CUET Mathematics 2024Algebra > Easy[457−3−300−3−2]\left[\begin{array}{ccc}4 & 5 & 7 \\ -3 & -3 & 0 \\ 0 & -3 & -2\end{array}\right]4−305−3−370−2[−2424−8−4−121]\left[\begin{array}{rrr}-2 & 4 & 2 \\ 4 & -8 & -4 \\ -1 & 2 & 1\end{array}\right]−24−14−822−41[552767−9−70]\left[\begin{array}{rrr}5 & 5 & 2 \\ 7 & 6 & 7 \\ -9 & -7 & 0\end{array}\right]57−956−7270[−248757−8−26]\left[\begin{array}{rrr}-2 & 4 & 8 \\ 7 & 5 & 7 \\ -8 & -2 & 6\end{array}\right]−27−845−2876✅ Correct Option: 2Related questions:If AAA is a square matrix and III is an identity matrix such that A2=AA^{2}=AA2=A, then A(I−2A)3+2A3A(I-2 A)^{3}+2 A^{3}A(I−2A)3+2A3 is equal to :The matrix [100010001]\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]100010001 is a : (A) scalar matrix (B) diagonal matrix (C) skew-symmetric matix (D) symmetric matrix Choose the correct answer from the options given below :For a square matrix An×nA_{n \times n}An×n (A) ∣adjA∣=∣A∣n−1|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{\mathrm{n}-1}∣adjA∣=∣A∣n−1 (B) ∣A∣=∣adjA∣n−1|\mathrm{A}|=|\operatorname{adj} \mathrm{A}|^{\mathrm{n}-1}∣A∣=∣adjA∣n−1 (C) A(adjA)=∣A∣\mathrm{A}(\operatorname{adj} \mathrm{A})=|\mathrm{A}|A(adjA)=∣A∣ (D) ∣A−1∣=1∣ A∣\left|\mathrm{A}^{-1}\right|=\frac{1}{|\mathrm{~A}|}A−1=∣ A∣1