If $B = left[ {begin{array}{*{20}{c}} 5&{2α }&1 0&2&1 α &3&{ - 1} end

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3 and 4 .Determinants and Matrices

hard

If $B = \left[ {\begin{array}{*{20}{c}}
5&{2\alpha }&1\\
0&2&1\\
\alpha &3&{ - 1}
\end{array}} \right]$ is the inverse of a $3 \times 3$ matrix $A$, then the sum of all values of $\alpha $ for which $det\, (A) + 1 = 0$, is

A

$0$

B

$-1$

C

$1$

D

$2$

(JEE MAIN-2019)

Solution

$\left| B \right| = 5\left( { – 5} \right) – 2\alpha \left( { – \alpha } \right) – 2\alpha $

$ = 2{\alpha ^2} – 2\alpha – 25$

$1 + \left| A \right| = 0$

${\alpha ^2} – \alpha – 12 = 0$

Sum $=1$

Standard 12

Mathematics

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