MAT223 Lecture 09

• 4
  • If $A$ and $B$ have $detA=-1$
  • $detB=3$
  • $det3A^2(B^t{)}^{-1}=1$
  • $det3A^{2}det(B^t{)}^{-1}=1$
  • $det3A^{2}det(B{)}^{-1}=1$
  • $det3A^2\frac{1}{detB}=1$
  • $3^{n}det{A}^2\frac{1}{detB}=1$
    • You can think of it as:
    • $3I_{n}A$
    • $\left[\begin{array}{ccc}3& 0& 0\\ 0& 3& 0\\ 0& 0& 3\end{array}\right]\left[\begin{array}{c}\dots \end{array}\right]$
  • $3^{n}detAdetA\frac{1}{detB}=1$
  • $3^n(-1)(-1)\frac{1}{3}=1$
  • $3^n\frac{1}{3}=1$
  • $3^n{3}^{-1}=1$
  • $3^{n-1}=1$
• 5
  • There exist $n\times n$ matrices $A,B$ so that $AB$ is invertible. But neither $A$ nor $B$ are invertible.
  • $detAB\ne 0$
  • $detA=0$
  • $detB=0$
  • $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$
  • $ad-bc=0$
  • $B=\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]$
  • $xw-yz=0$
  • $AB=\left[\begin{array}{cc}ax+bz& ay+bw\\ cx+dz& cy+dw\end{array}\right]$
  • …
  • If $detA=0$ and $detB=0⟹detAB=0$
  • So it's not possible.
• Week 5 stuff now.
  • The matrix $\left[\begin{array}{cc}2& 2\\ 1& 3\end{array}\right]$ has eigenvector $\left[\begin{array}{c}3\\ 3\end{array}\right]$
  • The eigenvalue is:
    • $A\overrightarrow{v}=\lambda \overrightarrow{v}$
    • $\left[\begin{array}{cc}2& 2\\ 1& 3\end{array}\right]\left[\begin{array}{c}3\\ 3\end{array}\right]=\lambda \left[\begin{array}{c}3\\ 3\end{array}\right]$
    • $\left[\begin{array}{c}12\\ 12\end{array}\right]=\lambda \left[\begin{array}{c}3\\ 3\end{array}\right]$
    • $12=3\lambda$
    • $\lambda =4$
• 2
  • $A=\left[\begin{array}{ccc}0& 0& 1\\ 0& 3& 4\\ 0& 0& -7\end{array}\right]$
  • $c_A(x)=det(x{I}_n-A)$
  • $c_A(x)=det(\left[\begin{array}{}\end{array}\right])$
• 3
  • $A=\left[\begin{array}{cc}2& -4\\ -1& -1\end{array}\right]$
  • $2$ eigenvalues
  • $c_A(x)=det(\left[\begin{array}{cc}x& 0\\ 0& x\end{array}\right]-\left[\begin{array}{cc}2& -4\\ -1& -1\end{array}\right])$
  • $c_A(x)=det\left[\begin{array}{cc}x-2& 4\\ 1& x+1\end{array}\right]$
  • $c_A(x)=(x-2)(x+1)-(4)(1)$
  • $c_A(x)=(x^2+x-2x-2)-4$
  • $c_A(x)=x^2+x-2x-2-4$
  • $c_A(x)=x^2-x-6=(x-3)(x+2)$
  • $p=-6$
  • $s=-1$
  • $f=2,-3$
  • $\lambda =-2,3$
• 4
  • Is $A=\left[\begin{array}{cc}-4& 2\\ -6& 3\end{array}\right]$ diagonalizble
  • $c_A(x)=det[\left[\begin{array}{cc}x& 0\\ 0& x\end{array}\right]-\left[\begin{array}{cc}-4& 2\\ -6& 3\end{array}\right]]$
  • $c_A(x)=det\left[\begin{array}{cc}x+4& -2\\ 6& x-3\end{array}\right]$
  • $c_A(x)=(x+4)(x-3)-(6)(-2)$
  • $c_A(x)=(x^2-3x+4x-12)+12$
  • $c_A(x)=x^2-3x+4x$
  • $c_A(x)=x^2+x=x(x+1)$
  • $x=0$
  • $x=-1$
  • If there is an Eigenvalue, there is always a Basic Eigenvector for that value.
  • $A$ is $n\times n$ and is diagonalizable if $A$ has $n$ Basic Eigenvectors.
  • If $\lambda$ is and Eigenvalue, you have at least on Basic Eigenvector for $\lambda$.
  • If $A$ is $n\times n$
  • We have $0\le x\le n$ where $x$ is the number of basic eigenvectors.