MAT223 Lecture 08

• 4
  • $det\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 2\\ 1& 1& -2\end{array}\right]$
  • $=(-1{)}^{1+1}det\left[\begin{array}{cc}1& 2\\ 1& -2\end{array}\right]+(-1{)}^{1+3}det\left[\begin{array}{cc}0& 1\\ 1& 2\end{array}\right]$
  • $=(-1{)}^{1+1}[(1)(-2)-(1)(2)]+(-1{)}^{1+3}[(0)(2)-(1)(1)]=-5.0$
• 5
  • Compute
  • $det\left[\begin{array}{cccc}2& 100& -3& 1\\ 0& -1& 23& 14\\ 0& 0& -3& 0\\ 0& 0& 0& 4\end{array}\right]$
  • $=(-1{)}^{4+4}(4)det\left[\begin{array}{ccc}2& 100& -3\\ 0& -1& 23\\ 0& 0& -3\end{array}\right]$
  • $=(-1{)}^{4+4}(4)[(-1{)}^{3+3}(-3)det\left[\begin{array}{cc}2& 100\\ 0& -1\end{array}\right]]$
  • $=(-1{)}^{4+4}(4)[(-1{)}^{3+3}(-3)[(2)(-1)-(0)(100)]]=24.0$
  • Easier method is Upper-Triangular Matrix
  • See that the determinant is the product of diagonal things.
    • $(2)(-1)(-3)(4)=24.0$
  • Transpose does not affect determinant.
• 6
  • If $detA=-2$ and $detB=6$
  • Then $det{A}^{-1}{B}^T=$
  • $det{A}^{-1}\cdot det{B}^T$
  • $\frac{1}{detA}\cdot detB$
  • $\frac{1}{-2}\cdot 6=-3.0$
• Reduce the matrix and then take the determinant to make it easier.
• Theorem 1:
  • Let $A$ be an $n\times n$ matrix:
  • If $A$ has a row of $0$s, then $detA=0$
  • If we exchange two rows of $A$, the determinant of the resulting matrix is $-detA$
  • If we multiply an entire row of $A$ by $k$, the determinant is $kdetA$
  • If you add a multiple of one row to another, the determinant is unchanged.
• Activity 2
  • 1
    • Let $A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$
    • $B$ is the matrix obtained by performing
      • Scale $R_1$ by $2$
      • Swap row 1 and 2
    • $B=\left[\begin{array}{cc}2& 4\\ 3& 4\end{array}\right]$
    • $B=\left[\begin{array}{cc}3& 4\\ 2& 4\end{array}\right]$
    • Verify $detB=-2detA$
    • $detA=(1)(4)-(2)(3)=-2.0$
    • $detB=(3)(4)-(4)(2)=4.0$
    • $4=(-2)(-2)$
    • True
  • 2
    • Let $A=\left[\begin{array}{ccc}1& 2& 3\\ 1& -1& 2\\ 1& 0& 5\end{array}\right]$
    • Transform into an Upper-Triangular Matrix only using adding and subtracting multiples of rows
      • $R_2=R_2-R_1$
      • $\left[\begin{array}{ccc}1& 2& 3\\ 0& -3& -1\\ 1& 0& 5\end{array}\right]$
      • $R_3=R_3-R_1$
      • $\left[\begin{array}{ccc}1& 2& 3\\ 0& -3& -1\\ 0& -2& 2\end{array}\right]$
      • $R_3=R_3-\frac{2}{3}{R}_2$
      • $\left[\begin{array}{ccc}1& 2& 3\\ 0& -3& -1\\ 0& 0& 2+\frac{2}{3}\end{array}\right]$
      • $\left[\begin{array}{ccc}1& 2& 3\\ 0& -3& -1\\ 0& 0& \frac{6}{3}+\frac{2}{3}\end{array}\right]$
      • $\left[\begin{array}{ccc}1& 2& 3\\ 0& -3& -1\\ 0& 0& \frac{8}{3}\end{array}\right]$
      • $(1)(-3)\left(\frac{8}{3}\right)=-8.0$
    • Transform into an Upper-Triangular Matrix using any row operation.
