Let be a real number. Consider the matrix A= ( ccc & 0 & 1 2 & 1 … - Vidyadip

MCQ

Let $\beta$ be a real number. Consider the matrix $A=\left(\begin{array}{ccc}\beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2\end{array}\right)$ If $A^7-(\beta-1) A^6-\beta A^5$ is a singular matrix, then the value of $9 \beta$ is

A
$2$
✓
$3$
C
$4$
D
$5$

✓

Answer

Correct option: B.

$3$

$A=\left(\begin{array}{ccc}\beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2\end{array}\right)|A|=-1$
$\Rightarrow\left|A^7-(\beta-1) A^6-\beta A^5\right|=0$
$\Rightarrow|A|^5\left|A^2-(\beta-1) A-\beta I\right|=0$
$\Rightarrow|A|^5\left|\left(A^2-\beta A\right)+A-\beta I\right|=0$
$\Rightarrow|A|^5|A(A-\beta I)+I(A-\beta I)|=0$
$|A|^5|(A+I)(A-\beta I)|=0$
$A+I=\left(\begin{array}{ccc}\beta+1 & 0 & 1 \\ 2 & 2 & -2 \\ 3 & 1 & -1\end{array}\right) $
$\Rightarrow|A+I|=-4, $
Here $|A| \neq 0 \ |A+I| \neq 0$
$A-\beta I$
$=\left(\begin{array}{ccc}0 & 0 & 1 \\ 2 & 1-\beta & -2 \\ 3 & 1 & -2-\beta\end{array}\right)$
$|A-\beta I|=2-3(1-\beta)=3 \beta-1=0 $
$\Rightarrow \beta=\frac{1}{3}$
$9 \beta=3$

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

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

$a\times (b\times c)+b\times (c\times a)+c\times (a\times b)=$

The coefficient of two consecutive terms in the expansion of ${(1 + x)^n}$ will be equal, if

The number of continuous functions $f:[0,1] \rightarrow R$ that satisfy $\int \limits_0^1 x f(x) d x=\frac{1}{3}+\frac{1}{4} \int \limits_0^1(f(x))^2 d x$ is

The value of $\int_{ - 2}^3 {|1 - {x^2}|dx} $ is

If ${u_n} = \int_0^{\pi /4} {{{\tan }^n}x\,dx,} $ then ${u_n} + {u_{n - 2}} = $

The sum of all the elements in the set $\{\mathrm{n} \in\{1,2, \ldots \ldots ., 100\} \mid$ $H.C.F.$ of $n$ and $2040$ is $1\,\}$ is equal to $.....$

The equation of circle which touches the axes of coordinates and the line $\frac{x}{3} + \frac{y}{4} = 1$ and whose centre lies in the first quadrant is ${x^2} + {y^2} - 2cx - 2cy + {c^2} = 0$, where $c$ is

Let one focus of the hyperbola $H: \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be at $(\sqrt{10}, 0)$ and the corresponding directrix be $x =\frac{9}{\sqrt{10}}$. If e and $l$ respectively are the eccentricity and the length of the latus rectum of H , then $9\left( e ^2+l\right)$ is equal to:

If $\frac{{ }^{11} C_1}{2}+\frac{{ }^{11} C_2}{3}+\ldots . .+\frac{{ }^{11} C_9}{10}=\frac{n}{m}$ with $\operatorname{gcd}(n, m)=1$, then $n+m$ is equal to

If $\theta $ and $\phi $ are eccentric angles of the ends of a pair of conjugate diameters of the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, then $\theta - \phi $ is equal to