what does c mean in linear algebra

linear independence for every finite subset {, ,} of B, if + + = for some , , in F, then = = =; spanning property for every vector v in V . In fact, they are both subspaces. Then in fact, both \(\mathrm{im}\left( T\right)\) and \(\ker \left( T\right)\) are subspaces of \(W\) and \(V\) respectively. via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. We formally define this and a few other terms in this following definition. Now assume that if \(T(\vec{x})=\vec{0},\) then it follows that \(\vec{x}=\vec{0}.\) If \(T(\vec{v})=T(\vec{u}),\) then \[T(\vec{v})-T(\vec{u})=T\left( \vec{v}-\vec{u}\right) =\vec{0}\nonumber \] which shows that \(\vec{v}-\vec{u}=0\). Lets try another example, one that uses more variables. Once \(x_3\) is chosen, we have a solution. We denote the degree of \(p(z)\) by \(\deg(p(z))\). In this case, we only have one equation, \[x_1+x_2=1 \nonumber \] or, equivalently, \[\begin{align}\begin{aligned} x_1 &=1-x_2\\ x_2&\text{ is free}. Since we have infinite choices for the value of \(x_3\), we have infinite solutions. One can probably see that free and independent are relatively synonymous. 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MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, 9.8: The Kernel and Image of a Linear Map, [ "article:topic", "kernel", "license:ccby", "showtoc:no", "authorname:kkuttler", "licenseversion:40", "source@https://lyryx.com/first-course-linear-algebra" ], https://math.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fmath.libretexts.org%2FBookshelves%2FLinear_Algebra%2FA_First_Course_in_Linear_Algebra_(Kuttler)%2F09%253A_Vector_Spaces%2F9.08%253A_The_Kernel_and_Image_of_a_Linear_Map, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), Kernel and Image of a Linear Transformation, 9.9: The Matrix of a Linear Transformation, Definition \(\PageIndex{1}\): Kernel and Image, Proposition \(\PageIndex{1}\): Kernel and Image as Subspaces, Example \(\PageIndex{1}\): Kernel and Image of a Transformation, Example \(\PageIndex{2}\): Kernel and Image of a Linear Transformation, Theorem \(\PageIndex{1}\): Dimension of Kernel + Image, Definition \(\PageIndex{2}\): Rank of Linear Transformation, Theorem \(\PageIndex{2}\): Subspace of Same Dimension, Corollary \(\PageIndex{1}\): One to One and Onto Characterization, Example \(\PageIndex{3}\): One to One Transformation, source@https://lyryx.com/first-course-linear-algebra. This leads us to a definition. A linear system is inconsistent if it does not have a solution. To find particular solutions, choose values for our free variables. Below we see the augmented matrix and one elementary row operation that starts the Gaussian elimination process. \end{aligned}\end{align} \nonumber \], \[\begin{align}\begin{aligned} x_1 &= 3\\ x_2 &=1 \\ x_3 &= 1 . Let \(T:\mathbb{P}_1\to\mathbb{R}\) be the linear transformation defined by \[T(p(x))=p(1)\mbox{ for all } p(x)\in \mathbb{P}_1.\nonumber \] Find the kernel and image of \(T\). . A vector space that is not finite-dimensional is called infinite-dimensional. First consider \(\ker \left( T\right) .\) It is necessary to show that if \(\vec{v}_{1},\vec{v}_{2}\) are vectors in \(\ker \left( T\right)\) and if \(a,b\) are scalars, then \(a\vec{v}_{1}+b\vec{v}_{2}\) is also in \(\ker \left( T\right) .