For a matrix

MATHEMATICS · Single correct · CBSE · Mathematics · Determinants · Determinants

Problem

For a 3×33\times 3 matrix AA, suppose ∣A∣=12|A|=12 and adj A=(4−2810−352−1).\text{adj}\,A=\begin{pmatrix}4 & -2 & 8\\ 1 & 0 & -3\\ 5 & 2 & -1\end{pmatrix}. Which of the following matrices is A−1A^{-1}? (a) (23−1343160−125613−16)\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} (b) (13112512−1601623−14−112)\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} (c) (13−16231120−1451216−112)\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} (d) (136−17211811440−1485144172−1144)\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}

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