Test: Properties Of Determinants - Commerce MCQ

Determinant = -2(-12 + 12) -2(-6 - 2) +1(-12 - 4)
= - 2(0) - 2(-8) + 1(16)
= 0 + 16 + 16
= 32

If you interchange (switch) two rows (or columns) of a matrix A to get B ,items then det(B) = –det(A). Hence, the determinant is -(-7)

= 7
# Rule 
If any two rows( or columns) of a determinant are interchanged, then the value of determinant is changed in sign only

R₁ ------> R₁ + R₂, taking (a+b+c) common from the first row, we get the resultant matrix.

|A|=3((4)-(4))-2((-4)-(2))-1((-8)-(-4))
|A|= -2(-6)-(-6)
|A|= -2(0)
=0

Let the symmetric matrix be

∴ |X| = bc – a²

Since the given matrix is real, to get maximum determinant value of a should be equal to 0.

b + c = 24

c = 24 – b

∴ |X| = b × (24 – b)

|X| = 24 b – b²

Differentiating with respect to b

|X|’ = 24 – 2b = 0

∴ b = 12

|X|’’ = - 2 < 0

At b = 12 it will have maximum value

c = 24 – 12 =12

Maximum value = 12 × 12 – 0 = 144

Taking 3 common from the column I in second determinant, we get the 14 * 3 = 42

A = {(5x+2, 2x, 2x) (5x+2, x+2, 2x) (5x+2, 2x, x+2)}
= (5x+2) {(1, 2x, 2x) (1, x+2, 2x) (1, 2x, x+2)}
= (5x+2) [(1(x+2)^2 - 4x^2) - 2x(x+2 - 2x) + 2x(2x - x - 2)]
= (5x+2) [(x+2 - 2x)(x+2+2x)-2x(2 - x) + 2x(x-2)]
= (5x+2) [(2-x)(3x+2) -2x(2-x) -2x(2-x)]
= (5x+2) (2-x)[3x+2-2x-2x]
= (5x+2) (2-x)²