Matrices & DeterminantseasyFree

Matrices & Determinants: Find Roots Equation Vmatrix Vmatrix

JEE Maths question with a full step-by-step solution.

Question
Find the roots of the equation
xα1βx1βγ1=0\begin{vmatrix} x & \alpha & 1 \\ \beta & x & 1 \\ \beta & \gamma & 1 \end{vmatrix} = 0
Aα\alpha and β\beta
Bβ\beta and γ\gammacorrect
Cγ\gamma and α\alpha
DAll of these
Solution
Step 1: Apply row operations R1R1R2R_1 \to R_1 - R_2 and R2R2R3R_2 \to R_2 - R_3:
xβαx00xγ0βγ1=0\begin{vmatrix} x-\beta & \alpha-x & 0 \\ 0 & x-\gamma & 0 \\ \beta & \gamma & 1 \end{vmatrix} = 0
Step 2: Expand along column 3
1xβαx0xγ=01 \cdot \begin{vmatrix} x-\beta & \alpha-x \\ 0 & x-\gamma \end{vmatrix} = 0
(xβ)(xγ)=0(x-\beta)(x-\gamma) = 0
Step 3: State the roots The roots are x=βx = \beta and x=γx = \gamma. Answer: (2)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Matrices & Determinants · hard
Let AA be a 3×33\times3 matrix such that AT[101]=[522]A^T\begin{bmatrix}1\\0\\1\end{bmatrix}=\begin{bmatrix}5\\2\\2\end{bmatrix}, AT[001]=[311]A^T\begin{bmatrix}0\\0\\1\end{bmatrix}=\begin{bmatrix}3\\1\\1\end{bmatrix}, A[101]=[344]A\begin{bmatrix}1\\0\\1\end{bmatrix}=\begin{bmatrix}3\\4\\4\end{bmatrix} and A[001]=[131]A\begin{bmatrix}0\\0\\1\end{bmatrix}=\begin{bmatrix}1\\3\\1\end{bmatrix}. If det(A)=1\det(A)=1, then det(adj(A2+A))\det\big(\mathrm{adj}(A^2+A)\big) is equal to
Matrices & Determinants · medium
Let MM be a 3×33\times3 matrix such that M(100)=(123)M\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}1\\2\\3\end{pmatrix}, M(010)=(012)M\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}0\\1\\2\end{pmatrix}, and M(001)=(111)M\begin{pmatrix}0\\0\\1\end{pmatrix}=\begin{pmatrix}-1\\1\\1\end{pmatrix}. If M(xyz)=(1711)M\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\7\\11\end{pmatrix}, then x+y+zx+y+z equals:
Matrices & Determinants · medium
If f:NZf:\mathbb{N} \to \mathbb{Z} is defined by f(n)=n152n23(2k+1)2k+13n33k(2k+1)3k(k+2)+1,kNf(n) = \begin{vmatrix} n & -1 & -5 \\ -2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3k(2k+1) & 3k(k+2)+1 \end{vmatrix}, \quad k \in \mathbb{N} and n=1kf(n)=98\displaystyle\sum_{n=1}^k f(n) = 98, then kk is equal to:
Matrices & Determinants · hard
If f(x)=x5sinx2x4tan3x1sec2xsin3xx45f(x) = \begin{vmatrix} x^5 & |\sin x| & 2x^4 \\ \tan^3 x & 1 & \sec 2x \\ \sin^3 x & x^4 & 5 \end{vmatrix}, then π/2π/2f(x)dx\displaystyle\int_{-\pi/2}^{\pi/2} f(x)\,dx is equal to:

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.