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Q.If A=[4132]A = \begin{bmatrix}4 & 1\\ 3 & 2\end{bmatrix} and I=[1001]I = \begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}, then (A2−6A)(A^2 - 6A) is equal to : (A) 3I3I (B) −5I-5I (C) 5I5I (D) −3I-3I

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A2−6A=[−500−5]=−5IA^2-6A = \begin{bmatrix}-5&0\\0&-5\end{bmatrix} = -5I.

For A=[abcd]A=\begin{bmatrix}a&b\\c&d\end{bmatrix}, A2=A⋅AA^2 = A\cdot A (row-by-column products) and 6A6A scales every entry by 6; subtract entrywise.

  1. A2=[4132][4132]=[16+34+212+63+4]=[196187]A^2 = \begin{bmatrix}4&1\\3&2\end{bmatrix}\begin{bmatrix}4&1\\3&2\end{bmatrix} = \begin{bmatrix}16+3 & 4+2\\ 12+6 & 3+4\end{bmatrix} = \begin{bmatrix}19&6\\18&7\end{bmatrix}.
  2. 6A=[2461812]6A = \begin{bmatrix}24&6\\18&12\end{bmatrix}.
  3. A2−6A=[19−246−618−187−12]=[−500−5]A^2-6A = \begin{bmatrix}19-24 & 6-6\\ 18-18 & 7-12\end{bmatrix} = \begin{bmatrix}-5&0\\0&-5\end{bmatrix}.
  4. This equals −5[1001]=−5I-5\begin{bmatrix}1&0\\0&1\end{bmatrix} = -5I.
✓Final answer

(B) −5I-5I

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