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Matrics — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsMatrics2 Marks
MCQ
$A =\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4\end{array}\right] ; I =\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$A^{-1}=\frac{1}{6}\left(A^2+c A+d I\right)$ where $c, d \in R$, then pair of values ( $c, d$ ) is
  • A
    $(6,11)$
  • B
    $(6,-11)$
  • ✓
    $(-6,11)$
  • D
    $(-6,-11)$
✓

Answer

Correct option: C.
$(-6,11)$
(C) $|A|=\left|\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4\end{array}\right|=6 \neq 0$
$\operatorname{adj} A=\left[\begin{array}{ccc}6 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 2 & 1\end{array}\right]$
$\therefore \quad A ^{-1}=\frac{1}{6}\left[\begin{array}{ccc}6 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 2 & 1\end{array}\right]$
$A^2=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4\end{array}\right]$
$=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & -1 & 5 \\ 0 & -10 & 14\end{array}\right]$
$\therefore \quad A ^2+ cA + dI$
$=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & -1 & 5 \\ 0 & -10 & 14\end{array}\right]+\left[\begin{array}{ccc} c & 0 & 0 \\ 0 & c & c \\ 0 & -2 c & 4 c \end{array}\right]+\left[\begin{array}{ccc} d & 0 & 0 \\ 0 & d & 0 \\ 0 & 0 & d\end{array}\right]$
$=\left[\begin{array}{ccc}1+ c + d & 0 & 0 \\ 0 & -1+ c + d & 5+ c \\ 0 & -10-2 c & 14+4 c + d \end{array}\right]$
Since $6 A^{-1}=A^2+c A+d I$
$\therefore\left[\begin{array}{ccc}6 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 2 & 1\end{array}\right]=\left[\begin{array}{ccc}1+ c + d & 0 & 0 \\ 0 & -1+ c + d & 5+ c \\ 0 & -10-2 c & 14+4 c + d \end{array}\right]$
∴ by equality of matrices,
$1+c+d=6$ and $5+c=-1$
$\therefore \quad c=-6$ and $d=11$

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