CUET Mathematics 2025 2 June Shift 1Algebra > Mediumcommon[1100]\begin{bmatrix}1 & 1\\0 & 0\end{bmatrix}[1010][−113−4]\begin{bmatrix}-1 & 1\\3 & -4\end{bmatrix}[−131−4][1−1−34]\begin{bmatrix}1 & -1\\-3 & 4\end{bmatrix}[1−3−14][1001]\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}[1001]✅ Correct Option: 3Related questions:6 Aug Shift 2If 3A=[12221−2x2y]3A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ x & 2 & y \end{bmatrix}3A=12x2122−2y and AAT=IAA^T = IAAT=I, then x+yx + yx+y is equal to3 June Shift 1If A=[123−4−5−2]A = \begin{bmatrix} 1 & 2 & 3 \\ -4 & -5 & -2 \end{bmatrix}A=[1−42−53−2], B=[2−34−52−1]B = \begin{bmatrix} 2 & -3 \\ 4 & -5 \\ 2 & -1 \end{bmatrix}B=242−3−5−1 and BA=[bij]BA = [b_{ij}]BA=[bij], then (b23−b31)(b_{23} - b_{31})(b23−b31) is equal to4 Aug Shift 1If A and B are square matrices of same order n, then identify correct statements from the statements given below: A. ∣adj A∣=∣A∣n−1|adj\ A| = |A|^{n-1}∣adj A∣=∣A∣n−1 B. ∣A⋅B∣=∣B∣⋅∣A∣|A \cdot B| = |B| \cdot |A|∣A⋅B∣=∣B∣⋅∣A∣ C. adj A′=(adj A)′adj\ A' = (adj\ A)'adj A′=(adj A)′ D. adj AB=(adj A)⋅(adj B)adj\ AB = (adj\ A) \cdot (adj\ B)adj AB=(adj A)⋅(adj B) E. ∣An∣=∣A∣n|A^n| = |A|^n∣An∣=∣A∣n Choose the correct answer from the options given below: