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$A$ is a $2 × 2$ matrix such that $A$ $\left[ {\begin{array}{*{20}{c}}1\\{ - 1} \end{array}} \right]$ $=$$\left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]$ and $A^2$ $\left[ {\begin{array}{*{20}{c}}1\\{ - 1}\end{array}} \right]$ $=$ $\left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]$ . The sum of the elements of $A$, is
$-1$
$0$
$2$
$5$
Solution
$A$ $\left[ {\begin{array}{*{20}{c}}1\\{ – 1}\end{array}} \right]$ $=$ $\left[{\begin{array}{*{20}{c}}{ – 1}\\2\end{array}} \right]$ ….$(1)$
$A^2$ $\left[ {\begin{array}{*{20}{c}}1\\{ – 1}\end{array}} \right]$ $=$ $\left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]$ ….$(2)$
Let $A$ be given by $A =$$\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.
