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Math
Posted 2 months ago

Question 3 (18 marks)
a) For matrix H=[224203061213002]H=\left[\begin{array}{rrrrr}2 & -2 & 4 & -2 & 0 \\ 3 & 0 & 6 & -1 & 2 \\ 1 & 3 & 0 & 0 & 2\end{array}\right], find its rank, column space, and null space. (14 marks)
b) Let u=[210],v1=[231],v2=[203]u=\left[\begin{array}{c}-2 \\ -1 \\ 0\end{array}\right], v_{1}=\left[\begin{array}{l}2 \\ 3 \\ 1\end{array}\right], v_{2}=\left[\begin{array}{c}-2 \\ 0 \\ 3\end{array}\right], and v3=[460]v_{3}=\left[\begin{array}{l}4 \\ 6 \\ 0\end{array}\right]. Show that V={v1,v2,v3}V=\left\{v_{1}, v_{2}, v_{3}\right\} is a basis for R3R^{3}, and then find the coordinate vector of uu relative to the basis VV. ( 4 marks)
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Answer from Sia
Posted 2 months ago
Solution by Steps
step 1
To find the rank of matrix HH, we first need to reduce it to its row echelon form (REF)
step 2
The given matrix HH is: H=[224203061213002] H = \begin{bmatrix} 2 & -2 & 4 & -2 & 0 \\ 3 & 0 & 6 & -1 & 2 \\ 1 & 3 & 0 & 0 & 2 \end{bmatrix} We perform row operations to convert it to REF
step 3
Subtract 32\frac{3}{2} times the first row from the second row: R2R232R1 R_2 \leftarrow R_2 - \frac{3}{2}R_1 [224200302213002] \begin{bmatrix} 2 & -2 & 4 & -2 & 0 \\ 0 & 3 & 0 & 2 & 2 \\ 1 & 3 & 0 & 0 & 2 \end{bmatrix}
step 4
Subtract 12\frac{1}{2} times the first row from the third row: R3R312R1 R_3 \leftarrow R_3 - \frac{1}{2}R_1 [224200302204212] \begin{bmatrix} 2 & -2 & 4 & -2 & 0 \\ 0 & 3 & 0 & 2 & 2 \\ 0 & 4 & -2 & 1 & 2 \end{bmatrix}
step 5
Subtract 43\frac{4}{3} times the second row from the third row: R3R343R2 R_3 \leftarrow R_3 - \frac{4}{3}R_2 [22420030220025323] \begin{bmatrix} 2 & -2 & 4 & -2 & 0 \\ 0 & 3 & 0 & 2 & 2 \\ 0 & 0 & -2 & -\frac{5}{3} & -\frac{2}{3} \end{bmatrix}
step 6
The matrix is now in row echelon form. The non-zero rows are: [22420030220025323] \begin{bmatrix} 2 & -2 & 4 & -2 & 0 \\ 0 & 3 & 0 & 2 & 2 \\ 0 & 0 & -2 & -\frac{5}{3} & -\frac{2}{3} \end{bmatrix} Thus, the rank of HH is 3
step 7
The column space of HH is the span of the pivot columns in the original matrix: Column space=span{[231],[203],[460]} \text{Column space} = \text{span} \left\{ \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ 3 \end{bmatrix}, \begin{bmatrix} 4 \\ 6 \\ 0 \end{bmatrix} \right\}
step 8
To find the null space, we solve Hx=0H \mathbf{x} = \mathbf{0}: [224203061213002][x1x2x3x4x5]=[000] \begin{bmatrix} 2 & -2 & 4 & -2 & 0 \\ 3 & 0 & 6 & -1 & 2 \\ 1 & 3 & 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} From the REF, we get: x3=56x4+13x5 x_3 = \frac{5}{6}x_4 + \frac{1}{3}x_5 x2=23x423x5 x_2 = -\frac{2}{3}x_4 - \frac{2}{3}x_5 x1=13x4+13x5 x_1 = \frac{1}{3}x_4 + \frac{1}{3}x_5 Thus, the null space is: Null space=span{[13235610],[13231301]} \text{Null space} = \text{span} \left\{ \begin{bmatrix} \frac{1}{3} \\ -\frac{2}{3} \\ \frac{5}{6} \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} \frac{1}{3} \\ -\frac{2}{3} \\ \frac{1}{3} \\ 0 \\ 1 \end{bmatrix} \right\}
Answer
Rank: 3, Column space: span of {[231],[203],[460]}\left\{ \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ 3 \end{bmatrix}, \begin{bmatrix} 4 \\ 6 \\ 0 \end{bmatrix} \right\}, Null space: span of {[13235610],[13231301]}\left\{ \begin{bmatrix} \frac{1}{3} \\ -\frac{2}{3} \\ \frac{5}{6} \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} \frac{1}{3} \\ -\frac{2}{3} \\ \frac{1}{3} \\ 0 \\ 1 \end{bmatrix} \right\}
Key Concept
Rank, Column Space, and Null Space of a Matrix
Explanation
The rank of a matrix is the number of linearly independent rows or columns. The column space is the span of the pivot columns, and the null space is the set of solutions to the homogeneous equation Hx=0H \mathbf{x} = \mathbf{0}.
