Please help me with this question.
The question is to find the rank of the matrix and then the basis of the null space, I first put the matrix A in reduced row echelon form and then I wrote the operations that I did at the side so it's clear how I got the values, I got the answer as rank = 2 because there r two pivot entry columns, but in the marking scheme they tell that the rank is 2 because there r two rows of 0, but this is something we never learnt in school , I was taught to put in rref(A) and make a entry of a column 1 and the rest 0. And I also want to ask is that whether we can multiply a row by a scalar multiple ? For example can I multiply row 2 by -1 to get 0 1 -3 4 ?
Best Answer
You have the correct rank of $A$, indeed $\text{rank}(A) = 2$. First, you have the row-reduced echelon form incorrect...it should be $$\begin{pmatrix} 1 & 0 & -5 & 1 \\ 0 & 1 & -3 & 4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \,.$$ In regards to your question about multiplying rows by scalars in the process of row reduction, this is perfectly fine. Writing the above in parametric form, we have $$\begin{cases} x_1 &= \,\,\,\, 5x_3-x_4 \\ x_2 &= \,\,\,\,3x_3 -4x_4 \end{cases}$$ with $x_3,x_4$ free. This is equivalent to the vector solution $$\vec{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = \begin{pmatrix} 5 \\3 \\1\\0 \end{pmatrix}x_3 + \begin{pmatrix} -1 \\ -4 \\0\\1 \end{pmatrix}x_4 \,\,.$$ Thus, the basis for the null space that you were seeking is $$ \mathscr{B} =\left\{ \begin{pmatrix} 5 \\3 \\1\\0 \end{pmatrix},\begin{pmatrix} -1 \\ -4 \\0\\1 \end{pmatrix} \right\} \,.$$