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Apply C_{2 }→ C_{2}_{ }+ C_{3,}
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 , R_{1} → R_{1}+R_{2}+R_{3},
Apply , C_{3}→ C_{3 }_{ }C_{1}, C_{2}^{→}C_{2}  C_{1},
=(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.
, Apply , C_{2} → C_{2} + C_{3}
= 0, (∵ C_{1} = C_{2})
⇒ (1x)(2x)(3x) = 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.
Apply , C1→C1  C3, C_{2}→C_{2}C_{3}
= 10  12 = 2
If A is a non singular matrix of order 3 , then adj(A^{3}) =
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 , C_{1} → C_{1}  C_{2}, C_{2} → C_{2}  C_{3},
Because here row 1 and 2 are identical
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 I_{3} is the identity matrix of order 3 , then 1_{3}^{−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’.
Only nonsingular matrices possess inverse.
Apply, C_{1}→ C_{1}+ C_{2}+C_{3}+C_{4},
Apply, R1 →R_{1}  R_{2},
Apply, R_{1}→_{ }R_{1 } R_{2},_{ }R_{2 }→ R_{2 } R_{3}
_{}
_{}
=(x+3a) (a x)^{3} (1) = (x+3a)(ax)^{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
Operate,
Apply R_{3}→R_{3} R_{1},_{ }R_{2}→_{ }R_{2} R_{1},
⇒ 6(5x^{2}  20) +15(2x4) = 0
⇒ (x 2)(x+1) = 0⇒x=2, 1
⇒6(5x^{2}  20) + 15(2x4) = 0
⇒(x2)(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.
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209 videos218 docs139 tests
