Unformatted text preview: 12' 13. 14. EXERCISES 3.4 . ‘ ' 41 We present a sequence of elementary row operations that lead to a row echelon form for
the augmented matrix of the given system. You may, however, use a different sequence of
operations, resulting in a different row echelon form. The solution of the system, however, must be the same.
3 2 —4 0 R1 — R3 1 1 0
1 1 3 0 —i 1 1 3
1 1 0 0 R3 —r R, 3 2 —4 0
0 R2_"_R1+R’Z
0 R3—i—3R1+R3 1 1 0 0 1 1 0 0
Hue003ORg—i—R3—>0140
0a1u40R3—»R2/3 0010
We now convert this augmented matrix to an equivalent system of equations,
m+y=m
y+4z=0,
220. When we substitute 2 = 0 into the second equation, we obtain 3; = 0. When we substitute
3; = 0 and z = 0 into the ﬁrst equation. we ﬁnd .1: = 0 aISo. We present a sequence of elementary row operations that lead to a rowr echelon form for
the augmented matrix of the given system. You may, however, use a different sequence of
operations, resulting in a different row echelon form. The solution of the system, however,
must be the same. 1 —1 2 0 1 m1 2 0
3 1 —4 0 R2 —r —3R1 + R2 —l 0 4 —10 0
2 2 wii 0 0 4 —10 0 R3 —) —R2 + Rs Rs—t—QRH‘Rs 1 —1 2 0 1 —1 2 0
—+ O 4 10 U Rz—iRQ/ﬂi —+ 0 1 —5/2 0
0 0 0 0 0 0 0 0
 We now convert this augmented matrix to an equivalent system of equations,
a: — y + 22: 2 0,
— a; — 0
y 2. '_ ‘
When we solve the second equation for y, we get 3; = gz. When we substitute this into the
ﬁrst equation, we obtain
5 z
$—§Z+22§—0 => $——§.
Thus, there is an inﬁnity of solutions for the system
a: = 3;, y = %, z arbitrary. We present a sequence of elementary row operations that lead to 'a row echelon form for
the augmented matrix of the given system. You may, however, use a different sequence of
operations, resulting in a different rowr echelon form. The solution of the system, however, must be the same.
3 1 —1 0 R1 —) R2 1 —2 3
1 —2 3 0 R2 —) R; —i 3 1 —1
5 H3 5 0 5 _3 5 0
0 Rem—3R1+R2
0 Has—salun a ____l.:l_f‘__l ...
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 Fall '11
 Dr.Cornell
 Math

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