Linear Combinations of Vectors - The Basics - MathBootCamps
}\) We know how to do this using Gaussian elimination; let's use our matrix \(B\) to find a different way: If \(A\mathbf x\) is defined, then the number of components of \(\mathbf x\) equals the number of rows of \(A\text{. Verify that \(PA\) is the matrix that results from interchanging the first and second rows. Quiz permutations & combinations, download emulator for T1-84 calculator, FOIL math pretest, Substitution Method of Algebra. on Wednesday? In this exercise, you will construct the inverse of a matrix, a subject that we will investigate more fully in the next chapter. GCD as Linear Combination Finder - math.oxford.emory.edu }\) Bicycles that are rented at one location may be returned to either location at the end of the day. What geometric effect does scalar multiplication have on a vector? be
If
Chapter 04.03: Lesson: Linear combination of matrices: Example The operations that we perform in Gaussian elimination can be accomplished using matrix multiplication. Form the vector \(\mathbf x_1\) and determine the number of bicycles at the two locations the next day by finding \(\mathbf x_2 = A\mathbf x_1\text{.}\). Use Sage to evaluate the product Item a yet again. How to use the linear combination method. }\), If a linear system of equations has 8 equations and 5 unknowns, then the dimensions of the matrix \(A\) in the corresponding equation \(A\mathbf x = \mathbf b\) is \(5\times8\text{.}\). gives. vectors:Compute
}\) We need to find weights \(a\) and \(b\) such that, Equating the components of the vectors on each side of the equation, we arrive at the linear system. Here zero (0) is the vector with in all coordinates holds if and only if \( a_1 + a_2 + a_3 + a_4 + + a_{n-1} + a_n = 0 \). The next activity introduces some properties of matrix multiplication. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. There are two operations we can perform with vectors: scalar multiplication and vector addition. Add this calculator to your site and lets users to perform easy calculations. }\) The effect is to translate the line \(a\mathbf v\) by the vector \(\mathbf w\text{,}\) as shown in Figure 2.1.3.
Definition
as
are all equal to each other. \end{equation*}, \begin{equation*} (x,y) = \{2,-3\}\text{.} Even though these vector operations are new, it is straightforward to check that some familiar properties hold. \end{equation*}, \begin{equation*} \left[\begin{array}{rrrr|r} \mathbf v_1 & \mathbf v_2 & \ldots & \mathbf v_n & \mathbf b \end{array}\right]\text{.} The vectors v and w are drawn in gray while the linear combination av + bw is in red. Just type matrix elements and click the button. Therefore, \(A\mathbf x\) will be 3-dimensional. This way, we've solved the system using linear combination! Try the plant spacing calculator. is a linear combination of
,
and
In particular, we saw that the vector \(\mathbf b\) is a linear combination of the vectors \(\mathbf v_1,\mathbf v_2,\ldots,\mathbf v_n\) if the linear system corresponding to the augmented matrix. \end{equation*}, \begin{equation*} A = \left[\begin{array}{rr} \mathbf v_1 & \mathbf v_2 \end{array} \right]\text{.} A
Compare the results of evaluating \(A(BC)\) and \((AB)C\) and state your finding as a general principle. Sketch below the line \(y=3x - 2\text{. Planning out your garden? }\) If so, describe all the ways in which you can do so. }\) If so, describe all the ways in which you can do so. }\), What are the dimensions of the matrix \(A\text{? If \(A\) is an \(m\times n\) matrix, then \(\mathbf x\) must be an \(n\)-dimensional vector, and the product \(A\mathbf x\) will be an \(m\)-dimensional vector. accordingly. }\) The information above tells us.
