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- https://math.libretexts.org/Bookshelves/Linear_Algebra/Map%3A_Linear_Algebra_(Waldron_Cherney_and_Denton)/03%3A_The_Simplex_Method/3.05%3A_Review_ProblemsMaximize \(f(x,y)=2x+3y\) subject to the constraints x\geq0\, ,\quad y\geq0\, ,\quad x+2y\leq2\, ,\quad 2x+y\leq2\, , a) sketching the region in the \(xy\)-plane defined by the constraints and then ch...Maximize \(f(x,y)=2x+3y\) subject to the constraints x\geq0\, ,\quad y\geq0\, ,\quad x+2y\leq2\, ,\quad 2x+y\leq2\, , a) sketching the region in the \(xy\)-plane defined by the constraints and then checking the values of \(f\) at its corners; and, b) the simplex algorithm (\(\textit{Hint:}\) introduce slack variables). Contributor David Cherney, Tom Denton, and Andrew Waldron (UC Davis)
- https://math.libretexts.org/Bookshelves/Linear_Algebra/Map%3A_Linear_Algebra_(Waldron_Cherney_and_Denton)/05%3A_Vector_SpacesThe two key properties of vectors are that they can be added together and multiplied by scalars, so we make the following definition.
- https://math.libretexts.org/Bookshelves/Linear_Algebra/Map%3A_Linear_Algebra_(Waldron_Cherney_and_Denton)/02%3A_Systems_of_Linear_EquationsThumbnail: 3 planes intersect at a point. (CC BY-SA 4.0; Fred the Oyster). Contributor David Cherney, Tom Denton, and Andrew Waldron (UC Davis)
- https://math.libretexts.org/Bookshelves/Linear_Algebra/Map%3A_Linear_Algebra_(Waldron_Cherney_and_Denton)/02%3A_Systems_of_Linear_Equations/2.06%3A_Review_ProblemsThe most important thing to remember is that the index \(j\) is a dummy variable, so that \(a_{j}^{2}x^{j}\equiva_{i}^2x^{i}\); this is called “relabeling dummy indices”. When dealing with products of...The most important thing to remember is that the index \(j\) is a dummy variable, so that \(a_{j}^{2}x^{j}\equiva_{i}^2x^{i}\); this is called “relabeling dummy indices”. When dealing with products of sums, you must remember to introduce a new dummy for each term; i.e., \(a_{i}x^{i}b_{i}y^{i} = \sum_{i}a_{i}x^{i}b_{i}y^{i}\) does not equal \(a_{i}x^{i}b_{j}y^{j} = \sum_{i}a_{i}x^{i}\sum_{j}b_{j}y^{j}\).
- https://math.libretexts.org/Bookshelves/Linear_Algebra/Map%3A_Linear_Algebra_(Waldron_Cherney_and_Denton)/03%3A_The_Simplex_Method/3.04%3A_Pablo_Meets_DantzigThus the so-called \(\textit {objective function}\) \(f=-s+95=-5x_1-10x_2\). (Notice that it makes no difference whether we maximize \(-s\) or \(-s+95\), we choose the latter since it is a linear func...Thus the so-called \(\textit {objective function}\) \(f=-s+95=-5x_1-10x_2\). (Notice that it makes no difference whether we maximize \(-s\) or \(-s+95\), we choose the latter since it is a linear function of \((x_1,x_2)\).) Now we can build an augmented matrix whose last row reflects the objective function equation \(5 x_1+10 x_2 +f=0\): The first row operation uses the \(1\) in the top of the first column to zero out the most negative entry in the last row:
- https://math.libretexts.org/Bookshelves/Linear_Algebra/Map%3A_Linear_Algebra_(Waldron_Cherney_and_Denton)/04%3A_Vectors_in_Space_n-VectorsHere \(a^2\) denotes the second component of the vector \(a\), rather than the number $a$ squared!}\) We emphasize that order matters: \[ {\mathbb{R}}^n :=\left\{ \begin{pmatrix}a^1 \\ \vdots\ \ \\ a^...Here \(a^2\) denotes the second component of the vector \(a\), rather than the number $a$ squared!}\) We emphasize that order matters: \[ {\mathbb{R}}^n :=\left\{ \begin{pmatrix}a^1 \\ \vdots\ \ \\ a^n\end{pmatrix} \middle\vert \, a^1,\dots, a^n \in \mathbb{R} \right\} \,.\] Thumbnail: The volume of this parallelepiped is the absolute value of the determinant of the 3-by-3 matrix formed by the vectors \(r_1\), \(r_2\), and \(r_3\). (CC BY-SA 3.0; Claudio Rocchini)
- https://math.libretexts.org/Bookshelves/Linear_Algebra/Map%3A_Linear_Algebra_(Waldron_Cherney_and_Denton)/07%3A_Matrices/7.05%3A_Inverse_MatrixA square matrix MM is invertible (or nonsingular) if there exists a matrix M⁻¹ such that M⁻¹M=I=M⁻¹M. If M has no inverse, we say M is Singular or non-invertible .
