Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 27

MATH152 April 2016

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[hide] Question A 27
The set of solutions of the homogeneous system $A{\textbf {x}}={\textbf {0}}$ can be written in parametric form ${\textbf {x}}=[0,1,0,1]t+[7,0,1,0]s.$ If $A$ is a $3\times 4$ matrix, what is the reduced row echelon form of $A$ ?

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[show] Hint 1
Number of variables in the parametric form tells us the number of pivot columns we have for the row reduced echelon form.

[show] Hint 2
Uniqueness of row reduced echelon form.

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[show] Solution 1
The null space of A is spanned by $[0,1,0,1]^{T},[7,0,1,0]^{T}$ By Rank–nullity theorem, the rank of A is 2. In the other words, the dimension of space spanned by column vectors are 2. The row reduced echelon form of a matrix is unique. There are two pivot columns. The range of A is the span of column space. From the kernels, we know the third column are seven multiples of the first column, and the fourth column are -1 multiple of the second column of A. Thus, the row reduced echelon form is ${\begin{bmatrix}1&0&-7&0\\0&1&0&-1\\0&0&0&0\\\end{bmatrix}}$

[show] Solution 2
Since the solution is $x=[0,1,0,1]^{T}s+[7,0,1,0]^{T}t$ , we can rewrite it as $x={\begin{bmatrix}7t\\s\\t\\s\end{bmatrix}}={\begin{bmatrix}7x_{3}\\x_{2}\\x_{3}\\x_{2}\end{bmatrix}}$ by changing variables. It means that when we transform the reduced echelon form to equations, we already have $x_{1}=7x_{3},\ \ \ \ x_{4}=x_{2}$ Or equivalently $x_{1}-7x_{3}=0,\ \ \ \ x_{2}-x_{4}=0$ So the echelon form is easily seen to be $\color {blue}{\begin{bmatrix}1&0&-7&0\\0&1&0&-1\\0&0&0&0\\\end{bmatrix}}$ Since when we transform following matrix system to equations, we can attain above equations we want: ${\begin{bmatrix}1&0&-7&0\\0&1&0&-1\\0&0&0&0\\\end{bmatrix}}\cdot {\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{bmatrix}}={\begin{bmatrix}0\\0\\0\\0\end{bmatrix}}$