Question

Write the value of x if $\left|\begin{aligned} & 2x\text{ 5} \\\ & \text{8 x} \\\ \end{aligned} \right|$ =$\left|\begin{aligned} & \text{6 -2} \\\ & \text{7 3} \\\ \end{aligned} \right|$ .

Answer 1

By using the concept, we can say that the determinant of a matrix $\left| \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right|$ is equal to $ad-bc$. From this question, it is clear that $\left|\begin{aligned} & 2x\text{ 5} \\\ & \text{8 x} \\\ \end{aligned} \right|$=$\left| \begin{aligned} & \text{6 -2} \\\ & \text{7 3} \\\ \end{aligned}\right|$. Let us assume this as equation (1). By using this concept, we should find the value of $\left|\begin{aligned} & 2x\text{ 5} \\\ & \text{8 x} \\\ \end{aligned} \right|$. Let us assume this as equation (2). Again, by using this concept, we should find the value of $\left| \begin{aligned} & \text{6 -2} \\\ & \text{7 3} \\\ \end{aligned}\right|$. Let us assume this as equation (3). Now we should substitute equation (2) and equation (3) in equation (1). In this way, we can find the value of x.

Complete step by step answer:
We know that the determinant of a matrix $\left( \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right)$ is equal to $ad-bc$.
From the question, we should solve the equation $\left| \begin{aligned} & 2x\text{ 5} \\\ & \text{8 x} \\\ \end{aligned} \right|$= $\left| \begin{aligned} & \text{6 -2} \\\ & \text{7 3} \\\ \end{aligned} \right|$.
Let us consider

& 2x\text{ 5} \\\ & \text{8 x} \\\ \end{aligned} \right|$$=$$\left| \begin{aligned} & \text{6 -2} \\\ & \text{7 3} \\\ \end{aligned} \right|$$.......(1) Now let us find the value of $$\left|\begin{aligned} & 2x\text{ 5} \\\ & \text{8 x} \\\ \end{aligned} \right|$$ $$\left| \begin{aligned} & 2x\text{ 5} \\\ & \text{8 x} \\\ \end{aligned} \right|$$= $( 2x)( x)-( 5)( 8 )$ $ \Rightarrow \left| \begin{aligned} & 2x\text{ 5} \\\ & \text{8 x} \\\ \end{aligned} \right| = 2{{x}^{2}}-40......(2) $ Now let find the value of $$\left| \begin{aligned} & \text{6 -2} \\\ & \text{7 3} \\\ \end{aligned} \right|$$. $ \left| \begin{aligned} & \text{6 -2} \\\ & \text{7 3} \\\ \end{aligned} \right|$= $( 6)( 3)-( -2 )( 7) $ $ \Rightarrow $ $$\left| \begin{aligned} & \text{6 -2} \\\ & \text{7 3} \\\ \end{aligned} \right|$$ = 18- (-14 ) $ \Rightarrow $$$\left| \begin{aligned} & \text{6 -2} \\\ & \text{7 3} \\\ \end{aligned} \right|$$ = 18+14 $ \therefore $$$\left| \begin{aligned} & \text{6 -2} \\\ & \text{7 3} \\\ \end{aligned} \right|$$ = 32.....(3) Now let us substitute equation (2) and equation (3) in equation (1), then we get $$\begin{aligned} & \Rightarrow 2{{x}^{2}}-40=32 \\\ & \Rightarrow 2{{x}^{2}}=32+40 \\\ & \Rightarrow 2{{x}^{2}}=72 \\\ & \Rightarrow {{x}^{2}}=\dfrac{72}{2} \\\ & \Rightarrow {{x}^{2}}=36 \\\ & \Rightarrow x=\pm 6.....(4) \\\ \end{aligned}$$ From equation (4), the values of x are equal to 6 and -6. **Note:** Some students have a misconception that the determinant of a matrix $$\left( \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right)$$ is equal to $$ad+bc$$. If this misconception is followed, then From the question, we should solve the equation $$\left| \begin{aligned} & 2x\text{ 5} \\\ & \text{8 x} \\\ \end{aligned} \right|$$=$$\left| \begin{aligned} & \text{6 -2} \\\ & \text{7 3} \\\ \end{aligned} \right|$$. Let us consider $$\left| \begin{aligned} & 2x\text{ 5} \\\ & \text{8 x} \\\ \end{aligned} \right|$$=$$\left| \begin{aligned} & \text{6 -2} \\\ & \text{7 3} \\\ \end{aligned} \right|.......(1)$$ Now let us find the value of $$\left| \begin{aligned} & 2x\text{ 5} \\\ & \text{8 x} \\\ \end{aligned} \right|$$. $ \left| \begin{aligned} & 2x\text{ 5} \\\ & \text{8 x} \\\ \end{aligned} \right| = ( 2x) ( x )+ ( 5)( 8) $ $\Rightarrow \left| \begin{aligned} & 2x\text{ 5} \\\ & \text{8 x} \\\ \end{aligned} \right|=2{{x}^{2}}+40......(2) $ Now let find the value of $$\left| \begin{aligned} & \text{6 -2} \\\ & \text{7 3} \\\ \end{aligned} \right|$$. $ \Rightarrow \left| \begin{aligned} & \text{6 -2} \\\ & \text{7 3} \\\ \end{aligned} \right|=( 6 )( 3 )+( -2 )( 7 ) $ $ \Rightarrow \left| \begin{aligned} & \text{6 -2} \\\ & \text{7 3} \\\ \end{aligned} \right|=18+( -14 ) $ $ \Rightarrow \left| \begin{aligned} & \text{6 -2} \\\ & \text{7 3} \\\ \end{aligned} \right|=18-14 $ $ \Rightarrow \left| \begin{aligned} & \text{6 -2} \\\ & \text{7 3} \\\ \end{aligned} \right|=4.....(3) $ Now let us substitute equation (2) and equation (3) in equation (1), then we get $$\begin{aligned} & \Rightarrow 2{{x}^{2}}+40=4 \\\ & \Rightarrow 2{{x}^{2}}=4-40 \\\ & \Rightarrow 2{{x}^{2}}=-36 \\\ & \Rightarrow {{x}^{2}}=\dfrac{-36}{2} \\\ & \Rightarrow {{x}^{2}}=-18 \\\ & \Rightarrow x=\sqrt{-18}.....(4) \\\ \end{aligned}$$ From equation (4), it is clear that we cannot have the real value of x. But we know that the value of x are equal to -6 and 6.