A = is a matrix satisfying the equation AA^{T} = 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)
Consider the system of linear equation
x^{1} + 2x^{2} + x^{3} = 3
2x^{1} + 3x^{2} + x^{3} = 3
3x^{1} + 5x^{2} + 2x^{3} = 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 x_{1} + x_{2} = 0
Subtract equations
Eq. (iii) – 2 × eq. (ii)
x_{1} + x_{2} = 5
Therefore, no solutions
If A = and A adj A = AA^{T}, then 5a +b is equal to
-1
5
4
5
B.
5
Given, A = and A adj A = AA^{T}, Clearly, A (adj A) = |A|I_{n}|
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 AA^{T} = A^{T}A, then BB^{T} is equal to
B^{-1}
(B^{-1})^{T}
B.
lIf A is non - singular matrix then |A| ≠0
AA^{T} = A^{T}A and B = A^{-1}A^{T}
BB^{T} = (A^{-1}A^{T})(A^{-1}A^{T})^{T}
= A^{-1}A^{T}A(A^{-1})^{T} [∵ (AB)^{T}= B^{T}A^{T}]
=A^{-1}AA^{T}(A^{-1})^{T} [∵ AA^{T} = A^{T}A]
=A^{T}(A^{-1})^{T} [ ∵A^{-1}A = l]
=A^{-1}A)^{T} [∵ (AB)^{T} = B^{T}A^{T}]
l^{T} = l
If A, then adj (3A2 + 12A) is equal to
C.