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Question: Solve the given determinant for x: \(\left| \begin{aligned} x-2 \;\;\;\; \;\;\;\; & 2x-3 & 3x-...

Solve the given determinant for x:
∣x−2                2x−33x−4 x−4                2x−93x−16 x−8                2x−273x−64 ∣=0\left| \begin{aligned} x-2 \;\;\;\; \;\;\;\; & 2x-3 & 3x-4 \\\ x-4 \;\;\;\; \;\;\;\; & 2x-9 & 3x-16 \\\ x-8 \;\;\;\; \;\;\;\; & 2x-27 & 3x-64 \\\ \end{aligned} \right|=0
(A). 74\dfrac{7}{4}
(B). 44
(C). 2813\dfrac{28}{13}
(D). 149\dfrac{14}{9}

Explanation

Solution

Hint: Apply the transformations on row 1 and row 2 of the given determinant. The transformation that we are going to do is to transform row 1 by subtracting row 2 from row 1 and to transform row 2 by subtracting row 3 from row 2 then expand along row 1 and then solve the equation to find the value of x.

Complete step-by-step solution -
The determinant equation which is given above:
∣x−2                2x−33x−4 x−4                2x−93x−16 x−8                2x−273x−64 ∣=0\left| \begin{aligned} x-2 \;\;\;\; \;\;\;\; & 2x-3 & 3x-4 \\\ x-4 \;\;\;\; \;\;\;\; & 2x-9 & 3x-16 \\\ x-8 \;\;\;\; \;\;\;\; & 2x-27 & 3x-64 \\\ \end{aligned} \right|=0
We are going to transform row 1 of the above determinant by subtracting row 2 from row 1 as follows:
∣2                612 x−4                2x−93x−16 x−8                2x−273x−64 ∣=0\left| \begin{aligned} 2 \;\;\;\; \;\;\;\;& 6 & 12\\\ x-4 \;\;\;\; \;\;\;\;& 2x-9 & 3x-16 \\\ x-8 \;\;\;\; \;\;\;\;& 2x-27 & 3x-64 \\\ \end{aligned} \right|=0
Now, we are going to transform row 2 by subtracting row 3 from row 2 as follows:
∣2                612 4                1848 x−8                2x−273x−64 ∣=0\left| \begin{aligned} 2 \;\;\;\; \;\;\;\;& 6 & 12 \\\ 4 \;\;\;\; \;\;\;\;& 18 & 48 \\\ x-8 \;\;\;\; \;\;\;\;& 2x-27 & 3x-64 \\\ \end{aligned} \right|=0
Expanding the above determinant along the first row we get,
2(18(3x−64)−48(2x−27))−6(4(3x−64)−48(x−8))+12(4(2x−27)−18(x−8))=0 ⇒2(54x−1152−96x+1296)−6(12x−256−48x+384)+12(8x−108−18x+144)=0 ⇒2(−42x+144)−6(−36x+128)+12(−10x+36)=0 ⇒−84x+288+216x−768−120x+432=0 ⇒12x−48=0 ⇒x=4 \begin{aligned} & 2\left( 18\left( 3x-64 \right)-48\left( 2x-27 \right) \right)-6\left( 4\left( 3x-64 \right)-48\left( x-8 \right) \right)+12\left( 4\left( 2x-27 \right)-18\left( x-8 \right) \right)=0 \\\ & \Rightarrow 2\left( 54x-1152-96x+1296 \right)-6\left( 12x-256-48x+384 \right)+12\left( 8x-108-18x+144 \right)=0 \\\ & \Rightarrow 2\left( -42x+144 \right)-6\left( -36x+128 \right)+12\left( -10x+36 \right)=0 \\\ & \Rightarrow -84x+288+216x-768-120x+432=0 \\\ & \Rightarrow 12x-48=0 \\\ & \Rightarrow x=4 \\\ \end{aligned}
From the above solution, we get the value of x is equal to 4.
Hence, the correct option is (b).

Note: You can check whether the value of x that we have got is correct or not by substituting the value of x in the given determinant and see whether by plugging the value of x will make the determinant value 0 or not.
Substituting the value of x = 4 in the given determinant we get,
∣x−2                2x−33x−4 x−4                2x−93x−16 x−8                2x−273x−64 ∣\left| \begin{aligned} x-2 \;\;\;\; \;\;\;\;& 2x-3 & 3x-4 \\\ x-4 \;\;\;\; \;\;\;\;& 2x-9 & 3x-16 \\\ x-8 \;\;\;\; \;\;\;\;& 2x-27 & 3x-64 \\\ \end{aligned} \right|
⇒∣2                58 0                −1−4 −4                −19−52 ∣ \Rightarrow \left| \begin{aligned} 2 \;\;\;\; \;\;\;\;& 5 & 8 \\\ 0 \;\;\;\; \;\;\;\;& -1 & -4 \\\ -4 \;\;\;\; \;\;\;\;& -19 & -52 \\\ \end{aligned} \right| \\\
Now, expanding the above determinant along second row to get the value of the determinant we get,
0−1(2(−52)+32)+4(−38+20) =−1(−104+32)+4(−18) =72−72=0 \begin{aligned} & 0-1\left( 2\left( -52 \right)+32 \right)+4\left( -38+20 \right) \\\ & =-1\left( -104+32 \right)+4\left( -18 \right) \\\ & =72-72 = 0 \\\ \end{aligned}