Question-4

If \(A\) is invertible, does the system \(\text{adj}(A)x=b\) have a solution? If yes, find the solution. Is it unique?

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We have:

\[ A\cdot\text{adj}(A)=\text{det}(A)\cdot I \]

Since \(A\) is invertible, \(A^{-1}\) exists. We can pre-multiply both sides of the above equation by \(A^{-1}\) to get:

\[ \begin{aligned} A^{-1}A\cdot\text{adj}(A) & =\text{det}(A)A^{-1}\\ \implies\text{adj}(A) & =\text{det}(A)A^{-1} \end{aligned} \]

Notice that \(\text{adj}(A)\) is just a non-zero multiple of \(A^{-1}\). So \(\text{adj}(A)\) is also invertible. Its inverse will be \(\cfrac{A}{\text{det}(A)}\). Now:

\[ \begin{aligned} \text{adj}(A)x & =b\\ \text{det}(A)A^{-1}x & =b\\ x & =\cfrac{Ab}{\text{det}(A)} \end{aligned} \]

Therefore, \(\cfrac{Ab}{\text{det}(A)}\) is the unique solution to the system \(\text{adj}(A)x=b\).