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If A=[1203]A = \begin{bmatrix}1 & 2\\ 0 & 3\end{bmatrix} then ∣A.adjA∣|A. adj A| is

Solution

✅ Correct Option: 3

Given A=[1203]A = \begin{bmatrix}1 & 2\\ 0 & 3\end{bmatrix}

For any matrix multiplication: ∣A⋅B∣=∣A∣×∣B∣|A \cdot B| = |A| \times |B|

Therefore: ∣A⋅adj(A)∣=∣A∣×∣adj(A)∣|A \cdot adj(A)| = |A| \times |adj(A)|


For a 2×2 matrix [abcd]\begin{bmatrix}a & b\\ c & d\end{bmatrix}, the determinant is ad−bcad - bc

∣A∣=(1)(3)−(2)(0)|A| = (1)(3) - (2)(0)

∣A∣=3−0|A| = 3 - 0

∣A∣=3|A| = 3


For an n×n matrix, ∣adj(A)∣=∣A∣n−1|adj(A)| = |A|^{n-1}

Since A is a 2×2 matrix, n = 2:

∣adj(A)∣=∣A∣2−1|adj(A)| = |A|^{2-1}

∣adj(A)∣=∣A∣1|adj(A)| = |A|^{1}

∣adj(A)∣=3|adj(A)| = 3


∣A⋅adj(A)∣=∣A∣×∣adj(A)∣|A \cdot adj(A)| = |A| \times |adj(A)|

∣A⋅adj(A)∣=3×3|A \cdot adj(A)| = 3 \times 3

∣A⋅adj(A)∣=9|A \cdot adj(A)| = 9

Therefore, ∣AadjA∣=9|A adj A| = 9

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