Find a nonzero vector in Nul A and a nonzero vector in Col A for the matrix A below.A=[235−9−8−9−11214−3−1727]In Nul A, locate a nonzero vector.A=[−3201]

Consider the following matrix, A=[235−9−8−9−11214−3−1727] Find a non-zero vector in Nul A. Use row reduced method. A=[13/25/2−9/2−8−9−11214−3−1727]R1→12R1 ≈[13/25/2−9/2039−150−9−2745]R2→R2+8R1,R3→R3−4R1 ≈[13/25/2−9/2039−150000]R3→R3+3R2 ≈[10−23013−50000]R1→R1−12R2 The matrix has 2 pivot columns and 2 free columns. Then, AX=0 [10−23013−50000][x1x2x3x4]=[000] x1−2x3+3x4=0⇒x1=2x3−3x4 x2+3x3−5x4=0⇒x2=−3x3+5x4 So, the general solution is, [10−23013−50000][x1x2x3x4]=[000] Then, a vector parametric solution is, {x1=2x3−3x4x2=−3x3+5x4x3 and x4 are free variables [x1x2x3x4]=[2x3−3x4−3x3+5x4x3x4]

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College
Algebra
Linear algebra
Matrix transformations

Amari Flowers

Answered question

2021-09-30

Find a nonzero vector in Nul A and a nonzero vector in Col A for the matrix A below.
$A = [\begin{matrix} 2 & 3 & 5 & - 9 \\ - 8 & - 9 & - 11 & 21 \\ 4 & - 3 & - 17 & 27 \end{matrix}]$
In Nul A, locate a nonzero vector.
$A = [\begin{matrix} - 3 \\ 2 \\ 0 \\ 1 \end{matrix}]$

Answer & Explanation

yagombyeR

Skilled2021-10-01Added 92 answers

Consider the following matrix,
$A = [\begin{matrix} 2 & 3 & 5 & - 9 \\ - 8 & - 9 & - 11 & 21 \\ 4 & - 3 & - 17 & 27 \end{matrix}]$
Find a non-zero vector in Nul A.
Use row reduced method.
$A = [\begin{matrix} 1 & 3 / 2 & 5 / 2 & - 9 / 2 \\ - 8 & - 9 & - 11 & 21 \\ 4 & - 3 & - 17 & 27 \end{matrix}] R_{1} \to \frac{1}{2} R_{1}$
$\approx [\begin{matrix} 1 & 3 / 2 & 5 / 2 & - 9 / 2 \\ 0 & 3 & 9 & - 15 \\ 0 & - 9 & - 27 & 45 \end{matrix}] R_{2} \to R_{2} + 8 R_{1}, R_{3} \to R_{3} - 4 R_{1}$
$\approx [\begin{matrix} 1 & 3 / 2 & 5 / 2 & - 9 / 2 \\ 0 & 3 & 9 & - 15 \\ 0 & 0 & 0 & 0 \end{matrix}] R_{3} \to R_{3} + 3 R_{2}$
$\approx [\begin{matrix} 1 & 0 & - 2 & 3 \\ 0 & 1 & 3 & - 5 \\ 0 & 0 & 0 & 0 \end{matrix}] R_{1} \to R_{1} - \frac{1}{2} R_{2}$
The matrix has 2 pivot columns and 2 free columns.
Then, AX=0
$[\begin{matrix} 1 & 0 & - 2 & 3 \\ 0 & 1 & 3 & - 5 \\ 0 & 0 & 0 & 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$
$x_{1} - 2 x_{3} + 3 x_{4} = 0 \Rightarrow x_{1} = 2 x_{3} - 3 x_{4}$
$x_{2} + 3 x_{3} - 5 x_{4} = 0 \Rightarrow x_{2} = - 3 x_{3} + 5 x_{4}$
So, the general solution is,
$[\begin{matrix} 1 & 0 & - 2 & 3 \\ 0 & 1 & 3 & - 5 \\ 0 & 0 & 0 & 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$
Then, a vector parametric solution is,
${\begin{cases} x_{1} = 2 x_{3} - 3 x_{4} \\ x_{2} = - 3 x_{3} + 5 x_{4} \\ x_{3} a n d x_{4} are free variables \end{cases}$
$[\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix}] = [\begin{matrix} 2 x_{3} - 3 x_{4} \\ - 3 x_{3} + 5 x_{4} \\ x_{3} \\ x_{4} \end{matrix}]$

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