Determinants
Sponsor Area
A =
is a matrix satisfying the equation AAT = 9I, Where I is 3 x 3 identity matrix, then the ordered pair (a,b) is equal to
-
(2,-1)
-
(-2,1)
-
(2,1)
-
(-2,-1)
D.
(-2,-1)
Given,
.
It is given that,
On comparing we get,
a+ 4 +2b = 0
a+ 2b = -4 ... (i)
2a + 2-2b = 0
a-b= -1 ... (ii)
a2 + 4 +b2 = 9 ... (iii)
On solving eqs. (i) and (ii) we get
a = - 2, b = - 1
Hence, (a,b) ≡ (-2,-1)
Sponsor Area
Consider the system of linear equation
x1 + 2x2 + x3 = 3
2x1 + 3x2 + x3 = 3
3x1 + 5x2 + 2x3 = 1
The system has
-
infinite number of solutions
-
exactly 3 solutions
-
a unique solution
-
no solution
D.
no solution
Subtracting the Eq. (ii) – Eq. (i)
We get x1 + x2 = 0
Subtract equations
Eq. (iii) – 2 × eq. (ii)
x1 + x2 = 5
Therefore, no solutions
If A =
and A adj A = AAT, then 5a +b is equal to
-
-1
-
5
-
4
-
5
B.
5
Given, A =
and A adj A = AAT, Clearly, A (adj A) = |A|In|
Now, substituting the value of b in Eq. (iii) we get
5a = 2
Hence, 5a + b = 2 +3 = 5
If A is a 3x3 non- singular matrix such that AAT = ATA, then BBT is equal to
- l +B
- l
-
B-1
-
(B-1)T
B.
lIf A is non - singular matrix then |A| ≠0
AAT = ATA and B = A-1AT
BBT = (A-1AT)(A-1AT)T
= A-1ATA(A-1)T [∵ (AB)T= BTAT]
=A-1AAT(A-1)T [∵ AAT = ATA]
=AT(A-1)T [ ∵A-1A = l]
=A-1A)T [∵ (AB)T = BTAT]
lT = l
If A,
then adj (3A2 + 12A) is equal to
C.
Sponsor Area
Mock Test Series
Mock Test Series