For example, three vectors in two-dimensional space: \( v (a_1, a_2), w (b_1, b_2), v (c_1, c_2) \), then write their coordinates as one matric with each row corresponding to the one of vectors. You will need to get assistance from your school if you are having problems entering the answers into your online assignment. \end{equation*}, \begin{equation*} P = \left[\begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array}\right]\text{.} Depending on whether the statement you got is true, like: If the statement is false, then the system has no solution. A theme that will later unfold concerns the use of coordinate systems. }\) Find the vector that is the linear combination when \(a = -2\) and \(b = 1\text{.}\). }\), The solution space to the equation \(A\mathbf x = \mathbf b\) is equivalent to the solution space to the linear system whose augmented matrix is \(\left[\begin{array}{r|r} A & \mathbf b \end{array}\right]\text{. Please follow the steps below on how to use the calculator: A linear equation of the form Ax + By = C. Here,xandyare variables, and A, B,and Care constants. is a linear combination of
same dimension (otherwise matrix addition would not be possible). \\ \end{aligned} \end{equation*}, \begin{equation*} A=\left[\begin{array}{rrrr} \mathbf v_1 & \mathbf v_2 & \ldots & \mathbf v_n \end{array}\right], \mathbf x = \left[\begin{array}{r} c_1 \\ c_2 \\ \vdots \\ c_n \end{array}\right], \end{equation*}, \begin{equation*} A\mathbf x = c_1\mathbf v_1 + c_2\mathbf v_2 + \ldots c_n\mathbf v_n\text{.} }\) What does this solution space represent geometrically? We then end up with a single equation in one variable, which we can easily solve. follows: Most of the times, in linear algebra we deal with linear combinations of
First, we see that scalar multiplication has the effect of stretching or compressing a vector. asBecause
such that
we can always use the least common multiple of a1 and a2. Can you find another vector \(\mathbf c\) such that \(A\mathbf x = \mathbf c\) is inconsistent? be
}\) Geometrically, this means that we begin from the tip of \(\mathbf w\) and move in a direction parallel to \(\mathbf v\text{. How easy was it to use our calculator?
Linear combinations - Statlect A subspace of R n is given by the span of a . There. be
In other words, the number of columns of \(A\) must equal the dimension of the vector \(\mathbf x\text{.}\). and
If \(A\) is a \(9\times5\) matrix, then \(A\mathbf x=\mathbf b\) is inconsistent for some vector \(\mathbf b\text{. For instance, if v 1 = [ 11, 5, 7, 0] T and v 1 = [ 2, 13, 0, 7] T, the set of all vectors of the form s v 1 + t v 2 for certain scalars 's' and 't' is the span of v1 and v2. solution:In
|D|=0, $$ A = (1, 1, 0), B = (2, 5, 3), C = (1, 2, 7) $$, $$ |D|= \left|\begin{array}{ccc}1 & 1 & 0\\2 & 5 & -3\\1 & 2 & 7\end{array}\right| $$, $$|D|= 1 \times \left|\begin{array}{cc}5 & -3\\2 & 7\end{array}\right| (1) \times \left|\begin{array}{cc}2 & -3\\1 & 7\end{array}\right| + (0) \times \left|\begin{array}{cc}2 & 5\\1 & 2\end{array}\right|$$, $$ |D|= 1 ((5) (7) (3) (2)) (1) ((2) (7) ( 3) (1)) + (0) ((2) (2) (5) (1)) $$, $$ |D|= 1 ((35) (- 6)) (1) ((14) ( 3)) + (0) ((4) (5)) $$, $$ |D|=1 (41) (1) (17) + (0) ( 1) $$. A linear combination of
}\), Suppose \(A\) is an \(m\times n\) matrix. Find the values of h for which the vectors are linearly dependent, if vectors \( h_1 = {1, 1, 0}, h_2 = {2, 5, -3}, h_3 = {1, 2, 7} \) in 3 dimensions, then find they are linear independent or not? different values of
}\) For instance. LinearCombination | Wolfram Function Repository }\) From there, we continue our walk using the horizontal and vertical changes prescribed by \(\mathbf w\text{,}\) after which we arrive at the sum \(\mathbf v + \mathbf w\text{. and
,
be the following
You are encouraged to evaluate Item a using this shortcut and compare the result to what you found while completing the previous activity.
To understand the sum \(\mathbf v + \mathbf w\text{,}\) we imagine walking from the origin with the appropriate horizontal and vertical changes given by \(\mathbf v\text{. \end{equation*}, \begin{equation*} \begin{aligned} A\mathbf x = \left[\begin{array}{rr} -2 & 3 \\ 0 & 2 \\ 3 & 1 \\ \end{array}\right] \left[\begin{array}{r} 2 \\ 3 \\ \end{array}\right] {}={} & 2 \left[\begin{array}{r} -2 \\ 0 \\ 3 \\ \end{array}\right] + 3 \left[\begin{array}{r} 3 \\ 2 \\ 1 \\ \end{array}\right] \\ \\ {}={} & \left[\begin{array}{r} -4 \\ 0 \\ 6 \\ \end{array}\right] + \left[\begin{array}{r} 9 \\ 6 \\ 3 \\ \end{array}\right] \\ \\ {}={} & \left[\begin{array}{r} 5 \\ 6 \\ 9 \\ \end{array}\right]. asIs
For the system of equations: a1x + b1y = c1 a2x + b2y = c2 we can always use the least common multiple of a1 and a2.