\) But \[T\left( a\vec{v}_{1}+b\vec{v}_{2}\right) =aT(\vec{v}_{1})+bT(\vec{v}_{2})=a\vec{0}+b\vec{0}=\vec{0}\nonumber \]. Accessibility StatementFor more information contact us atinfo@libretexts.org. Now suppose \(n=2\). Therefore, no solution exists; this system is inconsistent. Hence, every element in \(\mathbb{R}^2\) is identified by two components, \(x\) and \(y\), in the usual manner. In linear algebra, the rank of a matrix A is the dimension of the vector space generated (or spanned) by its columns. We can think as above that the first two coordinates determine a point in a plane. The statement \(\ker \left( T \right) =\left\{ \vec{0}\right\}\) is equivalent to saying if \(T \left( \vec{v} \right)=\vec{0},\) it follows that \(\vec{v}=\vec{0}\). Then \(T\) is one to one if and only if the rank of \(A\) is \(n\). If \(\Span(v_1,\ldots,v_m)=V\), then we say that \((v_1,\ldots,v_m)\) spans \(V\) and we call \(V\) finite-dimensional. Therefore, they are equal. Let \(V\) and \(W\) be vector spaces and let \(T:V\rightarrow W\) be a linear transformation. \[\left [ \begin{array}{rr|r} 1 & 1 & a \\ 1 & 2 & b \end{array} \right ] \rightarrow \left [ \begin{array}{rr|r} 1 & 0 & 2a-b \\ 0 & 1 & b-a \end{array} \right ] \label{ontomatrix}\] You can see from this point that the system has a solution. The numbers \(x_{j}\) are called the components of \(\vec{x}\). Introduction to linear independence (video) | Khan Academy We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. We start with a very simple example. You can verify that \(T\) represents a linear transformation. We dont particularly care about the solution, only that we would have exactly one as both \(x_1\) and \(x_2\) would correspond to a leading one and hence be dependent variables. Now suppose we are given two points, \(P,Q\) whose coordinates are \(\left( p_{1},\cdots ,p_{n}\right)\) and \(\left( q_{1},\cdots ,q_{n}\right)\) respectively. Thus by Lemma 9.7.1 \(T\) is one to one. We can describe \(\mathrm{ker}(T)\) as follows. \nonumber \]. We often write the solution as \(x=1-y\) to demonstrate that \(y\) can be any real number, and \(x\) is determined once we pick a value for \(y\). By setting \(x_2 = 1\) and \(x_4 = -5\), we have the solution \(x_1 = 15\), \(x_2 = 1\), \(x_3 = -8\), \(x_4 = -5\). How can we tell if a system is inconsistent? 2. We can verify that this system has no solution in two ways. In very large systems, it might be hard to determine whether or not a variable is actually used and one would not worry about it. Linear Algebra - Definition, Topics, Formulas, Examples - Cuemath If we were to consider a linear system with three equations and two unknowns, we could visualize the solution by graphing the corresponding three lines. This helps us learn not only the technique but some of its inner workings. We can then use technology once we have mastered the technique and are now learning how to use it to solve problems. (So if a given linear system has exactly one solution, it will always have exactly one solution even if the constants are changed.) A major result is the relation between the dimension of the kernel and dimension of the image of a linear transformation. It is asking whether there is a solution to the equation \[\left [ \begin{array}{cc} 1 & 1 \\ 1 & 2 \end{array} \right ] \left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{c} a \\ b \end{array} \right ]\nonumber \] This is the same thing as asking for a solution to the following system of equations. Suppose first that \(T\) is one to one and consider \(T(\vec{0})\). What exactly is a free variable? Lets look at an example to get an idea of how the values of constants and coefficients work together to determine the solution type. You may recall this example from earlier in Example 9.7.1. It is one of the most central topics of mathematics. Therefore, well do a little more practice. \end{aligned}\end{align} \nonumber \], \[\begin{align}\begin{aligned} x_1 &= 3-2\pi\\ x_2 &=5-4\pi \\ x_3 &= e^2 \\ x_4 &= \pi. These two equations tell us that the values of \(x_1\) and \(x_2\) depend on what \(x_3\) is. Notice that there is only one leading 1 in that matrix, and that leading 1 corresponded to the \(x_1\) variable. The idea behind the more general \(\mathbb{R}^n\) is that we can extend these ideas beyond \(n = 3.