To write the vector \( w \) as a linear combination of vectors \( u \) and \( v \), we need to find scalars \( a \) and \( b \) such that \( w = au + bv \). Given vectors: \[ u = \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \quad v = \begin{pmatrix} 2 \\ -1 \end{pmatrix}, \quad w = \begin{pmatrix} -5 \\ 1 \end{pmatrix} \] We want to solve the following system of equations for \( a \) and \( b \): \[ a(1) + b(2) = -5 \] \[ a(1) + b(-1) = 1 \] This can be written as two simultaneous equations: \[ a + 2b = -5 \] \[ a - b = 1 \] Solving for \( a \) and \( b \), we can subtract the second equation from the first to eliminate \( a \): \[ (a + 2b) - (a - b) = -5 - 1 \] \[ 3b = -6 \] \[ b = -2 \] Now we can substitute \( b \) into the second equation to find \( a \): \[ a - (-2) = 1 \] \[ a + 2 = 1 \] \[ a = -1 \] Therefore, the vector \( w \) can be written as a linear combination of \( u \) and \( v \) as follows: \[ w = -1u - 2v \] \[ w = -1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} - 2 \begin{pmatrix} 2 \\ -1 \end{pmatrix} \] \[ w = \begin{pmatrix} -1 \\ -1 \end{pmatrix} + \begin{pmatrix} -4 \\ 2 \end{pmatrix} \] \[ w = \begin{pmatrix} -5 \\ 1 \end{pmatrix} \] This confirms that \( w \) is indeed a linear combination of \( u \) and \( v \) with the coefficients \( a = -1 \) and \( b = -2 \).

Let's solve the given homework question step by step: (a) Write the vector \( w \) as a linear combination of \( u \) and \( v \). Given vectors: \[ u = \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \quad v = \begin{pmatrix} 2 \\ -1 \end{pmatrix}, \quad w = \begin{pmatrix} -5 \\ 1 \end{pmatrix} \] We want to find scalars \( a \) and \( b \) such that \( w = au + bv \). This gives us the following system of equations: \[ a(1) + b(2) = -5 \] \[ a(1) + b(-1) = 1 \] Solving this system, we get: \[ a + 2b = -5 \] \[ a - b = 1 \] Adding the two equations, we get: \[ 2a + b = -4 \] \[ a = -\frac{5}{2} \] Substituting \( a \) back into the second equation: \[ -\frac{5}{2} - b = 1 \] \[ b = -\frac{7}{2} \] So, \( w \) can be written as a linear combination of \( u \) and \( v \) with \( a = -\frac{5}{2} \) and \( b = -\frac{7}{2} \): \[ w = -\frac{5}{2}u - \frac{7}{2}v \] (b) Are the vectors \( u \), \( v \), and \( w \) linearly independent? Vectors are linearly independent if the only solution to the equation \( \alpha u + \beta v + \gamma w = 0 \) is \( \alpha = \beta = \gamma = 0 \). Since we were able to express \( w \) as a linear combination of \( u \) and \( v \), this means that there exists a non-trivial solution to the equation \( \alpha u + \beta v + \gamma w = 0 \) (specifically, \( \alpha = -\frac{5}{2} \), \( \beta = -\frac{7}{2} \), and \( \gamma = 1 \)). Therefore, \( u \), \( v \), and \( w \) are not linearly independent. (c) Show that span(\( \{u, v, w\} \)) = \( \mathbb{R}^2 \) by showing that an arbitrary vector in \( \mathbb{R}^2 \) can always be expressed as a linear combination of \( u \), \( v \), \( w \). Since \( u \) and \( v \) are linearly independent (you can verify this by checking that the matrix formed by \( u \) and \( v \) has a non-zero determinant), any vector in \( \mathbb{R}^2 \) can be expressed as a linear combination of \( u \) and \( v \). Therefore, the span of \( \{u, v\} \) is \( \mathbb{R}^2 \), and since \( w \) is also a linear combination of \( u \) and \( v \), the span of \( \{u, v, w\} \) is still \( \mathbb{R}^2 \). (d) What is the dimension of the vector space span(\( \{u, v, w\} \))? The dimension of a vector space is the number of vectors in its basis. Since \( u \) and \( v \) are linearly independent and span \( \mathbb{R}^2 \), they form a basis for \( \mathbb{R}^2 \). The vector \( w \) does not add any new dimension because it is a linear combination of \( u \) and \( v \). Therefore, the dimension of the vector space span(\( \{u, v, w\} \)) is 2, which is the same as the dimension of \( \mathbb{R}^2 \).