- https://math.libretexts.org/Bookshelves/Linear_Algebra/Map%3A_Linear_Algebra_(Waldron_Cherney_and_Denton)/08%3A_Determinants/8.03%3A_Review_Problemsm^{1}_{1} & m^{1}_{2} & m^{1}_{3}\\ For simplicity, assume that \(m_{1}^{1}\neq 0 \neq m^{1}_{1}m^{2}_{2}-m^{2}_{1}m^{1}_{2}\). b) Find elementary matrices \(R^{1}(\lambda)\) and \(R^{2}(\lambda)\) th...m^{1}_{1} & m^{1}_{2} & m^{1}_{3}\\ For simplicity, assume that \(m_{1}^{1}\neq 0 \neq m^{1}_{1}m^{2}_{2}-m^{2}_{1}m^{1}_{2}\). b) Find elementary matrices \(R^{1}(\lambda)\) and \(R^{2}(\lambda)\) that respectively multiply rows \(1\) and \(2\) of \(M\) by \(\lambda\) but otherwise leave \(M\) the same under left multiplication. Show that if \(M\) is a \(3\times 3\) matrix whose third row is a sum of multiples of the other rows (\(R_{3}=aR_{2}+bR_{1}\)) then \(\det M=0\).
- https://math.libretexts.org/Bookshelves/Linear_Algebra/Map%3A_Linear_Algebra_(Waldron_Cherney_and_Denton)/08%3A_Determinants/8.02%3A_Elementary_Matrices_and_DeterminantsNotice that because \(\det RREF(M) = \det (E_{1}E_{2}\cdots E_{k}M)\), by the theorem above, $$\det RREF(M)=\det (E_{1}) \cdots \det (E_{k}) \det M\, .$$ Since each \(E_{i}\) has non-zero determinant,...Notice that because \(\det RREF(M) = \det (E_{1}E_{2}\cdots E_{k}M)\), by the theorem above, $$\det RREF(M)=\det (E_{1}) \cdots \det (E_{k}) \det M\, .$$ Since each \(E_{i}\) has non-zero determinant, then \(\det RREF(M)=0\) if and only if \(\det M=0\).
- https://math.libretexts.org/Bookshelves/Linear_Algebra/Map%3A_Linear_Algebra_(Waldron_Cherney_and_Denton)/11%3A_Basis_and_Dimension/11.03%3A_Review_Problems(Hint: You can build up a basis for \(B^{n}\) by choosing one vector at a time, such that the vector you choose is not in the span of the previous vectors you've chosen. (Hint: Let \(\{w_{1}, \ldots, ...(Hint: You can build up a basis for \(B^{n}\) by choosing one vector at a time, such that the vector you choose is not in the span of the previous vectors you've chosen. (Hint: Let \(\{w_{1}, \ldots, w_{m}\}\) be a collection of \(n\) linearly independent vectors in \(V\), and let \(\{v_{1}, \ldots, v_{n}\}\) be a basis for \(V\).) a) \(L:V\rightarrow W\) where \(B=(v_{1},\ldots, v_{n})\) is a basis for \(V\) and \(B'=(L(v_{1}),\ldots, L(v_{n}))\) is a basis for \(W\).
- https://math.libretexts.org/Bookshelves/Linear_Algebra/Map%3A_Linear_Algebra_(Waldron_Cherney_and_Denton)/06%3A_Linear_Transformations/6.03%3A_Linear_Differential_OperatorsYour calculus class became much easier when you stopped using the limit definition of the derivative, learned the power rule, and started using linearity of the derivative operator.
