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If A=[0 1 3 1 2x2 3 1] and A^(-1)=[1//2-...

If `A=[0 1 3 1 2x2 3 1]` and `A^(-1)=[1//2-4 5//2-1//2 3-3//2 1//2y1//2]` ,find `x ,\ y` .

Text Solution

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Given,
`=[[0, 1 ,3],[ 1, 2,x],[2, 3, 1]]`
here,
`∴∣A∣=0[2−3x]−1[1−2x]+3[3−4]`
`∴∣A∣=−1[1−2x]+3[−1]`
...
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