Chapter 3: Problem 29
Find the area of the triangle with the given vertices. $$(0,0),(2,0),(0,3)$$
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Chapter 3: Problem 29
Find the area of the triangle with the given vertices. $$(0,0),(2,0),(0,3)$$
These are the key concepts you need to understand to accurately answer the question.
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Prove that the determinant of an invertible matrix \(A\) is equal to \(\pm 1\) when all of the entries of \(A\) and \(A^{-1}\) are integers. Getting Started: Denote det( \(A\) ) as \(x\) and \(\operatorname{det}\left(A^{-1}\right)\) as \(y\) Note that \(x\) and \(y\) are real numbers. To prove that \(\operatorname{det}(A)\) is equal to \(\pm 1,\) you must show that both \(x\) and \(y\) are integers such that their product \(x y\) is equal to 1 (i) Use the property for the determinant of a matrix product to show that \(x y=1\) (ii) Use the definition of a determinant and the fact that the entries of \(A\) and \(A^{-1}\) are integers to show that both \(x=\operatorname{det}(A)\) and \(y=\operatorname{det}\left(A^{-1}\right)\) are integers. (iii) Conclude that \(x=\operatorname{det}(A)\) must be either 1 or \(-1\) because these are the only integer solutions to the equation \(x y=1\)
Evaluate the determinant, in which the entries are functions. Determinants of this type occur when changes of variables are made in calculus. $$\left|\begin{array}{cc} e^{-x} & x e^{-x} \\ -e^{-x} & (1-x) e^{-x} \end{array}\right|$$
Determine whether the points are coplanar. $$(1,2,7),(-3,6,6),(4,4,2),(3,3,4)$$
Find (a) \(\left|\boldsymbol{A}^{T}\right|,(\mathbf{b})\left|\boldsymbol{A}^{2}\right|,(\mathbf{c})\left|\boldsymbol{A A}^{T}\right|,(\mathbf{d})|\boldsymbol{2} \boldsymbol{A}|,\) and \((\mathbf{e})\left|\boldsymbol{A}^{-\mathbf{1}}\right|\). $$A=\left[\begin{array}{rrr}1 & 5 & 4 \\\0 & -6 & 2 \\\0 & 0 & -3\end{array}\right]$$
Find (a) \(\left|\boldsymbol{A}^{T}\right|,(\mathbf{b})\left|\boldsymbol{A}^{2}\right|,(\mathbf{c})\left|\boldsymbol{A A}^{T}\right|,(\mathbf{d})|\boldsymbol{2} \boldsymbol{A}|,\) and \((\mathbf{e})\left|\boldsymbol{A}^{-\mathbf{1}}\right|\). $$A=\left[\begin{array}{rrr}4 & 1 & 9 \\\\-1 & 0 & -2 \\\\-3 & 3 & 0\end{array}\right]$$
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