Verify Property 2 for Delta=left|begin{array} | vedclass

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

Question

(N/A) The determinant is given by $\Delta = \left| \begin{array}{ccc} 2 & -3 & 5 \ 6 & 0 & 4 \ 1 & 5 & -7 \end{array} ight|$.Expanding along the f

Vedclass · Accepted Answer

Vedclass · Answer

(N/A) The determinant is given by $\Delta = \left| \begin{array}{ccc} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{array} \right|$.
Expanding along the first row:
$\Delta = 2(0 - 20) - (-3)(-42 - 4) + 5(30 - 0) = 2(-20) + 3(-46) + 5(30) = -40 - 138 + 150 = -28$.
Interchanging rows $R_{2}$ and $R_{3}$ i.e.,$R_{2} \leftrightarrow R_{3}$,we get $\Delta_{1} = \left| \begin{array}{ccc} 2 & -3 & 5 \\ 1 & 5 & -7 \\ 6 & 0 & 4 \end{array} \right|$.
Expanding $\Delta_{1}$ along the first row:
$\Delta_{1} = 2(20 - 0) - (-3)(4 + 42) + 5(0 - 30) = 2(20) + 3(46) + 5(-30) = 40 + 138 - 150 = 28$.
Clearly,$\Delta_{1} = -\Delta$,which is $28 = -(-28)$.
Hence,Property $2$ is verified.

Verify Property $2$ for $\Delta=\left|\begin{array}{ccc}2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7\end{array}\right|$

Similar Questions

The value of the determinant $\left| \begin{array}{ccc} 1 & 1 & 1 \\ b+c & c+a & a+b \\ b+c-a & c+a-b & a+b-c \end{array} \right|$ is

Using the property of determinants and without expanding,prove that:
$\left|\begin{array}{lll}1 & bc & a(b+c) \\ 1 & ca & b(c+a) \\ 1 & ab & c(a+b)\end{array}\right|=0$

If ${D_r} = \left| \begin{array}{ccc} {2^{r - 1}} & {2 \cdot 3^{r - 1}} & {4 \cdot 5^{r - 1}} \\ x & y & z \\ {2^n} - 1 & {3^n} - 1 & {5^n} - 1 \end{array} \right|$,then the value of $\sum\limits_{r = 1}^n {D_r} = $

If $a, b, c$ are all different and $\left| \begin{array}{ccc} a & a^3 & a^4 - 1 \\ b & b^3 & b^4 - 1 \\ c & c^3 & c^4 - 1 \end{array} \right| = 0$,then:

If $A$ is a square matrix of order $3 \times 3$,then $|KA|$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module