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MATHEMATICS · Single correct · CBSE · Mathematics · Determinants · Determinants
Problem For a 3 × 3 3\times 3 3 × 3 matrix A A A , suppose ∣ A ∣ = 12 |A|=12 ∣ A ∣ = 12 and adj A = ( 4 − 2 8 1 0 − 3 5 2 − 1 ) . \text{adj}\,A=\begin{pmatrix}4 & -2 & 8\\ 1 & 0 & -3\\ 5 & 2 & -1\end{pmatrix}. adj A = 4 1 5 − 2 0 2 8 − 3 − 1 .
Which of the following matrices is A − 1 A^{-1} A − 1 ?
(a) ( 2 3 − 1 3 4 3 1 6 0 − 1 2 5 6 1 3 − 1 6 ) \begin{pmatrix}\frac{2}{3} & -\frac{1}{3} & \frac{4}{3}\\ \frac{1}{6} & 0 & -\frac{1}{2}\\ \frac{5}{6} & \frac{1}{3} & -\frac{1}{6}\end{pmatrix} 3 2 6 1 6 5 − 3 1 0 3 1 3 4 − 2 1 − 6 1
(b) ( 1 3 1 12 5 12 − 1 6 0 1 6 2 3 − 1 4 − 1 12 ) \begin{pmatrix}\frac{1}{3} & \frac{1}{12} & \frac{5}{12}\\ -\frac{1}{6} & 0 & \frac{1}{6}\\ \frac{2}{3} & -\frac{1}{4} & -\frac{1}{12}\end{pmatrix} 3 1 − 6 1 3 2 12 1 0 − 4 1 12 5 6 1 − 12 1
(c) ( 1 3 − 1 6 2 3 1 12 0 − 1 4 5 12 1 6 − 1 12 ) \begin{pmatrix}\frac{1}{3} & -\frac{1}{6} & \frac{2}{3}\\ \frac{1}{12} & 0 & -\frac{1}{4}\\ \frac{5}{12} & \frac{1}{6} & -\frac{1}{12}\end{pmatrix} 3 1 12 1 12 5 − 6 1 0 6 1 3 2 − 4 1 − 12 1
(d) ( 1 36 − 1 72 1 18 1 144 0 − 1 48 5 144 1 72 − 1 144 ) \begin{pmatrix}\frac{1}{36} & -\frac{1}{72} & \frac{1}{18}\\ \frac{1}{144} & 0 & -\frac{1}{48}\\ \frac{5}{144} & \frac{1}{72} & -\frac{1}{144}\end{pmatrix} 36 1 144 1 144 5 − 72 1 0 72 1 18 1 − 48 1 − 144 1
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