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STD 12 - 3 and 4 . determinant and metrices — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 3 and 4 . determinant and metrices1 Mark
MCQ
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
    $k = a^2 + b^2 + c^2 + d^2$
  • ✓
    $ad-bc$
✓

Answer

Correct option: D.
$ad-bc$
d
We have $A^2 =$ $\left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)$ $\left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)$  = $\left({\begin{array}{*{20}{c}}{{a^2} + bc}&{ab + bd}\\{ac + cd}&{bc + {d^2}}\end{array}} \right)$

$\therefore$  $A^2 -(a + d)$ $A =$ $\left( {\begin{array}{*{20}{c}}{bc + ad}&0\\ 0&{bc + da}\end{array}} \right)$ $= (bc - ad) I$

As $A^2 - (a + d)A + kI = 0$, we get $(bc -ad)I + kI = 0 $

$==> k = ad - bc$

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