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I am given $A = \begin{pmatrix} a & b\\ c & d \end{pmatrix} $ and B = $ \begin{pmatrix} e & f\\ g & h \end{pmatrix}$ whose elements are non-zero reals.

If $BA = I$, where $I$ is the $2 \times 2$ identity matrix and $D$ is the value of the determinant of $B$, then find the value of $D$

Assume that four options are given for the correct answer (which is $\frac{d}{e}$) and only one is correct. How can I find the correct answer quickly?

ADDED:

The answer suggested in my module is $\frac{d}{e}$, so I am suppose to derive to that point.

2 Answers 2

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The determinant of $B$ is given by $eh-gf$ (see here). Presumably, the question is asking you to describe this quantity in terms of $a,b,c,d$. In that case, here are some facts you can combine to find the answer:

  • $\det(XY)=\det(X)\det(Y)$ for all matrices $X$ and $Y$

  • $\det(I)=1$

  • $\det(A)=ad-bc$

These facts together show that $D=\det(B)=\frac{1}{ad-bc}$. Now note that the entries of $B=A^{-1}$, in terms of the entries of $A$, are (see here) $B=\begin{pmatrix}\frac{d}{ad-bc}\,\,\,\, & \frac{-b}{ad-bc}\\ \stackrel{\vphantom{g}}{\frac{-a}{ad-bc}}\,\,\,\, & \frac{c}{ad-bc}\end{pmatrix}$ Thus, $\displaystyle e=\frac{d}{ad-bc}$, so that $\displaystyle\frac{d}{e}=ad-bc$, which is wrong. The correct answer is $\displaystyle\frac{e}{d}=\frac{1}{ad-bc}$.

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    You can't "reduce to" $d/e$ because that answer is wrong. Try an example: A = \left(\matrix{2 & 1\cr 1 & 3\cr}\right), B = \left(\matrix{3/5 & -1/5\cr -1/5 & 2/5\cr}\right), $D = 1/5 = e/d \ne d/e$. Or was $D$ supposed to be the determinant of $A$?2011-05-12
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HINT $\: $ Multiplicative maps $\rm\:d\:$ preserve products so inverses $\rm\ A\:B = 1\ \Rightarrow\ d(A)\:d(B) = d(1) = 1\:.\ $

Note: $\rm\ 1^2 = 1\ \Rightarrow\ d(1)^2 = d(1)\ $ so $\rm\ d(1) = 1\ $ if the target is a domain and $\rm\ d\not\equiv 0\:.$