Linear Algebra Exercises-n-Answers.pdf
Linear Algebra Exercises-n-Answers.pdf
Linear Algebra Exercises-n-Answers.pdf
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<strong>Answers</strong> to <strong>Exercises</strong> 11<br />
(d) This reduction<br />
⎛<br />
⎝ 2 1 −1 2<br />
⎞<br />
2 0 1 3⎠<br />
1 −1 0 0<br />
−ρ1+ρ2<br />
−→<br />
−(1/2)ρ 1 +ρ 3<br />
shows that the solution set is a singleton set.<br />
(e) This reduction is easy<br />
⎛<br />
⎝ 1 2 −1 0 3<br />
⎞<br />
2 1 0 1 4<br />
1 −1 1 1 1<br />
⎠ −2ρ1+ρ2<br />
−→<br />
−ρ 1+ρ 3<br />
⎛<br />
⎝ 2 1 −1 2<br />
⎞<br />
0 −1 2 1<br />
0 −3/2 1/2 −1<br />
⎛<br />
⎞<br />
{ ⎝ 1 1⎠}<br />
1<br />
⎛<br />
⎝ 1 2 −1 0 3<br />
⎞<br />
0 −3 2 1 −2<br />
0 −3 2 1 −2<br />
⎠ (−3/2)ρ2+ρ3<br />
−→<br />
⎠ −ρ2+ρ3<br />
−→<br />
⎛<br />
⎝ 2 0 1 −1 −1 2 2<br />
1<br />
⎞<br />
⎠<br />
0 0 −5/2 −5/2<br />
⎛<br />
⎝ 1 2 −1 0 3<br />
⎞<br />
0 −3 2 1 −2⎠<br />
0 0 0 0 0<br />
and ends with x and y leading, while z and w are free. Solving for y gives y = (2 + 2z + w)/3 and<br />
substitution shows that x + 2(2 + 2z + w)/3 − z = 3 so x = (5/3) − (1/3)z − (2/3)w, making the<br />
solution set<br />
⎛ ⎞ ⎞<br />
⎞<br />
(f) The reduction<br />
⎛<br />
⎝ 1 0 1 1 4<br />
⎞<br />
2 1 0 −1 2<br />
3 1 1 0 7<br />
⎛<br />
⎛<br />
5/3 −1/3 −2/3<br />
{ ⎜2/3<br />
⎟<br />
⎝ 0 ⎠ + ⎜ 2/3<br />
⎟<br />
⎝ 1 ⎠ z + ⎜ 1/3<br />
⎝ 0<br />
0 0<br />
1<br />
⎠ −2ρ1+ρ2<br />
−→<br />
−3ρ 1 +ρ 3<br />
⎛<br />
⎝ 1 0 1 1 4<br />
⎞<br />
0 1 −2 −3 −6<br />
0 1 −2 −3 −5<br />
shows that there is no solution — the solution set is empty.<br />
One.I.2.19<br />
(a) This reduction<br />
( )<br />
2 1 −1 1<br />
4 −1 0 3<br />
⎟<br />
⎠ w ∣ ∣ z, w ∈ R}.<br />
⎠ −ρ2+ρ3<br />
−→<br />
( )<br />
−2ρ 1 +ρ 2 2 1 −1 1<br />
−→<br />
0 −3 2 1<br />
⎛<br />
⎝ 1 0 1 1 4<br />
⎞<br />
0 1 −2 −3 −6⎠<br />
0 0 0 0 1<br />
ends with x and y leading while z is free. Solving for y gives y = (1−2z)/(−3), and then substitution<br />
2x + (1 − 2z)/(−3) − z = 1 shows that x = ((4/3) + (1/3)z)/2. Hence the solution set is<br />
⎛<br />
{ ⎝ 2/3<br />
⎞ ⎛<br />
−1/3⎠ + ⎝ 1/6<br />
⎞<br />
2/3⎠ z ∣ z ∈ R}.<br />
0 1<br />
(b) This application of Gauss’ method<br />
⎛<br />
⎝ 1 0 −1 0 1 ⎞ ⎛<br />
0 1 2 −1 3⎠ −ρ 1+ρ 3<br />
−→ ⎝ 1 0 −1 0 1 ⎞<br />
0 1 2 −1 3<br />
1 2 3 −1 7<br />
0 2 4 −1 6<br />
leaves x, y, and w leading. The solution set is<br />
⎛ ⎞ ⎛ ⎞<br />
1 1<br />
{ ⎜3<br />
⎟<br />
⎝0⎠ + ⎜−2<br />
⎟<br />
⎝ 1 ⎠ z ∣ z ∈ R}.<br />
0 0<br />
(c) This row reduction<br />
⎛<br />
⎞<br />
1 −1 1 0 0<br />
⎜0 1 0 1 0<br />
⎟<br />
⎝3 −2 3 1 0⎠<br />
0 −1 0 −1 0<br />
−3ρ 1+ρ 3<br />
−→<br />
⎛<br />
⎞<br />
1 −1 1 0 0<br />
⎜0 1 0 1 0<br />
⎟<br />
⎝0 1 0 1 0⎠<br />
0 −1 0 −1 0<br />
⎠ −2ρ 2+ρ 3<br />
−→<br />
−ρ 2+ρ 3<br />
−→<br />
ρ 2+ρ 4<br />
ends with z and w free. The solution set is<br />
⎛ ⎞ ⎛ ⎞ ⎛ ⎞<br />
0 −1 −1<br />
{ ⎜0<br />
⎟<br />
⎝0⎠ + ⎜ 0<br />
⎟<br />
⎝ 1 ⎠ z + ⎜−1<br />
⎟<br />
⎝ 0 ⎠ w ∣ z, w ∈ R}.<br />
0 0 1<br />
⎛<br />
⎝ 1 0 −1 0 1 ⎞<br />
0 1 2 −1 3⎠<br />
0 0 0 1 0<br />
⎛<br />
⎞<br />
1 −1 1 0 0<br />
⎜0 1 0 1 0<br />
⎟<br />
⎝0 0 0 0 0⎠<br />
0 0 0 0 0
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