031 Review Part 2, Problem 5

Let  $A$  and  $B$  be  $6\times 6$  matrices with  ${\text{det }}A=-10$  and  ${\text{det }}B=5.$  Use properties of determinants to compute:

(a)  ${\text{det }}3A$

(b)  ${\text{det }}(A^{T}B^{-1})$

Foundations:
Recall:
1. If the matrix  $B$  is identical to the matrix  $A$  except the entries in one of the rows of  $B$
are each equal to the corresponding entries of  $A$  multiplied by the same scalar  $c,$  then
${\text{det }}B=c({\text{det }}A).$
2.  ${\text{det }}(AB)=({\text{det }}A)({\text{det }}B)$
3. For an invertible matrix  $A,$  since  $AA^{-1}=I$  and  ${\text{det }}I=1,$  we have
${\text{det }}A^{-1}={\frac {1}{{\text{det }}A}}.$

Solution:

(a)

Step 1:
Every entry of the matrix  $3A$  is  $3$  times the corresponding entry of  $A.$
So, we multiply every row of the matrix  $A$  by  $3$  to get  $3A.$

Step 2:
Hence, we have
${\begin{array}{rcl}\displaystyle {{\text{det }}(3A)}&=&\displaystyle {3^{6}({\text{det }}A)}\\&&\\&=&\displaystyle {3^{6}(-10)}\\&&\\&=&\displaystyle {-7290.}\end{array}}$

(b)

Step 1:
Using properties of determinants, we have
${\begin{array}{rcl}\displaystyle {{\text{det }}(A^{T}B^{-1})}&=&\displaystyle {{\text{det }}(A^{T})\cdot {\text{det }}(B^{-1})}\\&&\\&=&\displaystyle {{\text{det }}(A)\cdot {\text{det }}(B^{-1}).}\end{array}}$

Step 2:
Continuing, we obtain
${\begin{array}{rcl}\displaystyle {{\text{det }}(A^{T}B^{-1})}&=&\displaystyle {{\text{det }}(A)\cdot {\frac {1}{{\text{det }}(B)}}}\\&&\\&=&\displaystyle {(-10)\cdot {\frac {1}{5}}}\\&&\\&=&\displaystyle {-2.}\end{array}}$

Final Answer:
(a)     ${\text{det }}(3A)=-7290$
(b)     ${\text{det }}(A^{T}B^{-1})=-2$

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