∵A=[a1100a12a220a13a23a33]⇒∣A∣=a11a22a33
⇒∣A∣=(J7,3−J4,3)⋅(J8,3−J5,3)⋅(J9,3−J6,3)
=∫01/2(x3−1x7−x4)dx⋅∫01/2(x3−1x8−x5)dx⋅∫01/2(x3−1x9−x6)dx
=∫01/2x4dx⋅∫01/2x5dx⋅∫01/2x6dx
=(210)⋅(2)181
Now
∣adjA−1∣=∣A−1∣2=∣A∣21=((210)⋅(218))2=(105)2⋅(2)38