031 Review Part 1, Problem 7

True or false: Let  $C=AB$  for  $4\times 4$  matrices  $A$  and  $B.$  If  $C$  is invertible, then  $A$  is invertible.

Solution:
If  $A$  is not invertible, then  ${\text{det }}A=0.$
Since  $C=AB,$  we have
${\begin{array}{rcl}\displaystyle {{\text{det}}C}&=&\displaystyle {{\text{det}}(AB)}\\&&\\&=&\displaystyle {({\text{det}}A)\cdot ({\text{det}}B)}\\&&\\&=&\displaystyle {0\cdot ({\text{det}}B)}\\&&\\&=&\displaystyle {0.}\end{array}}$
Since  ${\text{det }}C=0,$  we know  $C$  is not invertible, which is a contradiction.
So,  $A$  must be invertible and the statement is true.

Final Answer:
TRUE

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