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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
$A$ is a $2 × 2$ matrix such that $A$ $\left[ {\begin{array}{*{20}{c}}1\\{ - 1} \end{array}} \right]$ $=$$\left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]$ and $A^2$  $\left[ {\begin{array}{*{20}{c}}1\\{ - 1}\end{array}} \right]$ $=$ $\left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]$ . The sum of the elements of $A$, is
  • A
    $-1$
  • B
    $0$
  • C
    $2$
  • ✓
    $5$
✓

Answer

Correct option: D.
$5$
d
$A$ $\left[ {\begin{array}{*{20}{c}}1\\{ - 1}\end{array}} \right]$ $=$ $\left[{\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]$ ....$(1)$
$A^2$ $\left[ {\begin{array}{*{20}{c}}1\\{ - 1}\end{array}} \right]$ $=$ $\left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]$ ....$(2)$
Let $A$ be given by $A =$$\left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]$
The first equation gives $a - b = - 1$ ....$(3)$ and $c - d = 2$....$(4)$
For second equation, $A^2$ $\left[ {\begin{array}{*{20}{c}}1\\{ - 1}\end{array}} \right]$ $=$ $A\left( {A\left[ {\begin{array}{*{20}{c}}1\\{ - 1}\end{array}} \right]\,} \right)$ $=$$A\left( {\,\left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]\,} \right)$ $=$$\left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]$
This gives $- a + 2b = 1$ ....$(5)$ and $- c + 2d = 0$....$(6)$
$(3) + (5) ==> b = 0$ and $a = - 1$
$(4) + (6) ==> d = 2 $ and $c = 4$
so the sum $a + b + c + d = 5$ Ans.

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