5.2 Properties of Determinants

1. Compute the determinant by using row operations to put the matrix in to echelon form. a. $A=\left[\begin{array}{cc}2& 4\\ 1& -4\end{array}\right]$ $\left[\begin{array}{cc}2& 4\\ 1& -4\end{array}\right]\stackrel{R_2\to R_2-\frac{1}{2}{R}_1}{\to }\left[\begin{array}{cc}2& 4\\ 0& -6\end{array}\right]$ $\det (A)=(2)(-6)=-12$ b. $A=\left[\begin{array}{ccc}0& 1& -2\\ 1& 2& 4\\ 0& 2& 2\end{array}\right]$ $\left[\begin{array}{ccc}0& 1& -2\\ 1& 2& 4\\ 0& 2& 2\end{array}\right]\stackrel{R_1\leftrightarrow R_2}{\to }\left[\begin{array}{ccc}1& 2& 4\\ 0& 1& -2\\ 0& 2& 2\end{array}\right]\stackrel{R_3\to R_3-2R_2}{\to }\left[\begin{array}{ccc}1& 2& 4\\ 0& 1& -2\\ 0& 0& 6\end{array}\right]$ $\det (A)=(1)(1)(6)(-1)=-6$ (Note the -1 because of the row swap) c. $A=\left[\begin{array}{cccc}1& 1& 2& 1\\ 2& 2& 1& -2\\ 0& 1& 1& 2\\ -3& 3& 2& 0\end{array}\right]$ $\left[\begin{array}{cccc}1& 1& 2& 1\\ 2& 2& 1& -2\\ 0& 1& 1& 2\\ -3& 3& 2& 0\end{array}\right]\stackrel{R_2\to R_2-2R_1}{\to }\left[\begin{array}{cccc}1& 1& 2& 1\\ 0& 0& -3& -4\\ 0& 1& 1& 2\\ -3& 3& 2& 0\end{array}\right]\stackrel{R_4\to R_4+3R_1}{\to }\left[\begin{array}{cccc}1& 1& 2& 1\\ 0& 0& -3& -4\\ 0& 1& 1& 2\\ 0& 6& 8& 3\end{array}\right]\stackrel{R_2\leftrightarrow R_3}{\to }\left[\begin{array}{cccc}1& 1& 2& 1\\ 0& 1& 1& 2\\ 0& 0& -3& -4\\ 0& 6& 8& 3\end{array}\right]\stackrel{R_4\to R_4-6R_2}{\to }\left[\begin{array}{cccc}1& 1& 2& 1\\ 0& 1& 1& 2\\ 0& 0& -3& -4\\ 0& 0& 2& -9\end{array}\right]\stackrel{R_4\to R_4+\frac{2}{3}{R}_3}{\to }\left[\begin{array}{cccc}1& 1& 2& 1\\ 0& 1& 1& 2\\ 0& 0& -3& -4\\ 0& 0& 0& -\frac{35}{3}\end{array}\right]$ $\det (A)=(1)(1)(-3)(-35/3)(-1)=-35$ (Note the -1 because of the row swap) 2. Verify Theorem 5.13 for $A=\left[\begin{array}{cc}1& -3\\ 2& -5\end{array}\right]$. $\det (A)=(1)(-5)-(-3)(2)=-5+6=1$ a. $R_1\leftrightarrow R_2$ $B=\left[\begin{array}{cc}2& -5\\ 1& -3\end{array}\right]$ $\det (B)=(2)(-3)-(-5)(1)=-6+5=-1$ $\det (A)=-\det (B)$ b. $R_1\to 3R_1$ $B=\left[\begin{array}{cc}3& -9\\ 2& -5\end{array}\right]$ $\det (B)=(3)(-5)-(-9)(2)=-15+18=3$ $\det (A)=\frac{1}{3}\det (B)$ c. $R_2\to R_2-2R_1$ $B=\left[\begin{array}{cc}1& -3\\ 0& -1\end{array}\right]$ $\det (B)=(1)(-1)-(-3)(0)=-1$ $\det (A)=\det (B)$ 3. Compute the determinant of $A=\left[\begin{array}{ccccc}1& 1& 1& 1& 1\\ 1& 2& 2& 2& 2\\ 1& 2& 3& 3& 3\\ 1& 2& 3& 4& 4\\ 1& 2& 3& 4& 7\end{array}\right]$ $\left[\begin{array}{ccccc}1& 1& 1& 1& 1\\ 1& 2& 2& 2& 2\\ 1& 2& 3& 3& 3\\ 1& 2& 3& 4& 4\\ 1& 2& 3& 4& 7\end{array}\right]\stackrel{R_i\to R_i-R_1,i=2,3,4,5}{\to }\left[\begin{array}{ccccc}1& 1& 1& 1& 1\\ 0& 1& 1& 1& 1\\ 0& 1& 2& 2& 2\\ 0& 1& 2& 3& 3\\ 0& 1& 2& 3& 6\end{array}\right]\stackrel{R_i\to R_i-R_2,i=3,4,5}{\to }\left[\begin{array}{ccccc}1& 1& 1& 1& 1\\ 0& 1& 1& 1& 1\\ 0& 0& 1& 1& 1\\ 0& 0& 1& 2& 2\\ 0& 0& 1& 2& 5\end{array}\right]\stackrel{R_i\to R_i-R_3,i=4,5}{\to }\left[\begin{array}{ccccc}1& 1& 1& 1& 1\\ 0& 1& 1& 1& 1\\ 0& 0& 1& 1& 1\\ 0& 0& 0& 1& 1\\ 0& 0& 0& 1& 4\end{array}\right]\stackrel{R_5\to R_5-R_4}{\to }\left[\begin{array}{ccccc}1& 1& 1& 1& 1\\ 0& 1& 1& 1& 1\\ 0& 0& 1& 1& 1\\ 0& 0& 0& 1& 1\\ 0& 0& 0& 0& 3\end{array}\right]$ $\det (A)=(1)(1)(1)(1)(3)=3$ 4. Given $\det (A)=3$ and $\det (B)=-2$. a. $\det (AB)=\det (A)\det (B)=(3)(-2)=-6$ b. $\det (B^{-1}A)=\det (B^{-1})\det (A)=\frac{1}{\det (B)}\det (A)=\frac{1}{-2}(3)=-\frac{3}{2}$ c. $\det (B^{3}A{B}^{-3})=\det (B^3)\det (A)\det (B^{-3})=(\det (B){)}^3\det (A)(\frac{1}{(\det (B){)}^3})=\det (A)=3$ d. $\det ((AB{)}^T{A}^{-1})=\det (AB{)}^T\det (A^{-1})=\det (AB)\det (A^{-1})=\det (A)\det (B)\frac{1}{\det (A)}=\det (B)=-2$