Question

Solve for $x$, $\left( \begin{matrix} 1+x & 1-x & 1-x \\\ 1-x & 1+x & 1-x \\\ 1-x & 1-x & 1+x \\\ \end{matrix} \right)=0$

Answer 1

In the given question, we are given a determinant and we have to solve for $x$. We will consider the first row now. Taking each term of the first row we will multiply as per the determinant rule, that is, $(1+x)\left( \begin{matrix} 1+x & 1-x \\\ 1-x & 1+x \\\ \end{matrix} \right)$ then we have, $\left( 1-x \right)\left( \begin{matrix} 1-x & 1-x \\\ 1-x & 1+x \\\ \end{matrix} \right)$ and then $\left( 1-x \right)\left( \begin{matrix} 1-x & 1+x \\\ 1-x & 1-x \\\ \end{matrix} \right)$ and then equating it to 0. We will then solve further to get an expression using which we will find the value of $x$ from the obtained expression. Hence, we will have the value of the $x$.

Complete step by step solution:
According to the given question, we are given a question based on determinants. We are asked to find the value of $x$ using the given determinant.
The determinant we have is,

1+x & 1-x & 1-x \\\ 1-x & 1+x & 1-x \\\ 1-x & 1-x & 1+x \\\ \end{matrix} \right)=0$$ Here, we will first consider the first row, we have, $$\Rightarrow (1+x)\left( \begin{matrix} 1+x & 1-x \\\ 1-x & 1+x \\\ \end{matrix} \right)-\left( 1-x \right)\left( \begin{matrix} 1-x & 1-x \\\ 1-x & 1+x \\\ \end{matrix} \right)+\left( 1-x \right)\left( \begin{matrix} 1-x & 1+x \\\ 1-x & 1-x \\\ \end{matrix} \right)=0$$ Solving up the above expression we get, $$\Rightarrow (1+x)\left[ {{\left( 1+x \right)}^{2}}-{{\left( 1-x \right)}^{2}} \right]-\left( 1-x \right)\left[ \left( 1+x \right)\left( 1-x \right)-{{\left( 1-x \right)}^{2}} \right]+\left( 1-x \right)\left[ {{\left( 1-x \right)}^{2}}-\left( 1+x \right)\left( 1-x \right) \right]=0$$ We will now open use the appropriate identities and open the brackets one by one, so we get, We are using the identities of, $${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$$ $${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$$ and $$(a+b)(a-b)={{a}^{2}}-{{b}^{2}}$$ We get the new expression as, $$\Rightarrow (1+x)\left[ \left( 1+2x+{{x}^{2}} \right)-\left( 1-2x+{{x}^{2}} \right) \right]-\left( 1-x \right)\left[ \left( 1-{{x}^{2}} \right)-\left( 1-2x+{{x}^{2}} \right) \right]+\left( 1-x \right)\left[ \left( 1-2x+{{x}^{2}} \right)-\left( 1-{{x}^{2}} \right) \right]=0$$ Opening up the brackets in the above expression and applying the appropriate signs, we get, $$\Rightarrow (1+x)\left[ 1+2x+{{x}^{2}}-1+2x-{{x}^{2}} \right]-\left( 1-x \right)\left[ 1-{{x}^{2}}-1+2x-{{x}^{2}} \right]+\left( 1-x \right)\left[ 1-2x+{{x}^{2}}-1+{{x}^{2}} \right]=0$$ We will now cancel out the common terms and we get, $$\Rightarrow (1+x)\left[ 2x+2x \right]-\left( 1-x \right)\left[ -{{x}^{2}}+2x-{{x}^{2}} \right]+\left( 1-x \right)\left[ -2x+{{x}^{2}}+{{x}^{2}} \right]=0$$ Adding up the similar terms and subtracting the respective ones, we have the expression as, $$\Rightarrow (1+x)\left[ 4x \right]-\left( 1-x \right)\left[ -2{{x}^{2}}+2x \right]+\left( 1-x \right)\left[ -2x+2{{x}^{2}} \right]=0$$ Solving further, we get, $$\Rightarrow 4x+4{{x}^{2}}-\left( -2{{x}^{2}}+2x+2{{x}^{3}}-2{{x}^{2}} \right)+\left( -2x+2{{x}^{2}}+2{{x}^{2}}-2{{x}^{3}} \right)=0$$ $$\Rightarrow 4x+4{{x}^{2}}-\left( 2x+2{{x}^{3}}-4{{x}^{2}} \right)+\left( -2x+4{{x}^{2}}-2{{x}^{3}} \right)=0$$ Opening up the main brackets here and applying the sign wherever necessary, we get, $$\Rightarrow 4x+4{{x}^{2}}-2x-2{{x}^{3}}+4{{x}^{2}}-2x+4{{x}^{2}}-2{{x}^{3}}=0$$ Solving the above expression further, we have, $$\Rightarrow 12{{x}^{2}}-4{{x}^{3}}=0$$ Writing the above equation in the decreasing order of the degree, we get, $$\Rightarrow 4{{x}^{3}}-12{{x}^{2}}=0$$ Taking the common terms out, we get the new expression as, $$\Rightarrow 4{{x}^{2}}(x-3)=0$$ Separating the components and equating it to zero, we get, $$4{{x}^{2}}=0$$ we get, $$\Rightarrow x=0$$ And $$x-3=0$$, we get, $$\Rightarrow x=3$$ Therefore, the value of $$x=0,3$$. **Note:** The determinant should be solved in a very clear and correct way, else even the slightest mistake would mean the entire question getting wrong. Also, the second term in the expansion of the determinant should have the negative sign, do not ever miss this point.