Question
The matrix \[A = \left[ {\begin{array}{*{20}{c}}
1&3&2\\
1&{x - 1}&1\\
2&7&{x - 3}
\end{array}} \right]\] will have inverse for every real number $$x$$ except for
A.
$$x = \frac{{11 \pm \sqrt 5 }}{2}$$
B.
$$x = \frac{{9 \pm \sqrt 5 }}{2}$$
C.
$$x = \frac{{11 \pm \sqrt 3 }}{2}$$
D.
$$x = \frac{{9 \pm \sqrt 3 }}{2}$$
Answer :
$$x = \frac{{11 \pm \sqrt 5 }}{2}$$
Solution :
\[A = \left[ {\begin{array}{*{20}{c}}
1&3&2\\
1&{x - 1}&1\\
2&7&{x - 3}
\end{array}} \right]\]
$$\eqalign{
& \left| A \right| = 1\left[ {\left( {x - 1} \right)\left( {x - 3} \right) - 7} \right] - 3\left[ {\left( {x - 3} \right) - 2} \right] + 2\left[ {7 - 2\left( {x - 1} \right)} \right] \cr
& = {x^2} - 11x + 29 \cr} $$
If inverse will not exist then $$\left| A \right| = 0$$
$$\eqalign{
& {x^2} - 11x + 29 = 0 \cr
& x = \frac{{11 \pm \sqrt 5 }}{2} \cr} $$