Using the property of determinants and withou | vedclass

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

Question

(A) Let $\Delta = \left|\begin{array}{lll}2 & 7 & 65 \ 3 & 8 & 75 \ 5 & 9 & 86\end{array}ight|$.We can write the third column as $C_3 = 9C_2 + C_1

Vedclass · Accepted Answer

Vedclass · Answer

(A) Let $\Delta = \left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|$.
We can write the third column as $C_3 = 9C_2 + C_1$:
$\Delta = \left|\begin{array}{lll}2 & 7 & 9(7)+2 \\ 3 & 8 & 9(8)+3 \\ 5 & 9 & 9(9)+5\end{array}\right|$
Using the property of determinants,we can split this into two determinants:
$\Delta = \left|\begin{array}{lll}2 & 7 & 9(7) \\ 3 & 8 & 9(8) \\ 5 & 9 & 9(9)\end{array}\right| + \left|\begin{array}{lll}2 & 7 & 2 \\ 3 & 8 & 3 \\ 5 & 9 & 5\end{array}\right|$
In the first determinant,we can take $9$ as a common factor from the third column:
$\Delta = 9 \left|\begin{array}{lll}2 & 7 & 7 \\ 3 & 8 & 8 \\ 5 & 9 & 9\end{array}\right| + \left|\begin{array}{lll}2 & 7 & 2 \\ 3 & 8 & 3 \\ 5 & 9 & 5\end{array}\right|$
In the first determinant,the second and third columns are identical,so its value is $0$.
In the second determinant,the first and third columns are identical,so its value is $0$.
Therefore,$\Delta = 9(0) + 0 = 0$.

Using the property of determinants and without expanding,prove that:
$\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|=0$

Similar Questions

If $\Delta = \begin{vmatrix} a & b & c \\ x & y & z \\ p & q & r \end{vmatrix}$,then $\begin{vmatrix} ka & kb & kc \\ kx & ky & kz \\ kp & kq & kr \end{vmatrix}$ =

Let $a, b, c$ be such that $(b+c) \neq 0$ and $\left|\begin{array}{ccc} a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1 \end{array}\right|+\left|\begin{array}{ccc} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} a & (-1)^{n-1} b & (-1)^n c \end{array}\right|=0$. Then the value of $n$ is

$A$ value of $\theta$ in $\left(0, \frac{\pi}{2}\right)$ satisfying $\left|\begin{array}{ccc}1+\sin ^2 \theta & \cos ^2 \theta & 4 \sin 4 \theta \\ \sin ^2 \theta & 1+\cos ^2 \theta & 4 \sin 4 \theta \\ \sin ^2 \theta & \cos ^2 \theta & 1+4 \sin 4 \theta\end{array}\right|=0$ is

The determinant $\left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 6 \end{array} \right|$ is not equal to

Let $a, b, c$ be such that $b(a+c) \neq 0$. If $\left|\begin{array}{ccc}a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1\end{array}\right| + \left|\begin{array}{ccc}a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} a & (-1)^{n+1} b & (-1)^{n} c\end{array}\right|=0$, then the value of $n$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Using the property of determinants and without expanding,prove that:$\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|=0$