Question
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

Answer

c
$\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 $=1$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

The sum of all the natural numbers for which $log_{(4-x)}(x^2 -14x + 45)$ is defined is -
Let the complex numbers $\alpha$ and $\frac{1}{\bar{\alpha}}$ lie on the circles $\left|z-z_0\right|^2=4$ and $\left|z-z_0\right|^2=16$ respectively, where $z_0=1+i$. Then, the value of $100|\alpha|^2$ is $........$
Let ${A_n} = \left( {\frac{3}{4}} \right) - {\left( {\frac{3}{4}} \right)^2} + {\left( {\frac{3}{4}} \right)^3} - ..... + {\left( { - 1} \right)^{n - 1}}{\left( {\frac{3}{4}} \right)^n}$ and $B_n \,= 1 - A_n$ . Then, the least odd natural number $p$ , so that ${B_n} > {A_n}$, for all $n \geq p$ is
If  $\alpha  , \beta $ are two real numbers satisfying $\alpha^2 + \beta^2$ = $ 5$ and $3(\alpha^5 +  \beta^5) = 11$$(\alpha^3 + \beta^3)$, then $\alpha  \beta$ is 
Lei $\alpha|x|=|y| e^{x y-\beta}, \alpha, \beta \in N$ be the solution of the differential equation $x\ d y-y\ d x+x y(x\ d y+y\ d x)=0, y (1)=2$. Then $\alpha+\beta$ is equal to
The difference between the fourth term and the first term of a Geometrical Progresssion is $52.$ If the sum of its first three terms is $26,$ then the sum of the first six terms of the progression is
If $\log _{3} 2, \log _{3}\left(2^{x}-5\right), \log _{3}\left(2^{x}-\frac{7}{2}\right)$ are in an arithmetic progression, then the value of $x$ is equal to $.....$
The circle $x^2 + y^2 = 5 $ meets the parabola $y^2 = 4x$  at $P $ and $ Q.$  Then the length $ PQ$  is equal to :
Let $f: R \rightarrow R$ be such that for all $\mathrm{x} \in \mathrm{R}\left(2^{1+\mathrm{x}}+2^{1-\mathrm{x}}\right), f(\mathrm{x})$ and $\left(3 ^\mathrm{x}+3^{-\mathrm{x}}\right)$ are in $A.P.$, then the minimum value of $f(x)$ is
The sum, $\sum\limits_{n=1}^{7} \frac{n(n+1)(2 n+1)}{4}$ is equal to