Question 132

If A is a 3 x 3, square matrix and if A Adj $$A = \begin{bmatrix}4 & 0 & 0\\0 & 4&0 \\ 0 & 0 & 4\end{bmatrix}$$, then det(2A) =

Solution

given that A Adj $$A = \begin{bmatrix}4 & 0 & 0\\0 & 4&0 \\ 0 & 0 & 4\end{bmatrix}$$

means that

A Adj $$A = \begin{bmatrix}4 & 0 & 0\\0 & 4&0 \\ 0 & 0 & 4\end{bmatrix}$$ = |A|I = 4 $$\begin{bmatrix}1 & 0 & 0\\0 & 1&0 \\ 0 & 0 & 1\end{bmatrix}$$

 ∣A∣=4

det (2Adj A)  =  ∣A∣^(n−1) , where n is number of rows in a matrix

                     =  2$$\times$$4^(3-1)
                     =  2$$\times$$ 16   

                     = 32  Answer 


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