A = \begin{bmatrix}2 & 3 \\\4 & 5 \end{bmatrix} \Rightarrow \mid A^5 \mid =

Question 132

$$A = \begin{bmatrix}2 & 3 \\4 & 5 \end{bmatrix} \Rightarrow \mid A^5 \mid =$$

Solution

Upon successive multiplication, obtain the matrix $$A^5$$
Its terms are:
$$A_{11}=6140$$, $$A_{12}=8097$$, $$A_{21}=10796$$, $$A_{22}=14237$$

Cross multiplication and subtraction of terms results in determinant = -32

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AP ICET 2018 Question Paper Shift-2 (2nd May)

$$A = \begin{bmatrix}2 & 3 \\4 & 5 \end{bmatrix} \Rightarrow \mid A^5 \mid =$$

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