# Finding the determinant of a matrix

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?

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

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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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I am aware of this,but I can't see how these is going to help me here :( –  Quixotic May 12 '11 at 15:37
By the first fact, $\det(BA)=\det(B)\det(A)$. Because $BA=I$, we have... do you see the rest? –  Zev Chonoles May 12 '11 at 15:41
Yes,but the answer is $\frac{d}{e}$ ?! –  Quixotic May 12 '11 at 15:44
No... you want an answer that depends on $a,b,c,$ and $d$. Putting our three facts together, $$1=\det(I)=\det(BA)=\det(B)\det(A)=\det(B)\cdot(ad-bc)$$ Now divide. –  Zev Chonoles May 12 '11 at 15:45
@Zev Chonoles:I think you don't get me,the answer suggested in my modules is $\frac{d}{e}$ how to reduce to that? –  Quixotic May 12 '11 at 15:52

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 $\rm\ d\not\equiv 0\:.$

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