Let A be a square matrix of order n such that A-10 and 2A2-40- Then the order of A is

We know that, (KA)=K^n·(A) ⇒(2A)=2^n·(A) ⇒(2A)2=2^n 1· (A) ⇒40=2^n 1×10 ⇒2^n 1=4 ⇒ n=3 (order of matrix A) ...

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Standard XII

Mathematics

Inverse of a Matrix

Let A be a sq...

Question

Let $A$ be a square matrix of order $n$ such that $det (A) = 10$ and $\frac{det (2 A)}{2} = 40.$ Then the order of $A$ is

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Solution

The correct option is B $3$
We know that,
$det (K A) = K^{n} \cdot det (A)$
$\Rightarrow det (2 A) = 2^{n} \cdot det (A)$
$\Rightarrow \frac{det (2 A)}{2} = 2^{n - 1} \cdot det (A)$
$\Rightarrow 40 = 2^{n - 1} \times 10$
$\Rightarrow 2^{n - 1} = 4$
$\Rightarrow n = 3$ (order of matrix $A$ )

Suggest Corrections

Similar questions

Q. Let

A

be a square matrix of order

n

such that

det (A) = 10

and

\frac{det (2 A)}{2} = 40.

Then the order of

A

is

[1 mark]

Q. If

A

is a square matrix of order

3

and

A^{'}

denotes transpose of matrix

A, i f A = A^{'}

and

d e t A = 1

, then

d e t (A - I)

must be equal to

Q. Assertion :If A is skew symmetric matrix of order 3 then its determinant should be zero Reason: If A is square matrix, then

d e t A = d e t A^{'} = d e t (- A^{'})

Q. Let A be a square matrix of order 3 such that

|A|

= 11 and B be the matrix of confactors of elements of A. Then,

{|B|}^{2}

= ________________.

Q. Assertion :If

A

is skew symmetric matrix of order

3 \times 3

, then

d e t (A) = 0

| A | = 0

Reason: If

A

is a square matrix, then

d e t (A) = d e t (A)^{'} = d e t ({- A}^{^{'}})

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Inverse of a Matrix

Standard XII Mathematics

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Let A be a square matrix of order n such that det(A)=10 and det(2A)2=40. Then the order of A is

The correct option is B 3We know that, det(KA)=Kn⋅det(A) ⇒det(2A)=2n⋅det(A) ⇒det(2A)2=2n−1⋅det(A) ⇒40=2n−1×10 ⇒2n−1=4 ⇒n=3 (order of matrix A)

Let $A$ be a square matrix of order $n$ such that $det (A) = 10$ and $\frac{det (2 A)}{2} = 40.$ Then the order of $A$ is

The correct option is B $3$
We know that,
$det (K A) = K^{n} \cdot det (A)$
$\Rightarrow det (2 A) = 2^{n} \cdot det (A)$
$\Rightarrow \frac{det (2 A)}{2} = 2^{n - 1} \cdot det (A)$
$\Rightarrow 40 = 2^{n - 1} \times 10$
$\Rightarrow 2^{n - 1} = 4$
$\Rightarrow n = 3$ (order of matrix $A$ )