Solution by Steps
step 1
To show that V={v1,v2,v3}V = \{v_1, v_2, v_3\} is a basis for R3\mathbb{R}^3, we need to show that v1,v2,v3v_1, v_2, v_3 are linearly independent and span R3\mathbb{R}^3
step 2
The vectors are: v1=[231],v2=[203],v3=[460] v_1 = \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}, v_2 = \begin{bmatrix} -2 \\ 0 \\ 3 \end{bmatrix}, v_3 = \begin{bmatrix} 4 \\ 6 \\ 0 \end{bmatrix} We form the matrix AA with these vectors as columns: A=[224306130] A = \begin{bmatrix} 2 & -2 & 4 \\ 3 & 0 & 6 \\ 1 & 3 & 0 \end{bmatrix}
step 3
We find the determinant of AA to check for linear independence: det(A)=224306130 \det(A) = \begin{vmatrix} 2 & -2 & 4 \\ 3 & 0 & 6 \\ 1 & 3 & 0 \end{vmatrix} Expanding along the first row: det(A)=20630(2)3610+43013 \det(A) = 2 \begin{vmatrix} 0 & 6 \\ 3 & 0 \end{vmatrix} - (-2) \begin{vmatrix} 3 & 6 \\ 1 & 0 \end{vmatrix} + 4 \begin{vmatrix} 3 & 0 \\ 1 & 3 \end{vmatrix} =2(018)+2(06)+4(90) = 2(0 - 18) + 2(0 - 6) + 4(9 - 0) =3612+36=12 = -36 - 12 + 36 = -12 Since det(A)0\det(A) \neq 0, the vectors are linearly independent
step 4
Since v1,v2,v3v_1, v_2, v_3 are linearly independent and there are 3 vectors in R3\mathbb{R}^3, they span R3\mathbb{R}^3. Thus, VV is a basis for R3\mathbb{R}^3
step 5
To find the coordinate vector of uu relative to the basis VV, we express uu as a linear combination of v1,v2,v3v_1, v_2, v_3: u=c1v1+c2v2+c3v3 u = c_1 v_1 + c_2 v_2 + c_3 v_3 [210]=c1[231]+c2[203]+c3[460] \begin{bmatrix} -2 \\ -1 \\ 0 \end{bmatrix} = c_1 \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} -2 \\ 0 \\ 3 \end{bmatrix} + c_3 \begin{bmatrix} 4 \\ 6 \\ 0 \end{bmatrix}
step 6
This gives us the system of equations: 2c12c2+4c3=2 2c_1 - 2c_2 + 4c_3 = -2 3c1+6c3=1 3c_1 + 6c_3 = -1 c1+3c2=0 c_1 + 3c_2 = 0 Solving this system, we get: c1=13,c2=19,c3=118 c_1 = -\frac{1}{3}, c_2 = -\frac{1}{9}, c_3 = -\frac{1}{18} Thus, the coordinate vector of uu relative to the basis VV is: [u]V=[1319118] [u]_V = \begin{bmatrix} -\frac{1}{3} \\ -\frac{1}{9} \\ -\frac{1}{18} \end{bmatrix}
Answer
VV is a basis for R3\mathbb{R}^3, and the coordinate vector of uu relative to VV is [1319118]\begin{bmatrix} -\frac{1}{3} \\ -\frac{1}{9} \\ -\frac{1}{18} \end{bmatrix}
Key Concept
Basis and Coordinate Vectors
Explanation
A set of vectors forms a basis for a vector space if they are linearly independent and span the space. The coordinate vector of a vector relative to a basis is the unique combination of basis vectors that equals the vector.

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