Gujarati
Hindi
3 and 4 .Determinants and Matrices
medium

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

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.