CUET · MATHS · PYQ PAPER 2025
If \(A=\left[\begin{array}{ll}5 & 3 \\ 2 & 4\end{array}\right]\) then the matrix \(A^2-6 A+14 I\) (where \(I\) is an identity matrix of order 2 )
- A \(\left[\begin{array}{cc}15 & 6 \\ 9 & 8\end{array}\right]\)
- B \(\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\)
- C \(\left[\begin{array}{cc}15 & 9 \\ -12 & -6\end{array}\right]\)
- D \(\left[\begin{array}{cc}15 & 9 \\ 6 & 12\end{array}\right]\)
Answer & Solution
Correct Answer
(D) \(\left[\begin{array}{cc}15 & 9 \\ 6 & 12\end{array}\right]\)
Step-by-step Solution
Detailed explanation
\(Tr(A) = 5+4 = 9\) \(det(A) = 5(4) - 3(2) = 14\) \(A^2 - Tr(A)A + det(A)I = 0 \implies A^2 - 9A + 14I = 0\) \(A^2 = 9A - 14I\) \(A^2 - 6A + 14I = (9A - 14I) - 6A + 14I\) \(= 3A\) \(= 3 \left[\begin{array}{ll}5 & 3 \\ 2 & 4\end{array}\right]\)…
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