}\), Identify the matrix \(A\) and vector \(\mathbf b\) to express this system in the form \(A\mathbf x = \mathbf b\text{.}\). If \(A\text{,}\) \(B\text{,}\) and \(C\) are matrices such that the following operations are defined, it follows that. the value of the linear
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\end{equation*}, \begin{equation*} \mathbf x =\left[ \begin{array}{r} x_1 \\ x_2 \\ x_3 \end{array} \right] = \left[ \begin{array}{r} -x_3 \\ 5 + 2x_3 \\ x_3 \end{array} \right] =\left[\begin{array}{r}0\\5\\0\end{array}\right] +x_3\left[\begin{array}{r}-1\\2\\1\end{array}\right] \end{equation*}, \begin{equation*} \begin{alignedat}{4} 2x & {}+{} & y & {}-{} & 3z & {}={} & 4 \\ -x & {}+{} & 2y & {}+{} & z & {}={} & 3 \\ 3x & {}-{} & y & & & {}={} & -4 \\ \end{alignedat}\text{.} This means we have \(\mathbf x_1 = \twovec{1000}{0}\text{. can be rewritten as a linear combination of just \(\mathbf v_1\) and \(\mathbf v_2\text{. Show that \(\mathbf v_3\) can be written as a linear combination of \(\mathbf v_1\) and \(\mathbf v_2\text{. Linearity of matrix multiplication. Linear Algebra. Though we allow ourselves to begin walking from any point in the plane, we will most frequently begin at the origin, in which case we arrive at the the point \((2,1)\text{,}\) as shown in the figure. If \(A\) has a pivot in every row, then every equation \(A\mathbf x = \mathbf b\) is consistent. Suppose that \(\mathbf x_h\) is a solution to the homogeneous equation; that is \(A\mathbf x_h=\zerovec\text{. }\) What do you find when you evaluate \(I\mathbf x\text{?}\). Solving a system of linear equations: Solve the given system of m linear equations in n unknowns. }\) What is the product \(A\twovec{0}{1}\text{? A(cv) = cAv. Also, we can see that the \( e_1 and e_2 \) without problematic vector v are linearly independent vectors. What is matrix used for? problem:Therefore,
Matrix-vector multiplication and linear systems So far, we have begun with a matrix A and a vector x and formed their product Ax = b. zero vector is a linear combination of
Scalar Multiplication and Vector Addition. We think of \(\mathbf v\) as describing a walk we take in the plane where we move two units horizontally and one unit vertically.
the
If \(A=\left[\begin{array}{rrrr} \mathbf v_1& \mathbf v_2& \ldots\mathbf v_n \end{array}\right]\) and \(\mathbf x=\left[ \begin{array}{r} x_1 \\ x_2 \\ \vdots \\ x_n \\ \end{array}\right] \text{,}\) then the following are equivalent. To check for linear dependence, we change the values from vector to matrices. This lecture is about linear combinations of
give the zero vector as a result. In this article, we break down what dependent and independent variables are and explain how to determine if vectors are linearly independent? When the matrix \(A = \left[\begin{array}{rrrr} \mathbf v_1& \mathbf v_2& \ldots& \mathbf v_n\end{array}\right]\text{,}\) we will frequently write, and say that we augment the matrix \(A\) by the vector \(\mathbf b\text{.}\). Namely, put: m1 := LCM (a1, a2) / a1 m2 := LCM (a1, a2) / a2 and **multiply the first equation by m1 and the second equation by **-m 2 ****. and changing
Matrix Calculator A matrix, in a mathematical context, is a rectangular array of numbers, symbols, or expressions that are arranged in rows and columns. Did you face any problem, tell us! What is the general strategy for converting a point from standard Cartesian coordinates \((x,y)\) to the new coordinates \(\{a,b\}\text{?
