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Matrices and Determinants question

2020 · 8 Jan · Shift 2 · Q27
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Matrices and Determinants question

2020 · 8 Jan · Shift 2 · Q27

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If A=(2294)A = \left( {\begin{matrix} 2 & 2 \\ 9 & 4 \\ \end{matrix} } \right)A=(29​24​) and I=(1001)I = \left( {\begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} } \right)I=(10​01​) then 10A–1 is equal to :
  1. A
    6I – A
  2. B
    4I – A
  3. C
    A – 6I
  4. D
    A – 4I
View written solutionFree

Correct answer: C

  1. We need to find 10A−110A^{-1}10A−1 for A=(2294).A=\begin{pmatrix}2&2\\9&4\end{pmatrix}.A=(29​24​).

  2. First compute the determinant of AAA: ∣A∣=2⋅4−2⋅9=8−18=−10.|A|=2\cdot 4-2\cdot 9=8-18=-10.∣A∣=2⋅4−2⋅9=8−18=−10.

  3. For a 2×22\times 22×2 matrix (abcd)−1=1ad−bc(d−b−ca).\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}.(ac​bd​)−1=ad−bc1​(d−c​−ba​). So, A−1=1−10(4−2−92).A^{-1}=\frac{1}{-10}\begin{pmatrix}4&-2\\-9&2\end{pmatrix}.A−1=−101​(4−9​−22​).

  4. Multiply by 101010:

=-\begin{pmatrix}4&-2\\-9&2\end{pmatrix} =\begin{pmatrix}-4&2\\9&-2\end{pmatrix}.$$ 5. Now compare with the given options. First compute $A-6I$: $$A-6I=\begin{pmatrix}2&2\\9&4\end{pmatrix}-\begin{pmatrix}6&0\\0&6\end{pmatrix} =\begin{pmatrix}-4&2\\9&-2\end{pmatrix}.$$ This matches exactly with $10A^{-1}$. Check others quickly: - $6I-A=\begin{pmatrix}4&-2\\-9&2\end{pmatrix}$, not equal. - $4I-A=\begin{pmatrix}2&-2\\-9&0\end{pmatrix}$, not equal. - $A-4I=\begin{pmatrix}-2&2\\9&0\end{pmatrix}$, not equal. 6. Therefore, the correct option is $$\boxed{\text{C: }A-6I}.$$
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