WORD PROBLEMS ON SYMMETRIC AND SKEW SYMMETRIC

Problem 1 :

If A and B are symmetric matrices of same order, prove that

(i) AB + BA is a symmetric matrix.

(ii) AB - BA is a skew-symmetric matrix

Solution :

(i)  Since A and B are symmetric matrices, then

AT  =  A

BT  =  B

Let P  =  AB + BA

PT  =  (AB + BA)T

PT  =  (AB)T + (BA)T

PT  =  BTAT + ATBT

PT  =  BA + AB

PT  =  AB + BA

PT  =  P

So, AB + BA is a symmetric matrix.

(ii)  Let Q  =  AB - BA

QT  =  (AB - BA)T

QT  =  (AB)T - (BA)T

QT  =  BTAT - ATBT

QT  =  BA - AB

QT  =  -(AB - BA)

QT  =  -Q

So, AB - BA is skew symmetric matrix.

Problem 2 :

A shopkeeper in a Nuts and Spices shop makes gift packs of cashew nuts, raisins and almonds.

Pack I contains 100 gm of cashew nuts, 100 gm of raisins and 50 gm of almonds.

Pack-II contains 200 gm of cashew nuts, 100 gm of raisins and 100 gm of almonds.

Pack-III contains 250 gm of cashew nuts, 250 gm of raisins and 150 gm of almonds.

The cost of 50 gm of cashew nuts is $50, 50 gm of raisins is $10, and 50 gm of almonds is $60. What is the cost of each gift pack?

Solution :

By using the given items, we may construct a matrix. Then construct another matrix with cost per grams.

By multiplying the above matrices, we may get the cost of each pack.

Cost of 50 gm of cashew nuts is $50

Cost of 1 gm of cashew nuts is 50/50  =  1

Cost of 50 gm of raisins is $10

Cost 1 gm of raisins is 10/50  =  1/5

Cost 50 gm of almonds is $60

Cost 1 gm of almonds is 60/50  =  6/5

By multiplying the above matrices, we get

Cost of pack 1  =  100 + (100)⋅ (1/5) + 50 ⋅ (6/5)

   =  100 + 20  +  60

   =  180

Cost of pack 2  =  200 + (100)⋅ (1/5) + 100 ⋅ (6/5)

   =  200 + 20  +  120

   =  340

Cost of pack 2  =  250 + (250)⋅ (1/5) + 150 ⋅ (6/5)

   =  250 + 50 + 180

   =  480

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