MCQ
If $A =$ $\left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]$ (where $bc \ne 0$) satisfies the equations $x^2 + k = 0$, then
- A$a + d = 0$
- B$k = -|A|$
- C$k = |A|$
- ✓both $(A)$ and $(C)$
✓
Answer
Correct option: D.
both $(A)$ and $(C)$
d
We have $A^2 =$ $\left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]$ $\left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]$ $=$ $\left[{\begin{array}{*{20}{c}}{{a^2} + bc}&{ab + db}\\{ac + cd}&{bc + {d^2}} \end{array}} \right]$ $= 0$
We have $A^2 =$ $\left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]$ $\left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]$ $=$ $\left[{\begin{array}{*{20}{c}}{{a^2} + bc}&{ab + db}\\{ac + cd}&{bc + {d^2}} \end{array}} \right]$ $= 0$
As $A$ satisfies, $x^2 + k = 0, A^2 + kI = O$
==>$\left[ {\begin{array}{*{20}{c}}{{a^2} + bc + k}&{(a + d)b}\\{(a + d)c}&{bc + {d^2} + k}\end{array}} \right]$
==>$a^2 + bc + k = 0 = bc + d^2 + k = 0$ and $(a + d)b = (a + d) c = 0$
As $bc \ne 0, b \ne 0, c \ne 0$ ==> $a + d = 0$ ==> $a = -d$
Also, $k = -(a^2 + bc)$ $= -(d^2 + bc)$ $= - ( (-ad) + bc ) = |A|$
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