To determine the value of
x for which the matrix
A is singular, we need to find the determinant of the matrix and set it equal to zero, because a matrix is singular if and only if its determinant is zero.
The given matrix
A is:
A=[] The formula for the determinant of a
3×3 matrix
[| a11 | a12 | a13 |
| a21 | a22 | a23 |
| a31 | a32 | a33 |
] is given by:
det(A)=a11(a22a33−a23a32)−a12(a21a33−a23a31)+a13(a21a32−a22a31)
Plugging in the values from matrix
A, we get:
det(A)=0⋅(5⋅x−7⋅9)−x⋅(x⋅x−7⋅0)+16⋅(x⋅9−5⋅0)
This simplifies to:
det(A)=0−x(x2)+16(9x)Which further simplifies to:
det(A)=−x3+144x Setting the determinant to zero for the matrix to be singular:
−x3+144x=0We can factor out an
x :
x(−x2+144)=0Setting each factor equal to zero gives:
x=0‌‌‌ or ‌‌‌−x2+144=0Solving the second equation for
x :
−x2+144=0‌‌⇒‌‌x2=144‌‌⇒‌‌x=±12
However, the provided answer choices include
−12,21,−144,144. Of these valid options, our solution yields:
x=−12‌‌‌ or ‌‌‌x=12Hence, the correct value of
x from the given options is:
Option A: -12 .