This problem is a continuation of the previous problem. . }\), If the vector \(\mathbf e_1 = \left[\begin{array}{r} 1 \\ 0 \\ \vdots \\ 0 \end{array}\right]\text{,}\) what is the product \(A\mathbf e_1\text{? A(v + w) = Av + Aw. The length of the line segment represents the magnitude of the vector, and the arrowhead pointing in a specific direction represents the direction of the vector. By combining linear equations we mean multiplying one or both equations by suitably chosen numbers and then adding the equations together. The previous activity also shows that questions about linear combinations lead naturally to linear systems. What do you find when you evaluate \(A(\mathbf v+\mathbf w)\) and \(A\mathbf v + A\mathbf w\) and compare your results? }\) Explain why every four-dimensional vector can be written as a linear combination of the vectors \(\mathbf v_1\text{,}\) \(\mathbf v_2\text{,}\) \(\mathbf v_3\text{,}\) and \(\mathbf v_4\) in exactly one way. }\), Shown below are two vectors \(\mathbf v\) and \(\mathbf w\), Nutritional information about a breakfast cereal is printed on the box. The Span of Vectors Calculator is a calculator that returns a list of all linear vector combinations. Linearly Dependent Vectors -- from Wolfram MathWorld Vector Calculator - Symbolab the system is satisfied provided we set
So far, we have begun with a matrix \(A\) and a vector \(\mathbf x\) and formed their product \(A\mathbf x = \mathbf b\text{. Check out 35 similar linear algebra calculators . Though we wrote it as \(I_n\) in the activity, we will often just write \(I\) when the dimensions are clear. Let and be -dimensional vectors. Wolfram|Alpha Examples: Linear Algebra Accessibility StatementFor more information contact us atinfo@libretexts.org. ,
}\) State your finding as a general principle. \end{equation*}, \begin{equation*} \left[ \begin{array}{rrrr} \mathbf v_1 & \mathbf v_2 & \mathbf v_3 & \mathbf v_4 \end{array} \right] \end{equation*}, \begin{equation*} \mathbf v_1=\left[\begin{array}{r}0\\3\\2\\ \end{array}\right], \mathbf v_2=\left[\begin{array}{r}4\\-1\\0\\ \end{array}\right], \mathbf v_3=\left[\begin{array}{r}-3\\2\\-1\\ \end{array}\right], \mathbf v_4=\left[\begin{array}{r}1\\0\\1\\ \end{array}\right]\text{.} How many servings of each cereal have you eaten if you have consumed 342 calories, 385 milligrams of sodium, and 3.4 grams of protein. }\), What does this say about the pivots of the matrix \(A\text{? Matrix-vector multiplication and linear systems. If we deal with two linear equations in two variables, we want to combine these equations into one equation with a single variable. Example
Linearity of matrix multiplication. We can identify the point \((x,y)\) with the tip of the vector \(\left[\begin{array}{r}x\\y\end{array}\right]\text{,}\) drawn emanating from the origin. matrixis
if and only if the following three equations are simultaneously
The key idea is to combine the equations into a system of fewer and simpler equations. \end{equation*}, \begin{equation*} \mathbf x_k = \twovec{B_k}{C_k} \end{equation*}, \begin{equation*} \mathbf x_{k+1} = A\mathbf x_k \end{equation*}, \begin{equation*} A = \left[\begin{array}{rr} 0.8 & 0.5 \\ 0.2 & 0.5 \\ \end{array}\right]\text{.} Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. The vector \(\mathbf b\) is a linear combination of the vectors \(\mathbf v_1,\mathbf v_2,\ldots,\mathbf v_n\) if and only if the linear system corresponding to the augmented matrix, is consistent. \end{equation*}, \begin{equation*} AB = \left[\begin{array}{rrrr} A\mathbf v_1 & A\mathbf v_2 & \ldots & A\mathbf v_p \end{array}\right]\text{.}
From the source of Libre Text: Linear Independence and the Wronskian, determinant of the corresponding matrix, linear differential equations, Affine independence. Linear combinations and span (video) | Khan Academy
What can you say about the solution space to the equation \(A\mathbf x = \zerovec\text{?}\). To form the set of vectors \(a\mathbf v+\mathbf w\text{,}\) we can begin with the vector \(\mathbf w\) and add multiples of \(\mathbf v\text{. This equation will be a linear combination of these two variables and a constant. Note that all the matrices involved in a linear combination need to have the
\end{equation*}, \begin{equation*} AB = I = \left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \\ \end{array}\right]\text{.} Since we need the same number of vectors to add and since the vectors must be of the same dimension, two matrices must have the same dimensions as well if we wish to form their sum.