Given vector U = 2i + 3j and vector W = 5j, what is 3U + W?Group of answer choices11i - 9j6i + 14j11i + 9j6i - 14j

To show that the span of vectors \( u \), \( v \), and \( w \) is equal to \( \mathbb{R}^2 \), we need to demonstrate that any arbitrary vector in \( \mathbb{R}^2 \) can be expressed as a linear combination of \( u \), \( v \), and \( w \). Let's consider an arbitrary vector \( x \) in \( \mathbb{R}^2 \): \[ x = \begin{pmatrix} a \\ b \end{pmatrix} \] We need to find scalars \( \alpha \), \( \beta \), and \( \gamma \) such that: \[ x = \alpha u + \beta v + \gamma w \] Given vectors: \[ u = \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \quad v = \begin{pmatrix} 2 \\ -1 \end{pmatrix}, \quad w = \begin{pmatrix} -5 \\ 1 \end{pmatrix} \] Substituting the given vectors into the equation, we get: \[ \begin{pmatrix} a \\ b \end{pmatrix} = \alpha \begin{pmatrix} 1 \\ 1 \end{pmatrix} + \beta \begin{pmatrix} 2 \\ -1 \end{pmatrix} + \gamma \begin{pmatrix} -5 \\ 1 \end{pmatrix} \] This leads to the following system of equations: \[ \alpha + 2\beta - 5\gamma = a \] \[ \alpha - \beta + \gamma = b \] Since we have two equations and three unknowns, we can express one of the unknowns, say \( \gamma \), in terms of \( \alpha \) and \( \beta \), and then solve for \( \alpha \) and \( \beta \) to express any vector \( x \) as a linear combination of \( u \), \( v \), and \( w \). However, we have already established that \( w \) can be written as a linear combination of \( u \) and \( v \) (from the previous answer, \( w = -1u - 2v \)). This means that any linear combination of \( u \), \( v \), and \( w \) can be reduced to a linear combination of just \( u \) and \( v \). Since \( u \) and \( v \) are linearly independent, they span \( \mathbb{R}^2 \), and thus the span of \( u \), \( v \), and \( w \) is also \( \mathbb{R}^2 \). In conclusion, the span of \( u \), \( v \), and \( w \) is \( \mathbb{R}^2 \) because \( u \) and \( v \) are sufficient to express any vector in \( \mathbb{R}^2 \), and \( w \) does not add any new dimension to the span.

To determine if the vectors \( u \), \( v \), and \( w \) are linearly dependent, we need to check if there exists a non-trivial linear combination of these vectors that equals the zero vector. In other words, we need to find scalars \( a \), \( b \), and \( c \) (not all zero) such that: \[ a \cdot u + b \cdot v + c \cdot w = 0 \] Given vectors: \[ u = \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \quad v = \begin{pmatrix} 2 \\ -1 \end{pmatrix}, \quad w = \begin{pmatrix} -5 \\ 1 \end{pmatrix} \] We have already found that \( w \) can be written as a linear combination of \( u \) and \( v \) with specific scalars \( a = -1 \) and \( b = -2 \): \[ w = -1 \cdot u - 2 \cdot v \] This equation shows that there is a non-trivial linear combination of \( u \), \( v \), and \( w \) that equals the zero vector: \[ -1 \cdot u - 2 \cdot v + 1 \cdot w = 0 \] Since we have found a non-trivial solution where the scalars are not all zero, the vectors \( u \), \( v \), and \( w \) are linearly dependent. The fact that \( w \) can be expressed as a linear combination of \( u \) and \( v \) is sufficient to establish their linear dependence.

Let W1 = L{(1, 1, 0), (−1, 1, 0)} and W2 = L{(1, 0, 2), (−1, 0, 4)}. Show that W1 +W2 = R3. Give an example of a vector v ∈ R3 such that v can be written in twodifferent ways in the form v = v1 + v2, where v1 ∈ W1, v2 ∈ W2