Evaluate Delta = begin{vmatrix} 1 & a & bc 1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Applying $R_{2} ightarrow R_{2} - R_{1}$ and $R_{3} ightarrow R_{3} - R_{1}$,we get: $\Delta = \begin{vmatrix} 1 & a & bc \ 0 & b-a & ca-bc \

Vedclass · Accepted Answer

$(a-b)(b-c)(c-a)$

Vedclass · Answer

(B) Applying $R_{2} ightarrow R_{2} - R_{1}$ and $R_{3} ightarrow R_{3} - R_{1}$,we get: $\Delta = \begin{vmatrix} 1 & a & bc \ 0 & b-a & ca-bc \ 0 & c-a & ab-bc \end{vmatrix} = \begin{vmatrix} 1 & a & bc \ 0 & b-a & -c(b-a) \ 0 & c-a & -b(c-a) \end{vmatrix}$ Taking common factors $(b-a)$ from $R_{2}$ and $(c-a)$ from $R_{3}$: $\Delta = (b-a)(c-a) \begin{vmatrix} 1 & a & bc \ 0 & 1 & -c \ 0 & 1 & -b \end{vmatrix}$ Expanding along the first column: $\Delta = (b-a)(c-a) [1 \cdot (-b - (-c))] = (b-a)(c-a)(c-b)$ Rearranging terms to match the standard form: $\Delta = -(a-b)(c-a)(-(b-c)) = (a-b)(b-c)(c-a)$

Evaluate $\Delta = \begin{vmatrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix}$

Similar Questions

If $A = \begin{bmatrix} k & 2 \\ 2 & k \end{bmatrix}$ and $|A^3| = 125$,then the value of $k$ is

The solutions of the equation $\left| \begin{array}{ccc} 1 & 1 & x \\ p+1 & p+1 & p+x \\ 3 & x+1 & x+2 \end{array} \right| = 0$ are:

If the system of equations $ (k+1)^3 x + (k+2)^3 y = (k+3)^3 $, $ (k+1) x + (k+2) y = k+3 $, and $ x + y = 1 $ is consistent, then the value of $ k $ is:

Evaluate $\left|\begin{array}{rr}2 & 4 \\ -1 & 2\end{array}\right|$

The remainder when the determinant $\left|\begin{array}{lll} 2014^{2014} & 2015^{2015} & 2016^{2016} \\ 2017^{2017} & 2018^{2018} & 2019^{2019} \\ 2020^{2020} & 2021^{2021} & 2022^{2022} \end{array}\right|$ is divided by $5$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module