A linear equation is defined as an equation that is written fortwo different variables. When the number of rows is \(m\) and columns is \(n\text{,}\) we say that the dimensions of the matrix are \(m\times n\text{.
Legal. Let
and
follows:Let
From the source of Cornell University: Linear independence of values of G-functions, Alternative method using determinants, More vectors than dimensions, Natural basis vectors, Linear independence of functions, Space of linear dependencies. ,
What matrix \(S\) would scale the third row by -3? ResourceFunction [ "LinearCombination"] [ { u }, { vi }] returns a list of coefficients that express u as a linear combination of the basis vectors vi. Our goal in this section is to introduction matrix multiplication, another algebraic operation that connects linear systems and linear combinations. If the final statement is true, then the system has infinitely many solutions. Therefore, \(\mathbf b\) may be expressed as a linear combination of \(\mathbf v\) and \(\mathbf w\) in exactly one way. }\) Is it generally true that \(AB = BA\text{?}\). Define two
Sketch below the set of vectors having the form \(2\mathbf v + t\mathbf w\) where \(t\) is any scalar. In this way, we see that the third component of the product would be obtained from the third row of the matrix by computing \(2(3) + 3(1) = 9\text{.}\). matrices having dimension
Then, the linearly independent matrix calculator finds the determinant of vectors and provide a comprehensive solution. It is not generally true that \(AB = BA\text{. and
A more important operation will be matrix multiplication as it allows us to compactly express linear systems. How to know if a matrix is linearly independent? familiar with the concepts introduced in the lectures on
}\), Use the previous part of this problem to determine \(\mathbf x_2\text{,}\) \(\mathbf x_3\) and \(\mathbf x_4\text{.}\). \end{equation*}, \begin{equation*} A = \left[\begin{array}{rrr} 1 & 3 & 2 \\ -3 & 4 & -1 \\ \end{array}\right], B = \left[\begin{array}{rr} 3 & 0 \\ 1 & 2 \\ -2 & -1 \\ \end{array}\right]\text{.} So you scale them by c1, c2, all the way to cn, where everything from c1 to cn are all a member of the real numbers. be
GCD as Linear Combination Finder. We know that the matrix product \(A\mathbf x\) forms a linear combination of the columns of \(A\text{. \end{equation*}, \begin{equation*} S = \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 1 \\ \end{array}\right]\text{.} Over time, the city finds that 80% of bicycles rented at location \(B\) are returned to \(B\) with the other 20% returned to \(C\text{. }\) Write the vector \(\mathbf x_1\) and find the scalars \(c_1\) and \(c_2\) such that \(\mathbf x_1=c_1\mathbf v_1 + c_2\mathbf v_2\text{. Enter two numbers (separated by a space) in the text box below. ? Use the language of vectors and linear combinations to express the total amount of calories, sodium, and protein you have consumed.
We can then think of the usual Cartesian coordinate system in terms of linear combinations of the vectors, The point \((2,-3)\) is identified with the vector, we may define a new coordinate system, such that a point \(\{x,y\}\) will correspond to the vector, For instance, the point \(\{2,-3\}\) is shown on the right side of Figure 2.1.8. Let's ask how we can describe the vector \(\mathbf b=\left[\begin{array}{r} -1 \\ 4 \end{array} \right]\) as a linear combination of \(\mathbf v\) and \(\mathbf w\text{. When the coefficients of one variable are equal, one multiplier is equal to 1 and the other to -1. }\) You may do this by evaluating \(A(\mathbf x_h+\mathbf x_p)\text{.
