91Ó°ÊÓ

Find all solutions of \(A \mathbf{x}=\mathbf{0}\) for the matrices given. Express your answer in parametric form. \(A=\left(\begin{array}{rrr}1 & 0 & -2 \\ 0 & 1 & 3 \\ 0 & 0 & 0\end{array}\right)\)

For each system, perform each of the following tasks. All work is to be done by hand (pencil-and-paper calculations only). (i) Set up the augmented matrix for the system; then place the augmented matrix in row echelon form. (ii) If the system is inconsistent, so state, and explain why. Otherwise, proceed to the next item. (iii) Use back-solving to find the solution. Place the final solution in parametric form. \(x_{1}+2 x_{2}+2 x_{3}=2\) \(-x_{1}-x_{2}+2 x_{3}=4\) \(x_{1}+3 x_{2}+6 x_{3}=7\)

Consider the following collection of vectors, which you are to use. $$ \begin{array}{lll} \mathbf{u}_{1}=(1,-2)^{T}, & \mathbf{u}_{2}=(3,0)^{T}, & \mathbf{u}_{3}=(2,-4)^{T} \\ \mathbf{v}_{1}=(1,-4,4)^{T}, & \mathbf{v}_{2}=(0,-2,1)^{T}, & \mathbf{v}_{3}=(1,-2,3)^{T} \end{array} $$ In each exercise, if the given vector \(w\) lies in the span, provide a specific linear combination of the spanning vectors that equals the given vector; otherwise, provide a specific numerical argument why the given vector does not lie in the span. Is the vector \(\mathbf{w}=(-3,2,7)^{T}\) in the span \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) ?

For each system, perform each of the following tasks. All work is to be done by hand (pencil-and-paper calculations only). (i) Set up the augmented matrix for the system; then place the augmented matrix in row echelon form. (ii) If the system is inconsistent, so state, and explain why. Otherwise, proceed to the next item. (iii) Use back-solving to find the solution. Place the final solution in parametric form. \(x_{1}+x_{2}-2 x_{3}=2\) \(3 x_{1}-x_{2}-x_{3}=3\) \(5 x_{1}+x_{2}-5 x_{3}=7\)

Sketch the parallelogram spanned by the vectors \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) on graph paper. Estimate the area of your parallelogram using your sketch. Finally, compute the determinant of the matrix \(\left[\mathbf{v}_{1}, \mathbf{v}_{2}\right]\) and compare with your estimate. \(\mathbf{v}_{1}=\left(\begin{array}{r}-2 \\ 5\end{array}\right), \mathbf{v}_{2}=\left(\begin{array}{l}4 \\ 3\end{array}\right)\)

Find all solutions of \(A \mathbf{x}=\mathbf{0}\) for the matrices given. Express your answer in parametric form. \(A=\left(\begin{array}{rrr}1 & 0 & -5 \\ 0 & 1 & 2 \\ 0 & 0 & 0\end{array}\right)\)

Consider the following collection of vectors, which you are to use. $$ \begin{array}{lll} \mathbf{u}_{1}=(1,-2)^{T}, & \mathbf{u}_{2}=(3,0)^{T}, & \mathbf{u}_{3}=(2,-4)^{T} \\ \mathbf{v}_{1}=(1,-4,4)^{T}, & \mathbf{v}_{2}=(0,-2,1)^{T}, & \mathbf{v}_{3}=(1,-2,3)^{T} \end{array} $$ In each exercise, if the given vector \(w\) lies in the span, provide a specific linear combination of the spanning vectors that equals the given vector; otherwise, provide a specific numerical argument why the given vector does not lie in the span. Is the vector \(\mathbf{w}=(1,4,1)^{T}\) in the \(\operatorname{span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{3}\right\\}\) ?

Find all solutions of \(A \mathbf{x}=\mathbf{0}\) for the matrices given. Express your answer in parametric form. \(A=\left(\begin{array}{llll}1 & 0 & 2 & -2 \\ 0 & 1 & 3 & -1\end{array}\right)\)

Sketch the parallelogram spanned by the vectors \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) on graph paper. Estimate the area of your parallelogram using your sketch. Finally, compute the determinant of the matrix \(\left[\mathbf{v}_{1}, \mathbf{v}_{2}\right]\) and compare with your estimate. \(\mathbf{v}_{1}=\left(\begin{array}{l}5 \\ 5\end{array}\right), \mathbf{v}_{2}=\left(\begin{array}{r}-2 \\ 6\end{array}\right)\)

For each system, perform each of the following tasks. All work is to be done by hand (pencil-and-paper calculations only). (i) Set up the augmented matrix for the system; then place the augmented matrix in row echelon form. (ii) If the system is inconsistent, so state, and explain why. Otherwise, proceed to the next item. (iii) Use back-solving to find the solution. Place the final solution in parametric form. \(\begin{aligned} x_{1}+2 x_{2}-2 x_{3} &=6 \\ 2 x_{1}+4 x_{2}-4 x_{3} &=12 \\\ 3 x_{1}+6 x_{2}-6 x_{3} &=18 \end{aligned}\)