Test: Determinants - 1
25 Questions MCQ Test Mathematics (Maths) Class 12 | Test: Determinants - 1
| 1 Crore+ students have signed up on EduRev. Have you? Download the App |
, then Det. A is 
Apply C2 → C2 + C3,
because , the value of the determinant is zero only when , the two of its rows are identical., Which is possible only when Either x = 3 or x = 4 .
Apply , R1 → R1+R2+R3,
Apply , C3→ C3 - C1, C2→C2 - C1,
=(a+b+c)3
If A and B are invertible matrices of order 3 , then det (adj A) =
Let A be a non singular square matrix of order n . then , |adj.A| = 1
If A and B matrices are of same order and A + B = B + A, this law is known as
Commutative law, in mathematics, either of two laws relating to number operations of addition and multiplication, stated symbolically: a + b = b + a and ab = ba
The determinant of a matrix A and its transpose always same.
is
, Apply , C2 → C2 + C3
= 0, (∵ C1 = C2)
are
⇒ (1-x)(2-x)(3-x) = 0 ⇒x = 1,2,3
Let A be a square matrix of order 2 . then ,|adj.A| = |A|
, because , row 1 and row 3 are identical.
is equal to 
Apply , C1→C1 - C3, C2→C2-C3
= 10 - 12 = -2
If A is a non singular matrix of order 3 , then |adj(A3)| =
If A is anon singular matrix of order , then 
If A and B are any 2 × 2 matrices , then det. (A+B) = 0 implies
Det.(A+B) ≠ Det.A + Det.B.
If A B be two square matrices such that AB = O, then, 
Apply , C1 → C1 - C2, C2 → C2 - C3,
Because here row 1 and 2 are identical
, then equalsBecause , the determinant of a skew symmetric matrix of odd order is always zero and of even order is a non zero perfect square.
Because , the inverse of an identity matrix is an identity matrix.
If A and B are square matrices of same order and A’ denotes the transpose of A , then
By the property of transpose of a matrix ,(AB)’ = B’A’.
Only non-singular matrices possess inverse.
Apply, C1→ C1+ C2+C3+C4,
Apply, R1 →R1 - R2,
Apply, R1→ R1 - R2, R2 → R2 - R3
=(x+3a) (a -x)3 (1) = (x+3a)(a-x)3
If the entries in a 3 x 3 determinant are either 0 or 1 , then the greatest value of this determinant is :
Greatest value = 2
areOperate,
Apply R3→R3- R1, R2→ R2 -R1,
⇒ -6(5x2 - 20) +15(2x-4) = 0
⇒ (x- 2)(x+1) = 0⇒x=2, -1
⇒-6(5x2 - 20) + 15(2x-4) = 0
⇒(x-2)(x+1) = 0 ⇒ x=2, -1
In a third order determinant, each element of the first column consists of sum of two terms, each element of the second column consists of sum of three terms and each element of the third column consists of sum of four terms. Then it can be decomposed into n determinants, where n has value
N = 2 ×3 × 4 = 24.
|
209 videos|218 docs|139 tests
|
|
209 videos|218 docs|139 tests
|
Download free EduRev App
(Scan QR code)
|
|
|
© EduRev
|
Education Revolution
|
Follow Us
|
