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Solve and check (5) a. 5(z – 3) = 3(z + 2) b. 3 + 3−8 2 = 4−3 3 |
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Answer» Answer: Given system of linear equations are 2x+3y+3z=5 x−2y+z=−4 3x−y−2z=3. Represent it in matrix form ⎣ ⎢ ⎢ ⎡
2 3
3 −2 −1
3 1 −2
⎦ ⎥ ⎥ ⎤
⎣ ⎢ ⎢ ⎡
x y z
⎦ ⎥ ⎥ ⎤ = ⎣ ⎢ ⎢ ⎡
5 −4 3
⎦ ⎥ ⎥ ⎤ which is in the form of AX=B A= ⎣ ⎢ ⎢ ⎡
2 1 3
3 −2 −1
3 1 −2
⎦ ⎥ ⎥ ⎤
∣A∣=10+15+15=40 =0 ∴ A −1 exists To find adjoint of A A 11 =5,A 12 =5,A 13 =5 A 21 =3,A 22 =−13,A 23 =11 A 31 =9,A 32 =1,A =−7 ADJ(A)=co-factor ⎣ ⎢ ⎢ ⎡
5 3 9
5 −13 1
5 11 −7
⎦ ⎥ ⎥ ⎤
= ⎣ ⎢ ⎢ ⎡
5 5 5
3 −13 11
9 1 −7
⎦ ⎥ ⎥ ⎤
A −1 = ∣A∣ 1 Adj(A) = 40 1
⎣ ⎢ ⎢ ⎡
5 5 5
3 −13 11
9 1 −7
⎦ ⎥ ⎥ ⎤
X=A −1 B = 40 1
⎣ ⎢ ⎢ ⎡
5 5 5
3 −13 11
9 1 −7
⎦ ⎥ ⎥ ⎤
⎣ ⎢ ⎢ ⎡
5 −4 3
⎦ ⎥ ⎥ ⎤
X= 40 1
⎣ ⎢ ⎢ ⎡
25−12+27 25+52+3 25−44−21
⎦ ⎥ ⎥ ⎤
X= 40 1
⎣ ⎢ ⎢ ⎡
40 80 −40
⎦ ⎥ ⎥ ⎤
⎣ ⎢ ⎢ ⎡
x y z
⎦ ⎥ ⎥ ⎤ = ⎣ ⎢ ⎢ ⎡
1 2 −1
⎦ ⎥ ⎥ ⎤
Hence, x=1,y=2 and z=−1 |
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