Question

Find the value of X in the following determinant:

Answer 1

Given to find the value of x :-

$\left[\begin{array}{cc}x&-2&\\3&1-x&\\\end{array}\right]=0$

Solution :-

We know that to find the determinant by using  formula ad-bc

$\left[\begin{array}{cc}x&-2&\\3&1-x&\\\end{array}\right]=\left[\begin{array}{cc}a&b&\\c&d&\\\end{array}\right]$

By comparison we can say

• a = x
• b = -2
• c = 3
• d = 1-x

So,

$ad -bc =0$

$x(1-x) -(-2)(3) =0$

$x -x^2+6=0$

Take common " - "

$x^2-x-6=0$

By using Quadratic formula we get

$x = \dfrac{-b\pm\sqrt{b^2-4ac} }{2a}$

• a= 1
• b = -1
• c = -6

By compared with ax^2+bx+c [General form of Q.E]

$x = \dfrac{-(-1)\pm\sqrt{1-4(1)(-6)} }{2}$

$x =\dfrac{1\pm \sqrt{25} }{2}$

$x =\dfrac{1\pm5}{2}$

$x =\dfrac{1+5}{2} , \dfrac{1-5}{2}$

$x =\dfrac{6}{2} , \dfrac{-4}{2}$

$x = 3, -2$

So, the value of x is 3, -2

Answer 2

Answer:

Answers By Expert Tutors. I assume that the equation says that the determinant of the 3x3 matrix with an unknown, x, as the center element is equal to 740. To solve for x, work out the value of the determinant (which will depend on x) and set it equal to 740.