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3 and 4 .Determinants and Matrices
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If $A =$ $\left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)$ satisfies the equation $x^2 - (a + d) x + k = 0$, then
A
$k = bc$
B
$k = ad$
C
$k = a^2 + b^2 + c^2 + d^2$
D
$ad-bc$
Solution
We have $A^2 =$ $\left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)$ $\left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)$ = $\left({\begin{array}{*{20}{c}}{{a^2} + bc}&{ab + bd}\\{ac + cd}&{bc + {d^2}}\end{array}} \right)$
$\therefore$ $A^2 -(a + d)$ $A =$ $\left( {\begin{array}{*{20}{c}}{bc + ad}&0\\ 0&{bc + da}\end{array}} \right)$ $= (bc – ad) I$
As $A^2 – (a + d)A + kI = 0$, we get $(bc -ad)I + kI = 0 $
$==> k = ad – bc$
Standard 12
Mathematics
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