# If [a 1 0] = [1 0 0][a 1 0] [1 4 1] [b 1 0][0 a 1] [0 1 4] [0 b 1][0 0 a]and a>1, then (a,b)=?MATRICES This is algebra homework, I'm not sure how to do...

If

[a 1 0] = [1 0 0][a 1 0]

[1 4 1] [b 1 0][0 a 1]

[0 1 4] [0 b 1][0 0 a]

and a>1, then (a,b)=?

MATRICES

This is algebra homework, I'm not sure how to do it. Can you please help me?

*print*Print*list*Cite

### 2 Answers

You need to perform the multiplication of matrices to the right such that:

`((1,0,0),(b,1,0),(0,b,1))((a,1,0),(0,a,1),(0,0,a))` = `((a,1,0),(ab,a+b,1),(0,ab,b+a))`

You need to equate the terms that correspond to like positions in matrices both sides, such that:

`a = a`

`1 = 1`

`0 = 0`

`ab = 1`

`a + b = 4 => a = 4 - b`

You need to substitute `4 - b` for `a` in the product `ab = 1` such that:

`(4-b)b = 1 => 4b - b^2 - 1 = 0 => -b^2 + 4b - 1 = 0`

Multiplying by `-1` yields:

`b^2 - 4b + 1 = 0`

You need to use quadratic formula such that:

`b_(1,2) = (4+-sqrt(16 - 4))/2 => b_(1,2) = (4+-sqrt12)/2`

`b_(1,2) = 2+-sqrt3`

`a_(1,2) = 4 - (2+-sqrt3) => a_(1,2) = 2-+sqrt3`

**Hence, evaluating a and b yields `(2+sqrt3,2-sqrt3)` and `(2-sqrt3,2+sqrt3).` **

### User Comments

Given,

{{a, 1, 0} {{1, 0, 0} {{a, 1, 0}

{1, 4, 1} = {b, 1, 0} x {0, a, 1}

{0, 1, 4}} {0, b, 1}} {0, 0, a}}

=>{{a, 1, 0} {{a, 1, 0}

{1, 4, 1} = {ab, a+b, 1}

{0, 1, 4}} {0, ab, a+b}}

Equating element by element we get:

a = a, 1=1, 0=0

1 = ab, 4=a+b, 1=1

0 = 0, 1 =ab, 4=a+b

Useful equations are:

ab = 1

=> b = 1/a ------(1)

and

a + b =4

=> a + 1/a = 4

=> a^2 + 1 = 4 a

=> a^2 - 4a + 1 = 0

a = [4 + Sqrt{ 16 - 4 }]/2 or [4 - Sqrt{ 16 - 4 }]/2

=[2 + 2*Sqrt{3}] or [2 - 2*Sqrt{3}]

But given that

a> 1

so the only acceptable result out of the above two is:

a = [2+2*Sqrt{3}]

=2* [ 1 + Sqrt{3}]

**Answer: a = =2* [ 1 + Sqrt{3}]**