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$k = a^2 + b^2 + c^2 + d^2$
- ✓$ad-bc$
$\therefore$ $A^2 -(a + d)$ $A =$ $\left( {\begin{array}{*{20}{c}}{bc + ad}&0\\ 0&{bc + da}\end{array}} \right)$ $= (bc - ad) I$
As $A^2 - (a + d)A + kI = 0$, we get $(bc -ad)I + kI = 0 $
$==> k = ad - bc$
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${{b}_{1}}=b-\frac{b.a}{|a{{|}^{2}}}a,\,{{b}_{2}}=b+\frac{b.a}{|a{{|}^{2}}}a$ , ${{b}_{1}}=b-\frac{b.a}{|a{{|}^{2}}}a,\,{{b}_{2}}=b+\frac{b.a}{|a{{|}^{2}}}a$
, ${{c}_{2}}=c-\frac{c.a}{|a{{|}^{2}}}a-\frac{c.{{b}_{1}}}{|{{b}_{1}}{{|}^{2}}}{{b}_{1}}$
,${{c}_{3}}=c-\frac{c.a}{|a{{|}^{2}}}a-\frac{c.{{b}_{2}}}{|{{b}_{2}}{{|}^{2}}}{{b}_{2}}$,
${{c}_{4}}=a-\frac{c.a}{|a{{|}^{2}}}a$.
Then which of the following is a set of mutually orthogonal vectors is