      • $A=\left[\begin{array}{ccc}1& 2& 3\\ 1& -1& 2\\ 1& 0& 5\end{array}\right]$
      • $R_2=R_2-R_1$
      • $A=\left[\begin{array}{ccc}1& 2& 3\\ 0& -3& -1\\ 1& 0& 5\end{array}\right]$
      • $R_3=R_3-R_1$
      • $A=\left[\begin{array}{ccc}1& 2& 3\\ 0& -3& -1\\ 0& -2& 2\end{array}\right]$
      • $R_3=\frac{R_3}{-2}$
      • $\frac{A}{-2}=\left[\begin{array}{ccc}1& 2& 3\\ 0& -3& -1\\ 0& 1& -1\end{array}\right]$
      • $R_3=3R_3$
      • $\left(\frac{3}{-2}\right)A=\left[\begin{array}{ccc}1& 2& 3\\ 0& -3& -1\\ 0& 3& -3\end{array}\right]$
      • $R_3=R_3+R_2$
      • $\left(\frac{3}{-2}\right)detA=\left[\begin{array}{ccc}1& 2& 3\\ 0& -3& -1\\ 0& 0& -4\end{array}\right]$
      • $\left(\frac{3}{-2}\right)detA=(1)(-3)(-4)=12.0$
      • $\frac{12}{\frac{3}{-2}}=-8.0$
  • 3
    • Find the $detA=\left[\begin{array}{cccc}1& 2& -1& 3\\ 2& 6& -1& 6\\ 1& 2& -1& 4\\ -3& -6& 3& -7\end{array}\right]$
    • $detA=\left[\begin{array}{cccc}1& 2& -1& 3\\ 2& 6& -1& 6\\ 1& 2& -1& 4\\ -3& -6& 3& -7\end{array}\right]$
    • $R_2=R_2-2R_1$
    • $detA=\left[\begin{array}{cccc}1& 2& -1& 3\\ 0& 2& 1& 0\\ 1& 2& -1& 4\\ -3& -6& 3& -7\end{array}\right]$
    • $R_3=R_3-R_1$
    • $detA=\left[\begin{array}{cccc}1& 2& -1& 3\\ 0& 2& 1& 0\\ 0& 0& 0& 1\\ -3& -6& 3& -7\end{array}\right]$
    • $R_4=R_4+3R_1$
    • $detA=\left[\begin{array}{cccc}1& 2& -1& 3\\ 0& 2& 1& 0\\ 0& 0& 0& 1\\ 0& 0& 0& 2\end{array}\right]$
    • Upper-Triangular Matrix
    • $detA=(1)(2)(0)(2)=0$
  • 4
    • Come up with a strategy for computing the determinant of a $5\times 5$ matrix in a quick way.
      • Aim to move to Upper-Triangular Matrix
      • If you see you can get a row of $0$s quickly, then you can do that.
• If $A,B$ are invertible $n\times n$ matrices, which of the following other matrices must be invertible.
  • Invertible
    • $A^{-1}$
      • Inverse is invertible
    • $A^t$
    • $BA$
  • Not invertible
    • $A+B$
      • $A=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
      • $B=-A=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]$
      • $A+B=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
      • Not invertible
  • Determinant properties can tell us this:
    • Invertible $⟺detA\ne 0$
    • $BA$
      • $detBA$
      • $detBdetA$
      • Two nonzero can't give you a 0
    • $A^{-1}$
      • $det{A}^{-1}=\frac{1}{detA}$ which is not $0$
    • $A^t$
      • $det{A}^t=detA\ne 0$
• Activity 3
  • Assume that $A$ and $B$ are $n\times n$
  • If $A$ has two equal rows, then $A$ is not invertible.
    • True, without loss of generality suppose $R_1=R_2$
    • We can then do $R_2=R_2-R_1$
    • Resulting in $R_2=\{0,0,0,\dots ,0\}$
    • A row of $0$s results in the determinant being $0$ by theorem.
    • Invertibility requires that $detA\ne 0$
  • $detA+B=detA+detB$
    • False, this is not a property of determinants.
    • Counterexample:
      • Let $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$
      • $B=\left[\begin{array}{cc}e& f\\ g& h\end{array}\right]$
      • $A+B=\left[\begin{array}{cc}a+e& b+f\\ g+c& d+h\end{array}\right]$
      • $det(A+B)=[(a+e)(d+h)-(b+f)(g+c)]=ad+ah-bc-bg-cf+ed-fg+eh$
      • $detA=(a)(d)-(b)(c)$
      • $detB=(e)(h)-(f)(g)$
      • $ad+ah-bc-bg-cf+ed-fg+eh\ne ad-bc+eh-fg$
• 
• 3
  • If $A\overrightarrow{x}=\overrightarrow{b}$ has infinitely many solutions for some $\overrightarrow{b}$, then $A$ is not invertible.
    • False, counterexample:
      • $\overrightarrow{x}=\left[\begin{array}{c}0\\ 0\\ ⋮\\ 0\end{array}\right]$
      • So for any $A,\overrightarrow{b}$ we have infinite solution.
  • If $A$ and $B$ have $detA=-1,detB=3$
    • $det(3A^2(B^t{)}^{-1})=1$
    • $det(3A^2)det(B^t{)}^{-1}$
    • $det(3A^2)det(B{)}^{-1}$
    • $det(3A^2)\frac{1}{detB}$
    • $det(3(AA))\frac{1}{detB}$
    • $3^{n}det(AA)\frac{1}{detB}$
    • $3^{n}detAdetA\frac{1}{detB}$
    • $3^n(-1)(-1)\frac{1}{3}=3^{n-1}$