Determine the null space of each of the following matrices: [11−1222−31−1−10−5]

Step 1 Solve the given system Ax=0 using Gauss-jordan reduction method. Step 2 Obtain the augmented matrix as shown below: [11−12|022−31|0−1−10−5|0]∼[11−12|000−1−3|0−1−10−5|0](R2→R2−2R1) ∼[11−12|000−1−3|000−1−3|0](R3→R3+R1) ∼[11−12|000−1−3|00000|0](R3→R3−R2) ∼[1105|000−1−3|00000|0](R1→R1−R2) ∼[1105|00013|00000|0](R2→−R2) Step 3 Then, [110500130000][x1x2x3]=0 x1+x2+5x4=0 x3+3x4=0 Here the free variables are x2,x4 thus , on setting x2=α,x4=β we have , x1=−α−5β,x3=−3β Thus , the null space N(A) contain all the vectors of the form: (−α−5β,α,−3β,β)T=α(−1.1.0.0)T+β(−5,0,−3,1)T,α,β∈R

Key to "Determine the null space of each of the..."

Jaya Legge

Answered question

2021-02-21

Determine the null space of each of the following matrices:

[\begin{matrix} 1 & 1 & - 1 & 2 \\ 2 & 2 & - 3 & 1 \\ - 1 & - 1 & 0 & - 5 \end{matrix}]

Answer & Explanation

Talisha

Skilled2021-02-22Added 93 answers

Step 1
Solve the given system Ax=0 using Gauss-jordan reduction method.
Step 2
Obtain the augmented matrix as shown below:

[\begin{matrix} 1 & 1 & - 1 & 2 & | & 0 \\ 2 & 2 & - 3 & 1 & | & 0 \\ - 1 & - 1 & 0 & - 5 & | & 0 \end{matrix}] \sim [\begin{matrix} 1 & 1 & - 1 & 2 & | & 0 \\ 0 & 0 & - 1 & - 3 & | & 0 \\ - 1 & - 1 & 0 & - 5 & | & 0 \end{matrix}] (R_{2} \to R_{2} - 2 R_{1})

\sim [\begin{matrix} 1 & 1 & - 1 & 2 & | & 0 \\ 0 & 0 & - 1 & - 3 & | & 0 \\ 0 & 0 & - 1 & - 3 & | & 0 \end{matrix}] (R_{3} \to R_{3} + R_{1})

\sim [\begin{matrix} 1 & 1 & - 1 & 2 & | & 0 \\ 0 & 0 & - 1 & - 3 & | & 0 \\ 0 & 0 & 0 & 0 & | & 0 \end{matrix}] (R_{3} \to R_{3} - R_{2})

\sim [\begin{matrix} 1 & 1 & 0 & 5 & | & 0 \\ 0 & 0 & - 1 & - 3 & | & 0 \\ 0 & 0 & 0 & 0 & | & 0 \end{matrix}] (R_{1} \to R_{1} - R_{2})

\sim [\begin{matrix} 1 & 1 & 0 & 5 & | & 0 \\ 0 & 0 & 1 & 3 & | & 0 \\ 0 & 0 & 0 & 0 & | & 0 \end{matrix}] (R_{2} \to - R_{2})

Step 3
Then,

[\begin{matrix} 1 & 1 & 0 & 5 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = 0

x_{1} + x_{2} + 5 x_{4} = 0

x_{3} + 3 x_{4} = 0

Here the free variables are

x_{2}, x_{4}

thus , on setting

x_{2} = α, x_{4} = β

we have ,

x_{1} = - α - 5 β, x_{3} = - 3 β

Thus , the null space N(A) contain all the vectors of the form:

(- α - 5 β, α, - 3 β, β)^{T} = α (- 1.1 .0 .0)^{T} + β (- 5, 0, - 3, 1)^{T}, α, β \in R

Jeffrey Jordon

Expert2022-01-27Added 2605 answers

Answer is given below (on video)

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