Question

​Solve the following determinant equations:

(i) $\left|\begin{array}{ccc}x+a& b& c\\ a& x+b& c\\ a& b& x+c\end{array}\right|=0$

(ii) $\left|\begin{array}{ccc}x+a& x& x\\ x& x+a& x\\ x& x& x+a\end{array}\right|=0,a\ne 0$

(iii) $\left|\begin{array}{ccc}3x-8& 3& 3\\ 3& 3x-8& 3\\ 3& 3& 3x-8\end{array}\right|=0$

(iv) $\left|\begin{array}{ccc}1& x& x^2\\ 1& a& a^2\\ 1& b& b^2\end{array}\right|=0,a\ne b$

(v) $\left|\begin{array}{ccc}x+1& 3& 5\\ 2& x+2& 5\\ 2& 3& x+4\end{array}\right|=0$

(vi) $\left|\begin{array}{ccc}1& x& x^3\\ 1& b& b^3\\ 1& c& c^3\end{array}\right|=0,b\ne c$

(vii) $\left|\begin{array}{ccc}15-2x& 11-3x& 7-x\\ 11& 17& 14\\ 10& 16& 13\end{array}\right|=0$

(viii) $\left|\begin{array}{ccc}1& 1& x\\ p+1& p+1& p+x\\ 3& x+1& x+2\end{array}\right|=0$

(ix) $\left|\begin{array}{ccc}3& -2& \sin \left(3\theta \right)\\ -7& 8& \cos \left(2\theta \right)\\ -11& 14& 2\end{array}\right|=0$

(x) $\left|\begin{array}{ccc}4-x& 4+x& 4+x\\ 4+x& 4-x& 4+x\\ 4+x& 4+x& 4+x\end{array}\right|=0$

Answer 1

(x) Given: $\left|\begin{array}{ccc}4-x& 4+x& 4+x\\ 4+x& 4-x& 4+x\\ 4+x& 4+x& 4+x\end{array}\right|=0$



$$\begin{gathered} \mathrm{LHS}=\left|\begin{array}{ccc}4-x& 4+x& 4+x\\ 4+x& 4-x& 4+x\\ 4+x& 4+x& 4+x\end{array}\right| \\ \mathrm{Applying}{R}_2\to R_2-R_1 \\ =\left|\begin{array}{ccc}4-x& 4+x& 4+x\\ 4+x-4+x& 4-x-4-x& 4+x-4-x\\ 4+x& 4+x& 4+x\end{array}\right| \\ =\left|\begin{array}{ccc}4-x& 4+x& 4+x\\ 2x& -2x& 0\\ 4+x& 4+x& 4+x\end{array}\right| \\ \mathrm{Taking}\left(2x\right)\mathrm{common}\mathrm{from}{R}_2 \\ =\left(2x\right)\left|\begin{array}{ccc}4-x& 4+x& 4+x\\ 1& -1& 0\\ 4+x& 4+x& 4+x\end{array}\right| \\ \mathrm{Applying}{R}_3\to R_3-R_1 \\ =\left(2x\right)\left|\begin{array}{ccc}4-x& 4+x& 4+x\\ 1& -1& 0\\ 4+x-4+x& 4+x-4-x& 4+x-4-x\end{array}\right| \\ =\left(2x\right)\left|\begin{array}{ccc}4-x& 4+x& 4+x\\ 1& -1& 0\\ 2x& 0& 0\end{array}\right| \\ \mathrm{Expanding}\mathrm{through}{C}_3 \\ =\left(2x\right)\left[\left(4+x\right)\left(1\times 0+1\times 2x\right)\right] \\ =\left(2x\right)\left[\left(4+x\right)\left(2x\right)\right] \\ ={\left(2x\right)}^2\left(4+x\right) \\ \mathrm{Thus},\left|\begin{array}{ccc}4-x& 4+x& 4+x\\ 4+x& 4-x& 4+x\\ 4+x& 4+x& 4+x\end{array}\right|={\left(2x\right)}^2\left(4+x\right) \\ \mathrm{But}\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that},\left|\begin{array}{ccc}4-x& 4+x& 4+x\\ 4+x& 4-x& 4+x\\ 4+x& 4+x& 4+x\end{array}\right|=0 \\ \Rightarrow {\left(2x\right)}^2\left(4+x\right)=0 \\ \Rightarrow {\left(2x\right)}^2=0\mathrm{or}\left(4+x\right)=0 \\ \Rightarrow x=0\mathrm{or}x=-4 \end{gathered}$$

Hence, x = 0, −4.