Let a = , 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,
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| = 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
If A’ is the transpose of a square matrix A , then
The determinant of a matrix A and its transpose always same.
The value of the determinant is
, Apply , C2 → C2 + C3
= 0, (∵ C1 = C2)
The roots of the equation det. are
⇒ (1-x)(2-x)(3-x) = 0 ⇒x = 1,2,3
If A is a square matrix of order 2 , then det (adj A) = x
Let A be a square matrix of order 2 . then ,|adj.A| = |A|
If A is a symmetric matrix, then At =
, because , row 1 and row 3 are identical.
is equal to
Apply , C1→C1 - C3, C2→C2-C3
= 10 - 12 = -2
If A+B+C = π, then the value of
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
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
If , then equals
Because , the determinant of a skew symmetric matrix of odd order is always zero and of even order is a non zero perfect square.
If I3 is the identity matrix of order 3 , then 13−1 is
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’.
A square matrix A is invertible iff det A is equal to
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
The roots of the equation are
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.