Test: Matrices

For any matrix A, (A + A’) is always symmetric and (A - A') is always skew-symmetric.

where P = 1/2(A+ A'), a symmetric matrix and Q = 1/212(A - A'), is a skew-symmetric matrix.

Caluclation:

∴ Required symmetric matrix Q
= 1/2 (B-B')

 which is the required answer.

If A is any matrix, the (A^T)^T = A.

We have,

 

∴ (A^T)^T 

= A

Given:
A is a symmetric matrix,
⇒ AT = A or a_ij = a_ji

So, by property of symmetric matrices

Given:
Order of A is 4 × 3, the order of B is 4 × 5 and the order of C is 7 x 3
The transpose of the matrix obtained by interchanging the rows and columns of the original matrix.
So, order of AT is 3 × 4 and order of CT is 3 × 7
Now,
A^TB = {3 x 4} {4 × 5} = 3 x 5
⇒ Order of ATB is 3 x 5
Hence order of (A^TB)^T is 5 x 3
Now order of (A^TB)^T C^T = {5 x 3} {3 x 7} = 5 x 7
∴ Order of (A^TB)^T C^T is 5 x 7

⇒ [(2x - 9)x + 8 x 4x] = 0

⇒ [2x² - 9x + 32x] = 0

⇒ 2x² + 23x = 0

⇒ x(2x + 23) = 0

⇒ x = 0 or - 23/2

Given that A is involuntary matrix,
⇒ A² = I
Now,
(I − A) (I + A) = I² – IA + AI − A² 
⇒ I – A + A – I         (∵ A² = I)
⇒ 0
∴ (I − A) (I + A) is zero matrix.

From statement 1∶
AB = A
We cannot find anything from this statement.
From statement 2∶

We cannot find anything from this statement.
Combining statement 1 and 2∶

∴ We cannot find the value of n from both statements together.

Given: A² - 2A - I = 0
⇒ A.A - 2A = I
Post multiply by A^-1, we get
⇒ AAA^-1 - 2AA^-1 = IA^-1
⇒ AI - 2I = A^-1             [∵ AA ^-1 = A ^{- 1}A = I]
∴ A^-1 = A - 2
the inverse of A is A - 2

Now,

Hence Option 1st is correct answer.

Given