031 Review Part 2, Problem 5

Let  $A$  and  $B$  be  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 6\times 6}   matrices with  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \text{det }A=-10}   and  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \text{det }B=5.}   Use properties of determinants to compute:

(a)  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \text{det }3A}

(b)  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \text{det }(A^TB^{-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:

Step 2:

Final Answer:
(a)     ${\text{det }}(3A)=-7290.$
(b)

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