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Question: Find the value of x if it is given that \[\left| \begin{matrix} 3x & 7 \\\ -2 & 4 \\\ \e...

Find the value of x if it is given that 3x7 24 =87 64 \left| \begin{matrix} 3x & 7 \\\ -2 & 4 \\\ \end{matrix} \right|=\left| \begin{matrix} 8 & 7 \\\ 6 & 4 \\\ \end{matrix} \right| .

Explanation

Solution

We are given two determinants in the question. We can infer that their values are equal. We will use the formula ab cd =adbc\left| \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right|=ad-bc to find the value of the determinant. Since they are equal, so we will equate them at last to get the value of x.

Complete step by step answer:
We are given two determinants on either side of an equation in the question. Since there is an = sign, it means that the values on the right-hand side and the left-hand side are equal. We have been asked to find the value of x.
Now, we will solve to find the value of each of the given determinants one by one. Firstly, we know the value of any 2×22\times 2 determinant \left\\{ \left| \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right|,a,b,c,d\in R \right\\} is given as adbc.ad-bc.
We will consider the RHS. So, we have the first determinant as,

8 & 7 \\\ 6 & 4 \\\ \end{matrix} \right|$$ So, we can see that, we have a = 8, b = 7, c = 6 and d = 4. Now, the value of the determinant $$\left| \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right|$$ is given as $$ad-bc.$$ So, $${{D}_{1}}=8\times 4-6\times 7$$ Solving further, we will get, $$\Rightarrow {{D}_{1}}=32-42$$ $$\Rightarrow {{D}_{1}}=-10.....\left( i \right)$$ Now, we will consider the LHS and we will find the value of the determinant $$\left| \begin{matrix} 3x & 7 \\\ -2 & 4 \\\ \end{matrix} \right|.$$ $${{D}_{2}}=\left| \begin{matrix} 3x & 7 \\\ -2 & 4 \\\ \end{matrix} \right|$$ We have, a = 3x, b = 7, c = – 2 and d = 4. So, we will have, $$\Rightarrow {{D}_{2}}=ad-bc$$ Putting the value, we will get, $$\Rightarrow {{D}_{2}}=3x\times 4-7\times \left( -2 \right)$$ $$\Rightarrow {{D}_{2}}=12x+14......\left( ii \right)$$ Now, we are given, $$\Rightarrow \left| \begin{matrix} 3x & 7 \\\ -2 & 4 \\\ \end{matrix} \right|=\left| \begin{matrix} 8 & 7 \\\ 6 & 4 \\\ \end{matrix} \right|$$ So, using (i) and (ii), we get, $$\Rightarrow 12x+14=-10$$ Now solving for x, we have, $$\Rightarrow 12x=-10-14$$ $$\Rightarrow 12x=-24$$ Now, divide both the sides by 12 $$\Rightarrow \dfrac{12x}{12}=\dfrac{-24}{12}$$ $$\Rightarrow x=-2$$ **Therefore, we get the value of x as – 2.** **Note:** While solving, one should remember $$\left| \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right|$$ is a symbol for determinant and $$\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]$$ is a symbol for matrix. So, here we are supposed to find the determinant and then solve for x. If we consider it as a matrix and start comparing the terms to get the value of x, we will get the wrong solution. Here, while computing the determinant, take care not to make any mistake in the order of multiplying and subtracting the terms as it will affect the final answer.