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Id `[1//25 0x1//25]=[5 0-a5]^(-2)`, then the value of `x`is`a//125`b. `2a//125`c. `2a//25`d. none of theseA. `a//125`B. `2a//125`C. `2a//25`D. none of these |
Answer» Correct Answer - B Let `A=[(5,0),(-a,5)]` `implies` adj `(A)= [(5,0),(a,5)]` `implies A^(-1) =1/(|A|)[(5,0),(a,5)]=1/25 [(5,0),(a,5)]` `implies A^(-2)=(A^(-1))^(2)=1/25 [(5,0),(a,5)]1/25 [(5,0),(a,5)]` `=1/625 [(25,0),(10a, 25)]` `=[(1/25,0),((2a)/125,1/25)]` Now, `[(1//25,0),(x,1//25)]=[(1/25,0),((2a)/125,1/25)]` `implies x=2a//125` |
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