Let \(K\) be a commutative ring with identity, and let \(D\) be an alternating 2-linear function on \(2 \times 2\) matrices over \(K\). Show that \(D(A) = (\det A)D(I)\) for all \(A\). Now use this result (no computations with the entries allowed) to show that \(\det(AB) = (\det A)(\det B)\) for any \(2 \times 2\) matrices \(A\) and \(B\) over \(K\).
Solution
Let the rows of \(A\) be \(r_1,r_2\), and write \(e_1,e_2\) for the rows of \(I\). Because \(D\) is bilinear, \[ D(A)=D(a_{11}e_1+a_{12}e_2,\,a_{21}e_1+a_{22}e_2). \] Expanding and using alternation gives \[ D(A)=a_{11}a_{22}D(e_1,e_2)+a_{12}a_{21}D(e_2,e_1). \] Alternation also implies \(D(e_2,e_1)=-D(e_1,e_2)\), so \[ D(A)=(a_{11}a_{22}-a_{12}a_{21})D(I)=(\det A)D(I). \]
Now fix \(B\) and define \[ D_B(A)=\det(AB). \] Each row of \(AB\) is obtained by multiplying the corresponding row of \(A\) on the right by \(B\). Therefore \(D_B\) is 2-linear in the rows of \(A\); if the two rows of \(A\) are equal, then the two rows of \(AB\) are equal, so \(D_B\) is alternating. By the result just proved, \[ \det(AB)=D_B(A)=(\det A)D_B(I). \] But \(D_B(I)=\det B\). Hence \[ \det(AB)=(\det A)(\det B), \] without an entrywise computation of \(AB\).