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

MATH152 April 2016

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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 $A$?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

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

Hint 2
Uniqueness of row reduced echelon form.

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 ${\begin{bmatrix}1&0&-7&0\\0&1&0&-1\\0&0&0&0\\\end{bmatrix}}$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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: ${\begin{bmatrix}1&0&-7&0\\0&1&0&-1\\0&0&0&0\\\end{bmatrix}}\cdot {\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{bmatrix}}={\begin{bmatrix}0\\0\\0\\0\end{bmatrix}}$

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