\) This discussion regarding points in \(\mathbb{R}^n\) leads into a study of vectors in \(\mathbb{R}^n\). Linear algebra Definition & Meaning - Merriam-Webster The vectors \(v_1=(1,1,0)\) and \(v_2=(1,-1,0)\) span a subspace of \(\mathbb{R}^3\). -5-8w>19 - Solve linear inequalities with one unknown | Tiger Algebra Recall that to find the matrix \(A\) of \(T\), we apply \(T\) to each of the standard basis vectors \(\vec{e}_i\) of \(\mathbb{R}^4\). Here, the vector would have its tail sitting at the point determined by \(A= \left( d,e,f\right)\) and its point at \(B=\left( d+a,e+b,f+c\right) .\) It is the same vector because it will point in the same direction and have the same length. The next example shows the same concept with regards to one-to-one transformations. Compositions of linear transformations 1 (video) | Khan Academy Determinant, invertible matrices, and rank - Help with true/false However, the second equation of our system says that \(2x+2y= 4\). Lets find out through an example. Then, from the definition, \[\mathbb{R}^{2}= \left\{ \left(x_{1}, x_{2}\right) :x_{j}\in \mathbb{R}\text{ for }j=1,2 \right\}\nonumber \] Consider the familiar coordinate plane, with an \(x\) axis and a \(y\) axis. In practical terms, we could respond by removing the corresponding column from the matrix and just keep in mind that that variable is free. Linear Algebra - GeeksforGeeks Example: Let V = Span { [0, 0, 1], [2, 0, 1], [4, 1, 2]}. By picking two values for \(x_3\), we get two particular solutions. Answer by ntnk (54) ( Show Source ): You can put this solution on YOUR website! We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. In looking at the second row, we see that if \(k=6\), then that row contains only zeros and \(x_2\) is a free variable; we have infinite solutions. \[\left[\begin{array}{cccc}{1}&{1}&{1}&{5}\\{1}&{-1}&{1}&{3}\end{array}\right]\qquad\overrightarrow{\text{rref}}\qquad\left[\begin{array}{cccc}{1}&{0}&{1}&{4}\\{0}&{1}&{0}&{1}\end{array}\right] \nonumber \], Converting these two rows into equations, we have \[\begin{align}\begin{aligned} x_1+x_3&=4\\x_2&=1\\ \end{aligned}\end{align} \nonumber \] giving us the solution \[\begin{align}\begin{aligned} x_1&= 4-x_3\\x_2&=1\\x_3 &\text{ is free}.\\ \end{aligned}\end{align} \nonumber \]. Using this notation, we may use \(\vec{p}\) to denote the position vector of point \(P\). Group all constants on the right side of the inequality. Theorem 5.1.1: Matrix Transformations are Linear Transformations. Question 8. Describe the kernel and image of a linear transformation. So far, whenever we have solved a system of linear equations, we have always found exactly one solution. Next suppose \(T(\vec{v}_{1}),T(\vec{v}_{2})\) are two vectors in \(\mathrm{im}\left( T\right) .