Could someone check my working for this?(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

(Assuming that t is Real).

For a matrix, A =

1 0 3 t

0 1 t 0

1 0 2 -2t

2 0 -2 t^{2}

and a vector, b = [5 2t 1 t]^{T}

i) Find the rank of A and A|b for all values of t.

ii) Find conditions on t so that Ax=b is consistent, Ax=b has a unique solution.

iii) Find the general solution to Ax=b when t=-22, stating the dimension of the nullspace and a basis for it.

2. Relevant equations

Basic linear algebra?

3. The attempt at a solution

i)row reducing A|b ->

[1 0 3 t .........|5]

[0 1 t 0 .........|2t]

[0 0 1 3t .......|4]

[0 0 0 t(t+22) |t+22]

rank(A) = 4 when t(t+22) =/= 0, so t =/= -22 and t =/= 0 and rank(A) = 3 if t = -22 or t = 0

rank(A|b) = 4 when t =/= -22 and rank(A|b) = 3 if t = -22 or t = 0

ii)A|b is consistent when t =/= 0

and A|b has a unique solution when t =/= -22 (when t=-22, solution will be parametric) and t =/= 0 (t=0 is inconsistent)

iii)For t=-22, A|0 =

[1 0 3 -22 .|0]

[0 1 -22 0 .|0]

[0 0 1 -66 .|0]

[0 0 0 0 ....|0]

So, the dimension of the null space is 1 and:

-> x_{3}= 66x_{4}

-> x_{2}= 22x_{3}

-> x_{1}= 22x_{4}- 3x_{3}

let x_{4}= 1/22

-> x = [-8 66 3 1/22]^{T}is a basis for the null space of A.

Is the above correct/enough working for an exam question?

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# Homework Help: Linear algebra - general solutions to Ax=b

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