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