Linear Algebra

16 Chapter One. Linear Systems2.12 Example⎛ ⎞ ⎛ ⎞⎛⎜2⎞ ⎛ ⎞ ⎛ ⎞ ⎛3 2 + 3⎟ ⎜ ⎟ ⎜ ⎟ ⎜5⎞1 7⎟4⎝3⎠ + ⎝−1⎠ = ⎝3 − 1⎠ = ⎝2⎠ 7 · ⎜ ⎟⎝−1⎠ = 28⎜ ⎟⎝ −7⎠1 4 1 + 4 5−3 −21Notice that the definitions of vector addition and scalar multiplication agreewhere they overlap, for instance, ⃗v + ⃗v = 2⃗v.With the notation defined, we can now solve systems in the way that we willuse from now on.2.13 Example This system2x + y − w = 4y + w + u = 4x − z + 2w = 0reduces in this way.⎛⎞2 1 0 −1 0 4⎜⎟⎝0 1 0 1 1 41 0 −1 2 0 0−→⎠ −(1/2)ρ 1+ρ 3(1/2)ρ 2 +ρ 3−→⎛⎞2 1 0 −1 0 4⎜⎟⎝0 1 0 1 1 4⎠0 −1/2 −1 5/2 0 −2⎛⎞2 1 0 −1 0 4⎜⎟⎝0 1 0 1 1 4⎠0 0 −1 3 1/2 0The solution set is {(w + (1/2)u, 4 − w − u, 3w + (1/2)u, w, u) ∣ w, u ∈ R}. Wewrite that in vector form.⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞x 0 1 1/2y4−1−1{z=0+3w +1/2u ∣ w, u ∈ R}⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝w⎠⎝0⎠⎝ 1⎠⎝ 0⎠u 0 0 1Note how well vector notation sets off the coefficients of each parameter. Forinstance, the third row of the vector form shows plainly that if u is fixed then zincreases three times as fast as w. Another thing shown plainly is that settingboth w and u to zero gives that this vector⎛ ⎞ ⎛ ⎞x 0y4⎜ z ⎟ = ⎜0⎟⎜⎝wu⎟⎠⎜ ⎟⎝ ⎠is a particular solution of the linear system.00