Question 1

# 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)

Solution

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)

Question 2

## Consider the system of linear equationx1 + 2x2 + x3 = 32x1 + 3x2 + x3 = 3 3x1 + 5x2 + 2x3 = 1The system has infinite number of solutions exactly 3 solutions a unique solution no solution

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

Question 3

## If A =  and A adj A = AAT, then 5a +b is equal to -1 5 4 5

Solution

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

Question 4

## 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

Solution

B.

l

If 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]
lTl

Question 5

Solution

C.