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COMEDK · Maths · 21. Matrices

If \(A=\left[\begin{array}{ccc}1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\), then \(A \cdot \operatorname{adj}(A)\) is equal to

  1. A \(\left[\begin{array}{lll}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{array}\right]\)
  2. B \(\left[\begin{array}{lll}5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5\end{array}\right]\)
  3. C \(\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]\)
  4. D \(\left[\begin{array}{lll}8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8\end{array}\right]\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\left[\begin{array}{lll}8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8\end{array}\right]\)

Step-by-step Solution

Detailed explanation

We have, \[ \begin{aligned} A &=\left[\begin{array}{ccc} 1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right] \\ \therefore \quad|A| &=\left[\begin{array}{ccc} 1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right] \\ &=1[8-6]+3[6-4]=2+6=8 \end{aligned} \] Now, we know…