# Let vector x = <x_1, x_2> , vector y = <y_1, y_2, y_3> and vector z = <z_1, z_2>. Say that the systems x_1 = 2y_1 + 3y_2 - 5y_3 x_2 = -3y_1 - 4y_2 + 2y_3 and y_1 = 2z_1 - 3z_2 y_2...

Let vector x = <x_1, x_2> , vector y = <y_1, y_2, y_3> and vector z = <z_1, z_2>.

Say that the systems

x_1 = 2y_1 + 3y_2 - 5y_3

x_2 = -3y_1 - 4y_2 + 2y_3

and

y_1 = 2z_1 - 3z_2

y_2 = 3z_1 - z_2

y_3 = z_1 + z_2

Use the matrix product to write vector x in terms of vector y and vector y in terms of vector z.

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You need to replace `2z_1 - 3z_2` for `y_1, 3z_1 - z_2` for `y_2` and `z_1 + z_2` for `y_3` in equations `x_1 = 2y_1 + 3y_2 - 5y_3` and `x_2 = -3y_1 - 4y_2 + 2y_3` , such that:

`{(x_1 = 2(2z_1 - 3z_2) + 3(3z_1 - z_2) - 5(z_1 + z_2)),(x_2 = -3(2z_1 - 3z_2) - 4(3z_1 - z_2) + 2(z_1 + z_2)):}`

Performing the algebraic operations, yields:

`{(x_1 = 8z_1 - 14z_2),(x_2 = -16z_1 + 15z_2):}`

**You may convert now the system of equations in a matrices product to write `bar x` in terms of `bar z` , such that:**

`((x_1),(x_2)) = ((8,-14),(-16,15))*((z_1),(z_2))`

`[[x_1],[x_2]]=[[2,3,-5],[-3,-4,2]][[y_1],[y_2],[y_3]]` (i)

and

`[[y_1],[y_2],[y_3]]=[[2,-3],[3,-1],[1,1]][[z_1],[z_2]]` (ii)

substitute column matrix Y from (ii) in (i) ,we have

`[[x_1],[x_2]]=[[2,3,-5],[-3,-4,2]][[2,-3],[3,-1],[1,1]][[z_1],[z_2]] `

`=[[4+9-5,-6-3-5],[-6-12+2,9+4+2]][[z_1],[z_2]]`

`=[[8,-14],[-16,15]][[z_1],[z_2]]`

Ans.