Question

5. If A is a matrix such that A(adj A) =
20 0
0 20
then |A|=?​

Answer 1

♣ ANSWER:-

A(AdjA)=

⎝

⎜

⎜

⎛

4

0

0

0

4

0

0

0

4

⎠

⎟

⎟

⎞

=∣A∣I

⇒4

⎣

⎢

⎢

⎡

1

0

0

0

1

0

0

0

1

⎦

⎥

⎥

⎤

=∣A∣I

∴∣A∣=4

det (Adj A) =∣A∣

n−1

, where n is number of rows in a matrix

=∣4∣

3−1

=16.

hopefully it helps

Answer 2

HEY MATE HERE IS YOUR EXPLAINATION AND ANSWER

Consider the following identity A(adjA)=∣A∣I. Thus comparing it with the above equation gives us ∣A∣=10

Consider the following identity A(adjA)=∣A∣I. Thus comparing it with the above equation gives us ∣A∣=10Now ∣adjA∣=∣A∣

Consider the following identity A(adjA)=∣A∣I. Thus comparing it with the above equation gives us ∣A∣=10Now ∣adjA∣=∣A∣ n−1

Consider the following identity A(adjA)=∣A∣I. Thus comparing it with the above equation gives us ∣A∣=10Now ∣adjA∣=∣A∣ n−1 where n is the order of the square matrix.

Consider the following identity A(adjA)=∣A∣I. Thus comparing it with the above equation gives us ∣A∣=10Now ∣adjA∣=∣A∣ n−1 where n is the order of the square matrix.Here 'n' is 3, therefore, ∣adjA∣=∣A∣

Consider the following identity A(adjA)=∣A∣I. Thus comparing it with the above equation gives us ∣A∣=10Now ∣adjA∣=∣A∣ n−1 where n is the order of the square matrix.Here 'n' is 3, therefore, ∣adjA∣=∣A∣ 3−1

Consider the following identity A(adjA)=∣A∣I. Thus comparing it with the above equation gives us ∣A∣=10Now ∣adjA∣=∣A∣ n−1 where n is the order of the square matrix.Here 'n' is 3, therefore, ∣adjA∣=∣A∣ 3−1 =∣A∣

Consider the following identity A(adjA)=∣A∣I. Thus comparing it with the above equation gives us ∣A∣=10Now ∣adjA∣=∣A∣ n−1 where n is the order of the square matrix.Here 'n' is 3, therefore, ∣adjA∣=∣A∣ 3−1 =∣A∣ 2

Consider the following identity A(adjA)=∣A∣I. Thus comparing it with the above equation gives us ∣A∣=10Now ∣adjA∣=∣A∣ n−1 where n is the order of the square matrix.Here 'n' is 3, therefore, ∣adjA∣=∣A∣ 3−1 =∣A∣ 2 =10

Consider the following identity A(adjA)=∣A∣I. Thus comparing it with the above equation gives us ∣A∣=10Now ∣adjA∣=∣A∣ n−1 where n is the order of the square matrix.Here 'n' is 3, therefore, ∣adjA∣=∣A∣ 3−1 =∣A∣ 2 =10 2

Consider the following identity A(adjA)=∣A∣I. Thus comparing it with the above equation gives us ∣A∣=10Now ∣adjA∣=∣A∣ n−1 where n is the order of the square matrix.Here 'n' is 3, therefore, ∣adjA∣=∣A∣ 3−1 =∣A∣ 2 =10 2 =100.