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Question

Find x,y,z,w if [x+yxyy+z+w2wz]=[2195]

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Solution

By using the property of equality of matrices, we have

x+y=2

xy=1

By the method of elimination we have 2x=21=1 or x=12

and since x+y=212+y=2y=212=412=32

Again y+z+w=932+z+w=9 or z+w=932=1832=152

Solving z+w=152 and 2wz=5 we get

3w=152+5=15+102=252

w=256

Substituting w=256 in z+w=152 we get

z=152256=45256=206=103

x=12,y=32,w=256,z=103


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