Chapter 5.3, Problem 41PS

To determine

(a) To determine:

Guess which $2\text{ by 2}$ determinants to use from $A\text{ and }B$.

   $\det (AB)=\left(\begin{array}{cc}a& b\\ d& e\end{array}\right)\left(\begin{array}{cc}g& j\\ h& k\end{array}\right)+\left(\begin{array}{cc}b& c\\ e& f\end{array}\right)\left(\begin{array}{cc}h& k\\ i& l\end{array}\right)+\left(\begin{array}{cc}a& c\\ d& f\end{array}\right)\left(\begin{array}{cc}g& j\\ i& l\end{array}\right)$

Given information:

Determinant $A$is $2\times 3$matrix and determinant $B$is $3\times 2$

Concept used:

   ${(\det AB)}_{m\times n}={(\det A)}_{m\times l}{(\det B)}_{l\times n}$

The $2\text{ by 2}$ matrix $AB=\left(2\text{ by 3}\right)\left(3\text{ by 2}\right)$ has a "Cauchy-Benet formula" for $\det AB$.

Calculation:

Let us consider the following matrices

   $\begin{array}{l}A=\left[\begin{array}{l}\begin{array}{ccc}a& b& c\end{array}\\ \begin{array}{ccc}d& e& f\end{array}\end{array}\right]\\ B=\left[\begin{array}{l}\begin{array}{cc}g& j\end{array}\\ \begin{array}{cc}h& k\end{array}\\ \begin{array}{cc}i& l\end{array}\end{array}\right]\end{array}$

According to Cauchy-Binet formula:

   $\det (AB)=\left(\begin{array}{cc}a& b\\ d& e\end{array}\right)\left(\begin{array}{cc}g& j\\ h& k\end{array}\right)+\left(\begin{array}{cc}b& c\\ e& f\end{array}\right)\left(\begin{array}{cc}h& k\\ i& l\end{array}\right)+\left(\begin{array}{cc}a& c\\ d& f\end{array}\right)\left(\begin{array}{cc}g& j\\ i& l\end{array}\right)$

Conclusion:

   $\det (AB)=\left(\begin{array}{cc}a& b\\ d& e\end{array}\right)\left(\begin{array}{cc}g& j\\ h& k\end{array}\right)+\left(\begin{array}{cc}b& c\\ e& f\end{array}\right)\left(\begin{array}{cc}h& k\\ i& l\end{array}\right)+\left(\begin{array}{cc}a& c\\ d& f\end{array}\right)\left(\begin{array}{cc}g& j\\ i& l\end{array}\right)$

To determine

(b) To Check:

Test your formula when the rows of $A$ are $1,2,3\text{ and 1,4,7}$ with $B=A^T$.

Given information:

   $\begin{array}{l}A=\left[\begin{array}{l}\begin{array}{ccc}1& 2& 3\end{array}\\ \begin{array}{ccc}1& 4& 7\end{array}\end{array}\right]\\ B=A^T\end{array}$

Concept used:

The 2 by 2 matrix AB = (2 by 3)(3 by 2) has a "Cauchy-Binet formula" for det AB.

Calculation:

The $2$by $2$matrix $AB=\left(2\text{by}\text{3}\right)\left(3\text{ by 2}\right)$ has a "Cauchy-Benet formula" for $\det AB$.

Let us check the above statements with an example.

   $\begin{array}{l}A=\left[\begin{array}{l}\begin{array}{ccc}1& 2& 3\end{array}\\ \begin{array}{ccc}1& 4& 7\end{array}\end{array}\right]B=\left[\begin{array}{l}\begin{array}{cc}1& 1\\ 2& 4\end{array}\\ \begin{array}{cc}3& 7\end{array}\end{array}\right]\left(B=A^T\right)\\ AB=\left[\begin{array}{cc}14& 30\\ 30& 66\end{array}\right]\end{array}$

According to Cauchy-Benet formula,

   $\begin{array}{c}\det (AB)=\left(\begin{array}{cc}1& 2\\ 1& 4\end{array}\right)\left(\begin{array}{cc}1& 1\\ 2& 4\end{array}\right)+\left(\begin{array}{cc}2& 3\\ 4& 7\end{array}\right)\left(\begin{array}{cc}2& 4\\ 3& 7\end{array}\right)+\left(\begin{array}{cc}1& 3\\ 1& 7\end{array}\right)\left(\begin{array}{cc}1& 1\\ 3& 7\end{array}\right)\\ =(4-2)(4-2)+(14-12)(14-12)+(7-3)(7-3)\\ =2\times 2+2\times 2+4\times 4\\ =4+4+16\\ =24\end{array}$

Now again find the determinant of

   $AB=\left(\begin{array}{cc}14& 30\\ 30& 66\end{array}\right)$

   $\begin{array}{c}\det (AB)=14\times 66-30\times 30\\ =924-900\\ =24\end{array}$

Conclusion:

Thus, the $\det \text{of}AB$ is equal to the sum of ( $2\text{by}\text{2}$ determinants in $A$ ) & ( $2\text{by}\text{2}$

$2$ determinants in $B$ )