Chapter 8: Problem 22
Prove Theorem \(8.1:\left|A^{T}\right|=|A|\).
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Chapter 8: Problem 22
Prove Theorem \(8.1:\left|A^{T}\right|=|A|\).
These are the key concepts you need to understand to accurately answer the question.
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Compute the determinant of each of the following matrices: (a) \(A=\left[\begin{array}{lll}2 & 3 & 4 \\ 5 & 6 & 7 \\ 8 & 9 & 1\end{array}\right]\) (b) \(B=\left[\begin{array}{rrrr}4 & -6 & 8 & 9 \\ 0 & -2 & 7 & -3 \\ 0 & 0 & 5 & 6 \\ 0 & 0 & 0 & 3\end{array}\right]\) (c) \(\quad C=\left[\begin{array}{rrr}\frac{1}{2} & -1 & -\frac{1}{3} \\\ \frac{3}{4} & \frac{1}{2} & -1 \\ 1 & -4 & 1\end{array}\right]\)
Find the determinant of \(F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{3}\) defined by \\[ F(x, y, z)=(x+3 y-4 z, 2 y+7 z, x+5 y-3 z) \\]
Prove (a) \(\operatorname{det}\left(\mathbf{1}_{V}\right)=1\), where \(\mathbf{1}_{V}\) is the identity operator, (b) \(\operatorname{det}\left(T^{-1}\right)=\operatorname{det}(T)^{-1}\) when \(T\) is invertible.
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, \((\operatorname{adj} A) A=|A| I\)
Prove Theorem 8.13: Let \(F\) and \(G\) be linear operators on a vector space \(V\). Then (i) \(\operatorname{det}(F \circ G)=\operatorname{det}(F) \operatorname{det}(G)\), (ii) \(F\) is invertible if and only if \(\operatorname{det}(F) \neq 0\)
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