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Chapter 2: Problem 6
If \(A\) is singular, what can you say about the product \(A\) adj \(A ?\)
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A matrix \(A\) is said to be skew symmetric if \(A^{T}=-A .\) For example, \\[ A=\left(\begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array}\right) \\] is skew symmetric, since \\[ A^{T}=\left(\begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array}\right)=-A \\] If \(A\) is an \(n \times n\) skew-symmetric matrix and \(n\) is odd, show that \(A\) must be singular.
Consider the \(3 \times 3\) Vandermonde matrix \\[ V=\left(\begin{array}{lll} 1 & x_{1} & x_{1}^{2} \\ 1 & x_{2} & x_{2}^{2} \\ 1 & x_{3} & x_{3}^{2} \end{array}\right) \\] (a) Show that \(\operatorname{det}(V)=\left(x_{2}-x_{1}\right)\left(x_{3}-x_{1}\right)\left(x_{3}-x_{2}\right)\) [Hint: Make use of row operation III. (b) What conditions must the scalars \(x_{1}, x_{2},\) and \(x_{3}\) satisfy in order for \(V\) to be nonsingular?
Let \(\mathbf{x}, \mathbf{y},\) and \(\mathbf{z}\) be vectors in \(\mathbb{R}^{3} .\) Show each of the following: (a) \(\mathbf{x} \times \mathbf{x}=\mathbf{0}\) (b) \(\mathbf{y} \times \mathbf{x}=-(\mathbf{x} \times \mathbf{y})\) (c) \(\mathbf{x} \times(\mathbf{y}+\mathbf{z})=(\mathbf{x} \times \mathbf{y})+(\mathbf{x} \times \mathbf{z})\) (d) \(\mathbf{z}^{T}(\mathbf{x} \times \mathbf{y})=\left|\begin{array}{lll}x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \\ z_{1} & z_{2} & z_{3}\end{array}\right|\)
Evaluate each of the following determinants by inspection: (a) \(\left|\begin{array}{lll}0 & 0 & 3 \\ 0 & 4 & 1 \\ 2 & 3 & 1\end{array}\right|\) (b) \(\left|\begin{array}{rrrr}1 & 1 & 1 & 3 \\ 0 & 3 & 1 & 1 \\ 0 & 0 & 2 & 2 \\ -1 & -1 & -1 & 2\end{array}\right|\) (c) \(\left|\begin{array}{llll}0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0\end{array}\right|\)
Use Cramer's rule to solve each of the following systems: (a) \(\quad x_{1}+2 x_{2}=3\) (b) \(2 x_{1}+3 x_{2}=2\) \(3 x_{1}-x_{2}=1\) \(3 x_{1}+2 x_{2}=5\) (c) \(2 x_{1}+x_{2}-3 x_{3}=0\) \(4 x_{1}+5 x_{2}+x_{3}=8\) \(-2 x_{1}-x_{2}+4 x_{3}=2\) (d) \(\quad x_{1}+3 x_{2}+x_{3}=1\) \(2 x_{1}+x_{2}+x_{3}=5\) \(-2 x_{1}+2 x_{2}-x_{3}=-8\) (e) \(x_{1}+x_{2}\) \(=0\) \(x_{2}+x_{3}-2 x_{4}=1\) \(x_{1}+2 x_{3}+x_{4}=0\) \(x_{1}+x_{2}+x_{4}=0\)
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