11/6/14 Claire bought 5 kg of apples and 2 kg of bananas and paid altogether $22 Dale bought 4 kg of apples and 6 kg of bananas and paid altogether $33 Use inverse matrices to find the cost of 1 kg of bananas Claire bought 5 kg of apples and 2 kg of bananas and paid altogether $22 Dale bought 4 kg of apples and 6 kg of bananas and paid altogether $33 5a + 2b = 22 4a + 6b = 33 ⎡ 5 2 ⎤ ⎡ 22 ⎤
X
⎢
⎥=⎢
⎥
33
4
6
⎦
⎣
⎦ ⎣
1 11/6/14 apples bananas ⎡ 5 2 ⎤ ⎡ 22 ⎤
⎢
⎥=⎢ ⎥
kg ⎣ 4 6 ⎦ ⎣ 33 ⎦
kg As an equaIon: ⎡ 5 2 ⎤ ⎡ 22 ⎤
X⎢
⎥=⎢ ⎥
⎣ 4 6 ⎦ ⎣33 ⎦
⎡ a ⎤
⎥
⎣ b ⎦
* X will be a matrix with format ⎢
⎡ 6 −2 ⎤
⎡ 6 −2 ⎤
⎢
⎥
⎡ a ⎤ ⎡ 5 2 ⎤ 22 22
⎡ 22 ⎤ ⎢ 22 22 ⎥
⎥=⎢
⎥
⎢
⎥⎢
⎥⎢
⎥⎢
⎢
⎥
⎢
⎥
33
b
4
6
−4
5
−4
5
⎣
⎦
⎣
⎦⎣
⎦
⎢ 22 22 ⎥
⎢ 22 22 ⎥
⎣
⎦
⎣
⎦
To isolate a and b, muIply both sides of the equaIon by the inverse of ⎡ 5 2 ⎤
⎢
⎥
⎣ 4 6 ⎦
M
−1
1
=
[ adj M ]
M
−1
⎡ a b ⎤
1 ⎡ d −b ⎤
=
⎢
⎥
⎢
⎥
ad − bc ⎣ −c a ⎦
⎣ c d ⎦
2 11/6/14 M
−1
1
=
[ adj M ]
M
⎡ 5 2 ⎤
⎢
⎥
⎣ 4 6 ⎦
−1
=
−1
⎡ a b ⎤
1 ⎡ d −b ⎤
=
⎢
⎥
⎢
⎥
ad − bc ⎣ −c a ⎦
⎣ c d ⎦
1 ⎡ 6
⎢
22 ⎣ −4
⎡
⎢
−2 ⎤ = ⎢
⎥
5 ⎦ ⎢
⎢
⎣
6
22
−4
22
−2
22
5
22
⎤
⎥
⎥
⎥
⎥
⎦
⎡ a ⎤ ⎡ 5 2 ⎤ ⎡ 22 ⎤
⎢
⎥⎢
⎥=⎢
⎥
⎣ b ⎦ ⎣ 4 6 ⎦ ⎣ 33 ⎦
MulIply both sides of the equaIon by the inverse: ⎡
⎡ a ⎤⎢
⎢
⎥⎢
b
⎣
⎦⎢
⎢
⎣
6
22
−4
22
−2
22
5
22
⎤
⎡
⎥⎡
⎤ ⎢
⎥⎢ 5 2 ⎥ = ⎢
⎥⎣ 4 6 ⎦ ⎢
⎥
⎢
⎦
⎣
6
22
−4
22
−2
22
5
22
⎤
⎥⎡
⎤
⎥ ⎢ 22 ⎥
⎥ ⎣ 33 ⎦
⎥
⎦
3 11/6/14 ⎡
⎡ a ⎤⎢
⎢
⎥⎢
b
⎣
⎦⎢
⎢
⎣
6
22
−4
22
−2
22
5
22
⎤
⎡
⎥⎡
⎤ ⎢
⎥⎢ 5 2 ⎥ = ⎢
⎥⎣ 4 6 ⎦ ⎢
⎥
⎢
⎦
⎣
6
22
−4
22
−2
22
5
22
⎤
⎥⎡
⎤
⎥ ⎢ 22 ⎥
⎥ ⎣ 33 ⎦
⎥
⎦
⎡ 1 0 ⎤
⎥
⎦
The product of any matrix M and M-­‐1 is the idenIty matrix ⎢⎣ 0 1
(works as the mulIplicaIve I.D., 1, but for matrix mulIplicaIon. ⎡
⎡ a ⎤ ⎢
⎢
⎥=⎢
⎣ b ⎦ ⎢
⎢
⎣
6
22
−4
22
−2
22
5
22
⎤
⎥⎡
⎤
⎥ ⎢ 22 ⎥
⎥ ⎣ 33 ⎦
⎥
⎦
⎡
⎡ a ⎤ ⎢
⇒⎢
⎥=⎢
⎣ b ⎦ ⎢
⎢
⎣
66
22
77
22
⎤
⎥
⎥
⎥
⎥
⎦
⎡ a ⎤ ⎡ 3 ⎤
⇒⎢
⎥=⎢
⎥
b
⎣
⎦ ⎣ 3.5 ⎦
If matrices are equal, each element in one is equal to the corresponding element in the other, so b= 3.5 and a Kg of bananas costs $3.50. 4