\) Then if \(a,b\) are scalars, \[aT(\vec{v}_{2})+bT(\vec{v}_{2})=T\left( a\vec{v}_{1}+b\vec{v}_{2}\right)\nonumber \] and this last vector is in \(\mathrm{im}\left( T\right)\) by definition. Let \(T: \mathbb{R}^4 \mapsto \mathbb{R}^2\) be a linear transformation defined by \[T \left [ \begin{array}{c} a \\ b \\ c \\ d \end{array} \right ] = \left [ \begin{array}{c} a + d \\ b + c \end{array} \right ] \mbox{ for all } \left [ \begin{array}{c} a \\ b \\ c \\ d \end{array} \right ] \in \mathbb{R}^4\nonumber \] Prove that \(T\) is onto but not one to one. This page titled 1.4: Existence and Uniqueness of Solutions is shared under a CC BY-NC 3.0 license and was authored, remixed, and/or curated by Gregory Hartman et al. \end{aligned}\end{align} \nonumber \]. Our final analysis is then this. Then the image of \(T\) denoted as \(\mathrm{im}\left( T\right)\) is defined to be the set \[\left\{ T(\vec{v}):\vec{v}\in V\right\}\nonumber \] In words, it consists of all vectors in \(W\) which equal \(T(\vec{v})\) for some \(\vec{v}\in V\). For convenience in this chapter we may write vectors as the transpose of row vectors, or \(1 \times n\) matrices. Is \(T\) onto? If you're seeing this message, it means we're having trouble loading external resources on our website. Find the solution to the linear system \[\begin{array}{ccccccc}x_1&+&x_2&+&x_3&=&5\\x_1&-&x_2&+&x_3&=&3\\ \end{array} \nonumber \] and give two particular solutions. Similarly, a linear transformation which is onto is often called a surjection. The complex numbers are both a real and complex vector space; we have = and = So the dimension depends on the base field. Create the corresponding augmented matrix, and then put the matrix into reduced row echelon form. For Property~2, note that \(0\in\Span(v_1,v_2,\ldots,v_m)\) and that \(\Span(v_1,v_2,\ldots,v_m)\) is closed under addition and scalar multiplication. From Proposition \(\PageIndex{1}\), \(\mathrm{im}\left( T\right)\) is a subspace of \(W.\) By Theorem 9.4.8, there exists a basis for \(\mathrm{im}\left( T\right) ,\left\{ T(\vec{v}_{1}),\cdots ,T(\vec{v}_{r})\right\} .\) Similarly, there is a basis for \(\ker \left( T\right) ,\left\{ \vec{u} _{1},\cdots ,\vec{u}_{s}\right\}\). If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Intro to linear equation standard form | Algebra (video) | Khan Academy Let \(P=\left( p_{1},\cdots ,p_{n}\right)\) be the coordinates of a point in \(\mathbb{R}^{n}.\) Then the vector \(\overrightarrow{0P}\) with its tail at \(0=\left( 0,\cdots ,0\right)\) and its tip at \(P\) is called the position vector of the point \(P\). Step-by-step solution. The two vectors would be linearly independent. This is not always the case; we will find in this section that some systems do not have a solution, and others have more than one. Since the unique solution is \(a=b=c=0\), \(\ker(S)=\{\vec{0}\}\), and thus \(S\) is one-to-one by Corollary \(\PageIndex{1}\). This is the reason why it is named as a 'linear' equation. Definition 5.5.2: Onto. To find two particular solutions, we pick values for our free variables. These definitions help us understand when a consistent system of linear equations will have infinite solutions. 5.1: Linear Transformations - Mathematics LibreTexts In fact, \(\mathbb{F}_m[z]\) is a finite-dimensional subspace of \(\mathbb{F}[z]\) since, \[ \mathbb{F}_m[z] = \Span(1,z,z^2,\ldots,z^m). In those cases we leave the variable in the system just to remind ourselves that it is there. Let \(A\) be an \(m\times n\) matrix where \(A_{1},\cdots , A_{n}\) denote the columns of \(A.