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