Math, asked by hey69, 1 year ago

if A is a square matrix of order 3, such that  / adj.A / = 64  . then find / A' / .​

Answers

Answered by Shubu01
2

\huge{Hey Mate!!!}

☆☞ Here is ur answer ☜☆

✔✔ It can be explained as shown

✴ We have

A(adj.A) = (adj.A) = |A| * I

 | A(adj.A) | =  |  |A| * I  |

|A|  |adj.A | =  |A| | I | ..... ( since | kA | = kn|A| )

|A|  |adj.A |  =  |A|n * 1

| adj.A |  =  |A|n /  |A|

✔✔ This gives

| adj.A | = |A|n-1

where n is the order of the matrix.

 

✔✔ So in this question

| adj.A | = |A|n-1

64 = | A | 2

|A| = +8 or -8

HOPE IT HELPS!

Answered by dorgan399
0

 

A%2Aadj%28A%29+=+abs%28A%29I%5B3%5D,  

where abs%28A%29 is the determinant of A.  

===> abs%28A%29%2A%28adj%28A%29%29+=+abs%28A%29%5E3,  

since A is of order 3. (The determinant was applied on both sides.)  

===>adj%28A%29+=++abs%28A%29%5E2,  

===> abs%28A%29%5E2+=+36 ===> |A| = 6 or |A| = -6.  

===>  abs%28A%5ET%29+=+6 or  abs%28A%5ET%29+=+-6, since abs%28A%29+=+abs%28A%5ET%29.

Similar questions