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3 and 4 . determinant and metrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths3 and 4 . determinant and metrices1 Mark
MCQ
$A = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&1\\0&{ - 2}&4\end{array}} \right];\,\,I = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$${A^{ - 1}} = \frac{1}{6}[{A^2} + cA + dI]$ where $c,d \in R$, then pair of values $(c,d)$
  • A
    $(6, 11)$
  • B
    $(6, -11)$
  • ✓
    $(-6, 11)$
  • D
    $(-6, -11)$
✓

Answer

Correct option: C.
$(-6, 11)$
c
(c) Given $A = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&1\\0&{ - 2}&4\end{array}} \right],\,\,{A^{ - 1}} = \frac{1}{6}\left[ {\begin{array}{*{20}{c}}6&0&0\\0&4&{ - 1}\\0&2&1\end{array}} \right]$

${A^2} = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&1\\0&{ - 2}&4\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&1\\0&{ - 2}&4\end{array}} \right]\, = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&{ - 1}&5\\0&{ - 10}&{14}\end{array}} \right]$

$cA = \left[ {\begin{array}{*{20}{c}}c&0&0\\0&c&c\\0&{ - 2c}&{4c}\end{array}} \right]\,$; $dI = \left[ {\begin{array}{*{20}{c}}d&0&0\\0&d&0\\0&0&d\end{array}} \right]$

$\therefore $By ${A^{ - 1}} = \frac{1}{6}[{A^2} + cA + dI]$

$ 6 = 1 + c + d$,(By equality of matrices)

$\therefore $ $ (-6,11)$ satisfy the relation.

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