Question

If A = \(\rm \begin{bmatrix} -1 & 0 &1 \\ -2&3 & 4\\ 0& 2 & 1 \end{bmatrix}\), then [A{Adj(A)}A^-1]A is equal to
1. -I
2. 3A
3. -3A
4. I

Answer 1

Correct Answer - Option 4 : I

Concept:

• \(\frac{adjA}{|A|}= A^{-1}\)
•  AA-1 = I

Calculation:

 A = \(\rm \begin{bmatrix} -1 & 0 &1 \\ -2&3 & 4\\ 0& 2 & 1 \end{bmatrix}\),

\(|A|=\begin{vmatrix} -1 & 0 & 1\\ -2& 3 & 4\\ 0& 2 & 1 \end{vmatrix} = -1(3-8)-0(-2-0)+1(-4-0)=5-4=1\)

Here, we have to find the value of \(\rm [A{(adjA)}A^{-1}]A \)

⇒ \(\rm [A{(adjA)}A^{-1}]A = \frac{|A| [A{(adjA)}A^{-1}]A}{|A|}\)          (multiply and divide by |A| )

⇒ \(\rm [A{(adjA)}A^{-1}]A =|A|[AA^{-1}A^{-1}]A\)       (∴ \(\frac{adjA}{|A|}= A^{-1}\))

⇒ \(\rm [A{(adjA)}A^{-1}]A =|A|[IA^{-1}]A\)                (∴ AA^-1 = I)

⇒ \(\rm [A{(adjA)}A^{-1}]A =|A|A^{-1}A\)                    (∴ IA^-1 = A^-1)

⇒ \(\rm [A{(adjA)}A^{-1}]A =|A|I\)                              (∴ A-1A = I)

As we have calculated that |A| = 1

⇒ \(\rm [A{(adjA)}A^{-1}]A =1.I = I\) 

Hence, option 4 is correct.