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Matrices a level further maths

ja.19

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Can someone solve this ques and upload the picture of the solution

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Reply 1

joostan

Original post by ja.19

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Can someone solve this ques and upload the picture of the solution

Yes, but how would that help you?
Which part are you struggling on, and what have you tried?
For the first part the rank is the number of linearly independent rows, or equivalently columns of the matrix.

Reply 2

ja.19

The last part where we have to show that there is no vector that satisfies Ax=(0 2 -1 3)

Reply 3

joostan

Original post by ja.19

The last part where we have to show that there is no vector that satisfies Ax=(0 2 -1 3)

You don't need to show that no such vector can exist, only that it can't have all its coordinates positive.
So you suppose

\mathbf{x}=\mathbf{x_0}+\lambda \mathbf{e}

where:

\mathbf{x}=\left(\begin{array}{c} x \\ y \\ z \\ w \end{array}\right)

with

x,y,z,w>0

.
Sub in your values for

\mathbf{x_0},\mathbf{e}

and obtain a contradiction with the inequalities.

Spoiler

Reply 4

ja.19

Original post by joostan

You don't need to show that no such vector can exist, only that it can't have all its coordinates positive.
So you suppose

\mathbf{x}=\mathbf{x_0}+\lambda \mathbf{e}

where:

\mathbf{x}=\left(\begin{array}{c} x \\ y \\ z \\ w \end{array}\right)

with

x,y,z,w>0

.
Sub in your values for

\mathbf{x_0},\mathbf{e}

and obtain a contradiction with the inequalities.

Spoiler

Reply 5

ja.19

Can you help with part 3

Reply 6

joostan

Original post by ja.19

Can you help with part 3

Well if

\mathbf{x_0}

is the vector

(1,-3,1,-2)^T

. Then you know that if

\mathbf{v}

is a vector in the null space then:

H(\mathbf{x_0}+\lambda \mathbf{v})=H\mathbf{x_0}

, so you need to find minimise the magnitude of

\mathbf{x_0}+\lambda\mathbf{v}

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