Given: A=[2−112]
⇒∣A∣=2+1=3
B=[1101]⇒∣B∣=1
And C=ABAT
⇒∣C∣=∣ABAT∣
⇒∣C∣=∣A∣2∣B∣as ∣A∣=∣AT∣
⇒∣C∣=9
Now, solving X=AC2AT
.⇒∣X∣=∣A∣∣C2∣∣AT∣
⇒∣X∣=3×92×3
⇒∣X∣=729
If A=[2−112],B[1101],C=ABAT and X=ATC2A, then det X is equal to:
Held on 1 Feb 2024 · Verified 6 Jul 2026.
243
729
27
891
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