Chapter 8: Problem 31
Prove Theorem 8.7: Suppose \(A\) and \(B\) are similar matrices. Then \(|A|=|B|\).
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Chapter 8: Problem 31
Prove Theorem 8.7: Suppose \(A\) and \(B\) are similar matrices. Then \(|A|=|B|\).
These are the key concepts you need to understand to accurately answer the question.
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Write out \(g=g\left(x_{1}, x_{2}, x_{3}, x_{4}\right)\) explicitly where
\(g\left(x_{1}, x_{2}, \ldots,
x_{n}\right)=\prod_{i
Prove Theorem 8.9: \(A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I\) Let \(A=\left[a_{i j}\right]\) and let \(A(\operatorname{adj} A)=\left[b_{i j}\right] .\) The ith row of \(A\) is $$ \left(a_{i 1}, a_{i 2}, \ldots, a_{i n}\right) $$ Because adj \(A\) is the transpose of the matrix of cofactors, the \(j\) th column of adj \(A\) is the tranpose of the cofactors of the \(j\) th row of \(A\) : $$ \left(A_{j}, A_{j 2}, \ldots, A_{j n}\right)^{T} $$ Now \(b_{i j}\), the ij entry in \(A(\operatorname{adj} A)\), is obtained by multiplying expressions (1) and (2): $$ b_{i j}=a_{i 1} A_{j 1}+a_{i 2} A_{j 2}+\cdots+a_{i n} A_{j n} $$By Theorem \(8.8\) and Problem 8.33, $$ b_{i j}=\left\\{\begin{array}{cl} |A| & \text { if } i=j \\ 0 & \text { if } i \neq j \end{array}\right. $$ Accordingly, \(A(\operatorname{adj} A)\) is the diagonal matrix with each diagonal element \(|A| .\) In other words, \(A(\operatorname{adj} A)=|A| I\). Similarly, \((\operatorname{adj} A) A=|A| I\)
Evaluate: (a) \(\left|\begin{array}{rrrr}2 & -1 & 3 & -4 \\ 2 & 1 & -2 & 1 \\ 3 & 3 & -5 & 4 \\ 5 & 2 & -1 & 4\end{array}\right|\), (b) \(\left|\begin{array}{rrrr}2 & -1 & 4 & -3 \\ -1 & 1 & 0 & 2 \\ 3 & 2 & 3 & -1 \\ 1 & -2 & 2 & -3\end{array}\right|\) (c) \(\left|\begin{array}{rrrr}1 & -2 & 3 & -1 \\ 1 & 1 & -2 & 0 \\ 2 & 0 & 4 & -5 \\ 1 & 4 & 4 & -6\end{array}\right|\)
Show that \(g=g\left(x_{1}, \ldots, x_{n}\right)=(-1)^{n} V_{n-1}(x)\) where \(g=g\left(x_{i}\right)\) is the difference product in Problem 8.19 \(x=x_{n},\) and \(V_{n-1}\) is the Vandermonde determinant defined by
Prove Theorem 8.2.
(i) If \(A\) has a row (column) of zeros, then \(|A|=0\).
(ii) If \(A\) has two identical rows (columns), then \(|A|=0\).
(iii) If \(A\) is triangular, then \(|A|=\) product of diagonal elements. Thus,
\(|I|=1\).
(i) Each term in \(|A|\) contains a factor from every row, and so from the row
of zeros. Thus, each term of \(|A|\) is zero, and so \(|A|=0\)
(ii) Suppose \(1+1 \neq 0\) in \(K\). If we interchange the two identical rows of
\(A\), we still obtain the matrix \(A\). Hence, by Problem \(8.23,|A|=-|A|\), and so
\(|A|=0\)
Now suppose \(1+1=0\) in \(K\). Then sgn \(\sigma=1\) for every \(\sigma \in S_{n}\).
Because \(A\) has two identical rows, we can arrange the terms of \(A\) into pairs
of equal terms. Because each pair is 0 , the determinant of \(A\) is zero.
(iii) Suppose \(A=\left[a_{i j}\right]\) is lower triangular; that is, the
entries above the diagonal are all zero: \(a_{i j}=0\) whenever \(i
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