It is computed as
The previous section introduced vectors and linear combinations and demonstrated how they provide a means of thinking about linear systems geometrically. by
\end{equation*}, \begin{equation*} \mathbf v_1 = \left[\begin{array}{r} 4 \\ 0 \\ 2 \\ 1 \end{array} \right], \mathbf v_2 = \left[\begin{array}{r} 1 \\ -3 \\ 3 \\ 1 \end{array} \right], \mathbf v_3 = \left[\begin{array}{r} -2 \\ 1 \\ 1 \\ 0 \end{array} \right], \mathbf b = \left[\begin{array}{r} 0 \\ 1 \\ 2 \\ -2 \end{array} \right]\text{,} \end{equation*}, \begin{equation*} \begin{alignedat}{4} 3x_1 & {}+{} & 2x_2 & {}-{} x_3 & {}={} & 4 \\ x_1 & & & {}+{} 2x_3 & {}={} & 0 \\ -x_1 & {}-{} & x_2 & {}+{} 3x_3 & {}={} & 1 \\ \end{alignedat} \end{equation*}, \begin{equation*} \mathbf v_1 = \left[\begin{array}{r} 0 \\ -2 \\ 1 \\ \end{array} \right], \mathbf v_2 = \left[\begin{array}{r} 1 \\ 1 \\ -1 \\ \end{array} \right], \mathbf v_3 = \left[\begin{array}{r} 2 \\ 0 \\ -1 \\ \end{array} \right], \mathbf b = \left[\begin{array}{r} -1 \\ 3 \\ -1 \\ \end{array} \right]\text{.} For example, v = (2, -1), then also take \( e_1 = (1, 0), e_2 = (0, 1) \). column vectors (or row vectors), that is, matrices that have only one column
the answer to our question is affirmative. }\) Before computing, first explain how you know this product exists and then explain what the dimensions of the resulting matrix will be. Their product will be defined to be the linear combination of the columns of \(A\) using the components of \(\mathbf x\) as weights. \end{equation*}, \begin{equation*} \mathbf v_1 = \left[\begin{array}{r} 0 \\ -2 \\ 1 \\ \end{array} \right], \mathbf v_2 = \left[\begin{array}{r} 1 \\ 1 \\ -1 \\ \end{array} \right], \mathbf v_3 = \left[\begin{array}{r} 1 \\ -1 \\ -2 \\ \end{array} \right], \mathbf b = \left[\begin{array}{r} 0 \\ 8 \\ -4 \\ \end{array} \right]\text{.} At the same time, there are a few properties that hold for real numbers that do not hold for matrices.
}\), Use the Linearity Principle expressed in Proposition 2.2.3 to explain why \(\mathbf x_h+\mathbf x_p\) is a solution to the equation \(A\mathbf x = \mathbf b\text{. For math, science, nutrition, history . You can easily check that any of these linear combinations indeed
For our matrix \(A\text{,}\) find the row operations needed to find a row equivalent matrix \(U\) in triangular form. \end{equation*}, \begin{equation*} \mathbf v = \left[\begin{array}{r} 2 \\ 1 \end{array}\right], \mathbf w = \left[\begin{array}{r} 1 \\ 2 \end{array}\right] \end{equation*}, \begin{equation*} \begin{aligned} a\left[\begin{array}{r}2\\1\end{array}\right] + b\left[\begin{array}{r}1\\2\end{array}\right] & = \left[\begin{array}{r}-1\\4\end{array}\right] \\ \\ \left[\begin{array}{r}2a\\a\end{array}\right] + \left[\begin{array}{r}b\\2b\end{array}\right] & = \left[\begin{array}{r}-1\\4\end{array}\right] \\ \\ \left[\begin{array}{r}2a+b\\a+2b\end{array}\right] & = \left[\begin{array}{r}-1\\4\end{array}\right] \\ \end{aligned} \end{equation*}, \begin{equation*} \begin{alignedat}{3} 2a & {}+{} & b & {}={} & -1 \\ a & {}+{} & 2b & {}={} & 4 \\ \end{alignedat} \end{equation*}, \begin{equation*} \left[ \begin{array}{rr|r} 2 & 1 & -1 \\ 1 & 2 & 4 \end{array} \right] \sim \left[ \begin{array}{rr|r} 1 & 0 & -2 \\ 0 & 1 & 3 \end{array} \right]\text{,} \end{equation*}, \begin{equation*} -2\mathbf v + 3 \mathbf w = \mathbf b\text{.}
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linear combination matrix calculator