\) Then, for a vector \(\vec{x}=\left [ \begin{array}{c} x_{1} \\ \vdots \\ x_{n} \end{array} \right ]\) in \(\mathbb{R}^n\), \[A\vec{x}=\sum_{k=1}^{n}x_{k}A_{k}\nonumber \]. Now consider the image. We start by putting the corresponding matrix into reduced row echelon form. There is no solution to such a problem; this linear system has no solution. Then \(\ker \left( T\right) \subseteq V\) and \(\mathrm{im}\left( T\right) \subseteq W\). Therefore, \(x_3\) and \(x_4\) are independent variables. Let \(\vec{z}\in \mathbb{R}^m\). We also could have seen that \(T\) is one to one from our above solution for onto. The notation \(\mathbb{R}^{n}\) refers to the collection of ordered lists of \(n\) real numbers, that is \[\mathbb{R}^{n} = \left\{ \left( x_{1}\cdots x_{n}\right) :x_{j}\in \mathbb{R}\text{ for }j=1,\cdots ,n\right\}\nonumber \] In this chapter, we take a closer look at vectors in \(\mathbb{R}^n\). We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. It is common to write \(T\mathbb{R}^{n}\), \(T\left( \mathbb{R}^{n}\right)\), or \(\mathrm{Im}\left( T\right)\) to denote these vectors. \[\left[\begin{array}{cccc}{1}&{1}&{1}&{1}\\{1}&{2}&{1}&{2}\\{2}&{3}&{2}&{0}\end{array}\right]\qquad\overrightarrow{\text{rref}}\qquad\left[\begin{array}{cccc}{1}&{0}&{1}&{0}\\{0}&{1}&{0}&{0}\\{0}&{0}&{0}&{1}\end{array}\right] \nonumber \]. By Proposition \(\PageIndex{1}\), \(A\) is one to one, and so \(T\) is also one to one. [2] Then why include it? (lxn) matrix and (nx1) vector multiplication. It turns out that the matrix \(A\) of \(T\) can provide this information. (We can think of it as depending on the value of 1.) If is a linear subspace of then (). How do we recognize which variables are free and which are not? A system of linear equations is consistent if it has a solution (perhaps more than one). Let \(T: \mathbb{M}_{22} \mapsto \mathbb{R}^2\) be defined by \[T \left [ \begin{array}{cc} a & b \\ c & d \end{array} \right ] = \left [ \begin{array}{c} a - b \\ c + d \end{array} \right ]\nonumber \] Then \(T\) is a linear transformation. Linear Algebra Book: Linear Algebra (Schilling, Nachtergaele and Lankham) 5: Span and Bases 5.1: Linear Span Expand/collapse global location . 1: What is linear algebra - Mathematics LibreTexts This form is also very useful when solving systems of two linear equations. Thus \[\vec{z} = S(\vec{y}) = S(T(\vec{x})) = (ST)(\vec{x}),\nonumber \] showing that for each \(\vec{z}\in \mathbb{R}^m\) there exists and \(\vec{x}\in \mathbb{R}^k\) such that \((ST)(\vec{x})=\vec{z}\). For what values of \(k\) will the given system have exactly one solution, infinite solutions, or no solution? \[\begin{align}\begin{aligned} x_1 &= 4\\ x_2 &=1 \\ x_3 &= 0 . For example, 2x+3y=5 is a linear equation in standard form. GATE-CS-2014- (Set-2) Linear Algebra. \[\overrightarrow{PQ} = \left [ \begin{array}{c} q_{1}-p_{1}\\ \vdots \\ q_{n}-p_{n} \end{array} \right ] = \overrightarrow{0Q} - \overrightarrow{0P}\nonumber \]. Similarly, by Corollary \(\PageIndex{1}\), if \(S\) is onto it will have \(\mathrm{rank}(S) = \mathrm{dim}(\mathbb{M}_{22}) = 4\). Then \(z^{m+1}\in\mathbb{F}[z]\), but \(z^{m+1}\notin \Span(p_1(z),\ldots,p_k(z))\). The rank of \(A\) is \(2\). In previous sections we have only encountered linear systems with unique solutions (exactly one solution). The image of \(S\) is given by, \[\mathrm{im}(S) = \left\{ \left [\begin{array}{cc} a+b & a+c \\ b-c & b+c \end{array}\right ] \right\} = \mathrm{span} \left\{ \left [\begin{array}{rr} 1 & 1 \\ 0 & 0 \end{array} \right ], \left [\begin{array}{rr} 1 & 0 \\ 1 & 1 \end{array} \right ], \left [\begin{array}{rr} 0 & 1 \\ -1 & 1 \end{array} \right ] \right\}\nonumber \]. 