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If `A=[a b c d]`(where `b c!=0`) satisfies the equations `x^2+k=0,t h e n``a+d=0`b. `K=-|A|`c. `k=|A|`d. none of theseA. `a+d=0`B. `k=-|A|`C. `k=|A|`D. none of these |
Answer» Correct Answer - C We have, `A^(2)=[(a,b),(c,d)][(a,b),(c,d)]=[(a^(2)+bc,ab+db),(ac+cd,bc+d^(2))]` As A satisfies `x^(2)+k=0`, we have `A^(2)+kI=O` or `[(a^(2)+bc+k,(a+d)b),((a+d)c,bc+d^(2)+k)]=[(0,0),(0,0)]` or `a^(2)+bc+k=0, bc+d^(2)+k=0` and `(a+d)b=(a+d) c=0` As `bc ne 0, b ne 0, c ne 0`, so `a+d=0` or `a=-d` Also, `k=-(a^(2)+bc)` `=-(d^(2)+bc)` `=-((-ad)+bc)` `=|A|` |
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