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Standard XII

Mathematics

Skew Symmetric Matrix

If A is a sin...

Question

If A is a singular matrix of order (nxn), then rank
(A) is,

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\leq n

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< n

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= n

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Solution

The correct option is C $< n$
Rank of square matrix is always less than or equal to order of matrix.
So, A is singular matrix. so, $| A | = 0$
$∴$ rank of $A \leq n$ .

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Similar questions

Q. If A is a non-singular matrix of order (nxn), then rank (A) is ?

Q. If

A

is a non-singular square matrix of order

n,

then

a d j (a d j (A))

is equal to

Q. If

A

is a square matrix,

B

is a singular matrix of same order, then for a positive integer

n, (A^{- 1} B A)^{n}

equals

Q. Assertion :If

A = ⎡ ⎢ ⎣ \begin{matrix} 2 & 0 & - 1 5 & 1 & 0 0 & 1 & 3 \end{matrix} ⎤ ⎥ ⎦

then

a d j (a d j A) = A

Reason:

| a d j \cdot (a d j \cdot A) | =

{| A |}^{(n - 1)^{2}}, A

is a non singular matrix of order n

If A, B are square matrices of order 3, A is non-singular and AB = O, then B is a
(a) null matrix
(b) singular matrix
(c) unit-matrix
(d) non-singular matrix

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If A is a singular matrix of order (nxn), then rank(A) is,

The correct option is C <n Rank of square matrix is always less than or equal to order of matrix.So, A is singular matrix. so, |A|=0∴ rank of A≤n.

If A is a singular matrix of order (nxn), then rank
(A) is,

The correct option is C $< n$
Rank of square matrix is always less than or equal to order of matrix.
So, A is singular matrix. so, $| A | = 0$
$∴$ rank of $A \leq n$ .