1.4: Existence and Uniqueness of Solutions - Mathematics LibreTexts Consider \(n=3\). Prove that if \(T\) and \(S\) are one to one, then \(S \circ T\) is one-to-one. \nonumber \]. The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. I'm having trouble with some true/false questions in my linear algebra class and was hoping someone could help me out. Accessibility StatementFor more information contact us atinfo@libretexts.org. AboutTranscript. Therefore the dimension of \(\mathrm{im}(S)\), also called \(\mathrm{rank}(S)\), is equal to \(3\). Now we have seen three more examples with different solution types. When we learn about s and s, we will see that under certain circumstances this situation arises. If \(T\) and \(S\) are onto, then \(S \circ T\) is onto. This leads to a homogeneous system of four equations in three variables. Hence, if \(v_1,\ldots,v_m\in U\), then any linear combination \(a_1v_1+\cdots +a_m v_m\) must also be an element of \(U\). Suppose \(p(x)=ax^2+bx+c\in\ker(S)\). It follows that \(T\) is not one to one. If you are graphing a system with a quadratic and a linear equation, these will cross at either two points, one point or zero points. By convention, the degree of the zero polynomial \(p(z)=0\) is \(-\infty\). Now we want to know if \(T\) is one to one. By removing vectors from the set to create an independent set gives a basis of \(\mathrm{im}(T)\). A particular solution is one solution out of the infinite set of possible solutions. Using Theorem \(\PageIndex{1}\) we can show that \(T\) is onto but not one to one from the matrix of \(T\). Take any linear combination c 1 sin ( t) + c 2 cos ( t), assume that the c i (atleast one of which is non-zero) exist such that it is zero for all t, and derive a contradiction. First, we will consider what Rn looks like in more detail. Look also at the reduced matrix in Example \(\PageIndex{2}\). We have been studying the solutions to linear systems mostly in an academic setting; we have been solving systems for the sake of solving systems. In other words, \(\vec{v}=\vec{u}\), and \(T\) is one to one. By Proposition \(\PageIndex{1}\) \(T\) is one to one if and only if \(T(\vec{x}) = \vec{0}\) implies that \(\vec{x} = \vec{0}\). Legal. Putting the augmented matrix in reduced row-echelon form: \[\left [\begin{array}{rrr|c} 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & 1 & 0 \end{array}\right ] \rightarrow \cdots \rightarrow \left [\begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right ].\nonumber \]. The only vector space with dimension is {}, the vector space consisting only of its zero element.. Properties. Find the position vector of a point in \(\mathbb{R}^n\). T/F: It is possible for a linear system to have exactly 5 solutions. \[\begin{array}{ccccc}x_1&+&2x_2&=&3\\ 3x_1&+&kx_2&=&9\end{array} \nonumber \]. By setting up the augmented matrix and row reducing, we end up with \[\left [ \begin{array}{rr|r} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right ]\nonumber \], This tells us that \(x = 0\) and \(y = 0\). 1. We generally write our solution with the dependent variables on the left and independent variables and constants on the right. Hence \(\mathbb{F}^n\) is finite-dimensional. We have \[\begin{align}\begin{aligned} x_1 + 2x_3 &= 2 \\ x_2-x_3&=3 \end{aligned}\end{align} \nonumber \] or, equivalently, \[\begin{align}\begin{aligned} x_1 &= 2-2x_3 \\ x_2&=3+x_3\\x_3&\text{ is free.} Let \(T:V\rightarrow W\) be a linear transformation where \(V,W\) are vector spaces. \[\left[\begin{array}{ccc}{1}&{1}&{1}\\{2}&{2}&{2}\end{array}\right]\qquad\overrightarrow{\text{rref}}\qquad\left[\begin{array}{ccc}{1}&{1}&{1}\\{0}&{0}&{0}\end{array}\right] \nonumber \], Now convert the reduced matrix back into equations.

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