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• Notes

Given that $\left|\begin{array}{lll}{a} & {b} & {c} \\ {d} & {e} & {f} \\ {g} & {h} & {i}\end{array}\right|=-6,$ find:

(i) $\left|\begin{array}{lll}{d} & {e} & {f} \\ {g} & {h} & {i} \\ {a} & {b} & {c}\end{array}\right|=-\left|\begin{array}{lll}{d} & {e} & {f} \\ {a} & {b} & {c} \\ {g} & {h} & {i}\end{array}\right|$

$=(-)(-)\left|\begin{array}{ll}{a} & {b} & {c} \\ {d} & {e} & {f} \\ {g} & {h} & {i}\end{array}\right|=(-)(-)(-6)=-6$

$(i i)\left|\begin{array}{ccc}{3 a} & {3 b} & {3 c} \\ {-d} & {-e} & {-f} \\ {4 g} & {4 h} & {4 i}\end{array}\right|=3\left|\begin{array}{ccc}{a} & {b} & {c} \\ {-d} & {-e} & {-f} \\ {4 g} & {4 h} & {4 i}\end{array}\right|$

$=(-)(3)(4)\left|\begin{array}{ll}{a} & {b} & {c} \\ {d} & {e} & {f} \\ {g} & {h} & {i}\end{array}\right|=(-)(3)(4)(-6)=72$

Find det(A) if $A=\left[\begin{array}{lll}{2} & {2} & {8} \\ {1} & {2} & {3} \\ {0} & {4} & {8}\end{array}\right]$

$A=\left[\begin{array}{lll}{2} & {2} & {8} \\ {1} & {2} & {3} \\ {0} & {4} & {8}\end{array}\right] R_{1} \leftrightarrow R_{2}$

$|A|=-\left|\begin{array}{lll}{1} & {2} & {3} \\ {2} & {2} & {8} \\ {0} & {4} & {8}\end{array}\right| R_{2} \rightarrow R_{2}-2 R_{1}$

$|A|=-\left|\begin{array}{ccc}{1} & {2} & {3} \\ {0} & {-2} & {2} \\ {0} & {4} & {8}\end{array}\right|_{R_{3} \rightarrow R_{3}+2 R_{2}}$

$|A|=-\left|\begin{array}{ccc}{1} & {2} & {3} \\ {0} & {-2} & {2} \\ {0} & {0} & {12}\end{array}\right|=(-)(1 \times-2 \times 12)=24$

$|A|=(-)\left|\begin{array}{ccc}{1} & {2} & {3} \\ {0} & {-2} & {2} \\ {0} & {4} & {8}\end{array}\right|$

$|A|=(-)(-2)\left|\begin{array}{lll}{1} & {2} & {3} \\ {0} & {1} & {-1} \\ {0} & {4} & {8}\end{array}\right|_{R_{3} \rightarrow R_{3}-4 R_{2}}$

$|A|=(-)(-2)\left|\begin{array}{ccc}{1} & {2} & {3} \\ {0} & {1} & {-1} \\ {0} & {0} & {12}\end{array}\right|$

$=(-)(-2)(1 \times 1 \times 12)=24$

Show that if $\text {det}(A B)=0,$ then $\text {det}(A)=0$ or $\text {det}(b)=0$

$det(A B)=det(A) \cdot det(B)$

$0=det(A) \cdot det(B)$

\$det(A) \text { or } det(B)=0$

Is $det(A B)=det(B A) ?$ Justify your answer

$det(A B)=det(A) \cdot det(B)=det(B) \cdot det(A)=det(B A)$

\$det(A B)=det(B A)$