An important property that the determinant satisfies is the following: \[\det(AB) = \det(A)\det(B)\] where \(A\) and \(B\) are \(n \times n\) matrices. A immediate and useful consequence is \[\det(A^k) = \det(A)^k\] for all natural numbers \(k\).
Let \(A, B\in \mathbb{R}^{3\times 3}\) be such that \(\det(A) = 3\) and \(\det(B) = -2\). Then \(\det(AB) = \det(A)\det(B) = -6\).
Let \(A = \begin{bmatrix} -2 & 1 \\ -1 & 1\end{bmatrix}\). Then \(\det(A^{100}) = \det(A)^{100} = (-1)^{100} = 1.\)
We now prove that if \(A\) and \(B\) are \(2\times 2\) matrices then \(\det(AB) = \det(A)\det(B)\). Proving the result for the general case is somewhat challenging and is left as an exercise.
Suppose that \(A = \begin{bmatrix} a & b\\ c & d \end{bmatrix}\) and \(B = \begin{bmatrix} s & t \\ u & v \end{bmatrix}\). Then \(AB = \begin{bmatrix} as + bu & at + bv \\ cs + du & ct + dv \end{bmatrix}\). Hence, \begin{eqnarray*} \det(AB) - \det(A)\det(B)& = & \begin{vmatrix} as + bu & at + bv \\ cs + du & ct + dv \end{vmatrix} - \begin{vmatrix} a & b \\ c & d \end{vmatrix} \begin{vmatrix} s & t \\ u & v \end{vmatrix} \\ & = & (as+bu)(ct+dv) - (cs + du)(at+bv) - (ad-bc)(sv - ut) \\ & = & (acst + adsv+bctu + bduv) - (acst + bcsv + adtu + bduv) - (adsv - adtu - bcsv + bctu) \\ & = & acst + adsv + bctu + bduv - acst - bcsv - adtu - bduv - adsv + adtu + bcsv - bctu \\ & = & 0. \end{eqnarray*} Hence, \(\det(AB) = \det(A)\det(B)\).
Compute the determinants of each of the following products:
\(\begin{bmatrix} 2 & 3 \\ 1 & 2\end{bmatrix} \begin{bmatrix} 3 & -1 \\ 0 & 1\end{bmatrix}\)
\(\begin{bmatrix} 1 & 0 & i \\ 0 & 2 & 1 \\ i & 1 &-1\end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{bmatrix}\)
Let \(A = \begin{bmatrix} 1 & 2-i \\ 0 & 1+i\end{bmatrix}\). Give the value of \(\det(A^{16})\).
Let \(A\) and \(B\) be \(3\times 3\) matrices. Prove that \(\det(AB) = \det(A)\det(B)\).
Let \(A\) and \(B\) be \(n\times n\) matrices such that \(B\) is obtained from \(A\) by adding a constant multiple of a row of \(A\) to another row of \(A\). Prove that \(\det(B) = \det(A)\). (Hint: Express \(B\) as \(MA\) where \(M\) is an elementary matrix representing the row operation performed to obtain \(B\). What must \(\det(M)\) be?)
Let \(A\) and \(B\) be \(n\times n\) matrices such that \(B\) is obtained from \(A\) by interchanging two rows of \(A\). Prove that \(\det(B) = -\det(A)\).