Suppose that A is row equivalent to B. Find bases for the null space of A and the column space of A.A=[12−511−324−51521204536−519−2]B=[12045005−780000−900000]

Since B is a row echelon form of A, we see that the first, third and fifth columns of A are its pivot columns. Thus a basis fo Col A is {[1213],[−5−50−5],[−325−2]} To find a basis for Nul A, we find the general solution of Ax=0 in terms of the free variables. Since it is row equivalent to B we can simply get reduced row echelon form of B: [12045005−780000−900000]R2→15R2R3→1−9R3 [12045005−7500000100000] R1→R1−5R3R2→R2−85R3[12045001−7500000100000] to get: x1=−2x2−4x4,x3=75x4, x5=0, with x2 and x4 free. So x=[x1x2x3x4x5]=x2[−21000]+x4[−407/510] And a basis for Nul A is {[−21000],[−407/510]} Result: Basis for Col A is: {[1213],[−5−50−5],[−325−2]} Basis for Nul A is {[−21000],[−407/510]}

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Algebra
Linear algebra
Matrix transformations

djeljenike

Answered question

2021-09-13

Suppose that A is row equivalent to B. Find bases for the null space of A and the column space of A.
$A = [\begin{matrix} 1 & 2 & - 5 & 11 & - 3 \\ 2 & 4 & - 5 & 15 & 2 \\ 1 & 2 & 0 & 4 & 5 \\ 3 & 6 & - 5 & 19 & - 2 \end{matrix}]$
$B = [\begin{matrix} 1 & 2 & 0 & 4 & 5 \\ 0 & 0 & 5 & - 7 & 8 \\ 0 & 0 & 0 & 0 & - 9 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}]$

Answer & Explanation

BleabyinfibiaG

Skilled2021-09-14Added 118 answers

Since B is a row echelon form of A, we see that the first, third and fifth columns of A are its pivot columns. Thus a basis fo Col A is
${[\begin{matrix} 1 \\ 2 \\ 1 \\ 3 \end{matrix}], [\begin{matrix} - 5 \\ - 5 \\ 0 \\ - 5 \end{matrix}], [\begin{matrix} - 3 \\ 2 \\ 5 \\ - 2 \end{matrix}]}$
To find a basis for Nul A, we find the general solution of Ax=0 in terms of the free variables. Since it is row equivalent to B we can simply get reduced row echelon form of B:
$[\begin{matrix} 1 & 2 & 0 & 4 & 5 \\ 0 & 0 & 5 & - 7 & 8 \\ 0 & 0 & 0 & 0 & - 9 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}] \begin{matrix} R_{2} \to \frac{1}{5} R_{2} \\ R_{3} \to \frac{1}{- 9} R_{3} \end{matrix} [\begin{matrix} 1 & 2 & 0 & 4 & 5 \\ 0 & 0 & 5 & \frac{- 7}{5} & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}]$
$\begin{matrix} R_{1} \to R_{1} - 5 R_{3} \\ R_{2} \to R_{2} - \frac{8}{5} R_{3} \end{matrix} [\begin{matrix} 1 & 2 & 0 & 4 & 5 \\ 0 & 0 & 1 & \frac{- 7}{5} & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}]$
to get: $x_{1} = - 2 x_{2} - 4 x_{4}, x_{3} = \frac{7}{5} x_{4}, x_{5} = 0$ , with $x_{2}$ and $x_{4}$ free. So
$x = [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{matrix}] = x_{2} [\begin{matrix} - 2 \\ 1 \\ 0 \\ 0 \\ 0 \end{matrix}] + x_{4} [\begin{matrix} - 4 \\ 0 \\ 7 / 5 \\ 1 \\ 0 \end{matrix}]$
And a basis for Nul A is
${[\begin{matrix} - 2 \\ 1 \\ 0 \\ 0 \\ 0 \end{matrix}], [\begin{matrix} - 4 \\ 0 \\ 7 / 5 \\ 1 \\ 0 \end{matrix}]}$
Result: Basis for Col A is:
${[\begin{matrix} 1 \\ 2 \\ 1 \\ 3 \end{matrix}], [\begin{matrix} - 5 \\ - 5 \\ 0 \\ - 5 \end{matrix}], [\begin{matrix} - 3 \\ 2 \\ 5 \\ - 2 \end{matrix}]}$
Basis for Nul A is
${[\begin{matrix} - 2 \\ 1 \\ 0 \\ 0 \\ 0 \end{matrix}], [\begin{matrix} - 4 \\ 0 \\ 7 / 5 \\ 1 \\ 0 \end{matrix}]}$

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