-1 \end{pmatrix}.\]
  2. Let \(\mathbf{0}\) be the \(2 \times 2\) matrix with all zero entries.

    1. Is there a matrix \(A \ne \mathbf 0\) for which \(AA = \mathbf 0\)? Justify your answer.

    Yes, there are many examples. These matrices are called nilpotent matrices. For example,

    \[A = \begin{pmatrix}0&1\\0&0\end{pmatrix}.\]
    1. Is there a matrix \(B \ne \mathbf 0\) and \(B \ne I\) for which \(BB = B\)? Justify your answer.

    Yes, there are many examples. These matrices are called idempotent matrices. For example,

    \[B = \begin{pmatrix}.5&.5\\.5&.5\end{pmatrix}.\]
  3. Find the inverse \(R_\theta^{-1}\) of the matrix \(R_\theta\):

    \[R_\theta = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}\]

    \(R_\theta\) is a \(2 \times 2\) matrix, so we can just use the formula for the inverse.

    \[R_\theta^{-1} = \frac{1}{\cos^2\theta + \sin^2\theta} \begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix} = \begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}.\]