Question

If inverse of $A = \begin{bmatrix} \alpha & 1 & 2 \\[6pt] 0 & 3 & 1 \\[6pt] 1 & 1 & 1 \end{bmatrix}$ exist, then the value of $\alpha$ cannot be

• A. 5
• B. -5
• C. $\tfrac{5}{2}$
• D. $\tfrac{-5}{2}$

Answer 1

Answer: (C)

$\tfrac{5}{2}$

Answer 2

To have an inverse, $\det(A)\neq 0$. Compute the determinant by cofactor expansion along the first row:

$$\det(A) = \alpha\begin{vmatrix}3 & 1\\[3pt]1 & 1\end{vmatrix} - 1\begin{vmatrix}0 & 1\\[3pt]1 & 1\end{vmatrix} + 2\begin{vmatrix}0 & 3\\[3pt]1 & 1\end{vmatrix}.$$

Evaluate each minor:

$$\begin{vmatrix}3 & 1\\1 & 1\end{vmatrix} =3\cdot1 -1\cdot1=2,\quad \begin{vmatrix}0 & 1\\1 & 1\end{vmatrix} =0\cdot1 -1\cdot1=-1,\quad \begin{vmatrix}0 & 3\\1 & 1\end{vmatrix} =0\cdot1 -3\cdot1=-3.$$

Thus,

$$\det(A) = \alpha\cdot2 \;-\;1\cdot(-1)\;+\;2\cdot(-3) =2\alpha +1 -6 =2\alpha -5.$$

For $\det(A)\neq0$, we require $2\alpha -5 \neq 0$, i.e.\ $\alpha \neq \tfrac{5}{2}$.