College Algebra, 2013a

582 Systems of Equations and Matrices
Solution.
([ 2 −1
3A −
3 5
{ ([ 2 −1
3A + −
3 5
([ 2 −1
3A +(−1)
3 5
{
3A + (−1)
[ 2 −1
3 5
[ 2 −1
3A +(−1)
3 5
[ (−1)(2) (−1)(−1)
3A +
(−1)(3) (−1)(5)
[ −2 1
(3 + (−5))A +
−3 −5
] )
+5A
] )}
+5A
] )
+5A
]
+(−1)(5A)
}
=
=
=
=
]
+(−1)(5A) =
]
+((−1)(5))A =
[ −2 1
3A +
−3 −5
[ ] −2 1
3A +(−5)A +
−3 −5
] ( [ ])
−2 1
+ −
−3 −5
]
+(−5)A =
=
=
(−2)A +0 2×2 =
(−2)A =
(−2)A =
[ ] −4 2
+ 1 [ ]
9 12
6 −2 3 −3 39
[ ] [ (
−4 2
1
(
+ 3)
(9)
1
) ]
3 (12)
(
6 −2
1
) (
3 (−3) 1
)
3 (39)
[ ] [ ]
−4 2 3 4
+
6 −2 −1 13
[ −1
] 6
5 11
[ −1
] 6
5 11
[ −1
] 6
5 11
[ −1
] 6
5 11
[ −1
] 6
5 11
[ ] ( [ ])
−1 6 −2 1
+ −
5 11 −3 −5
[ ] [ ]
−1 6 −2 1
−
5 11 −3 −5
[ −1 − (−2)
] 6 − 1
5 − (−3) 11 − (−5)
[ 1
] 5
8 16
[
( )
−
1
2 ((−2)A) = −
1 1 5
2 8 16
[ (
(( ) )
−
1
−
1
2 (−2) A = ( 2) (
−
1
2 (8) −
1
2
[ ]
−
1
2
− 5 2
1A =
A =
]
) ( ]
(1) −
1
2)
(5)
)
(16)
−4 − 16
2
[ ]
−
1
2
− 5 2
−4 −8
The reader is encouraged to check our answer in the original equation.