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Question: If \(a\ne 6,b,c\) satisfy \(\left| \begin{matrix} a & 2b & 2c \\\ 3 & b & c \\\ 4 & a...

If a≠6,b,ca\ne 6,b,c satisfy ∣a2b2c 3bc 4ab ∣=0\left| \begin{matrix} a & 2b & 2c \\\ 3 & b & c \\\ 4 & a & b \\\ \end{matrix} \right|=0, then abc=abc=
A. a+b+ca+b+c
B. 0
C. b3{{b}^{3}}
D. ab+b−cab+b-c

Explanation

Solution

To solve this question we will use the properties of matrix. We will first simplify the given matrix by using the operations of matrix and determinants then we put the values obtained equal to zero, then we will find the value of abcabc by simplifying the obtained equation.

Complete step by step solution:
We have been given that a≠6,b,ca\ne 6,b,c satisfy ∣a2b2c 3bc 4ab ∣=0\left| \begin{matrix} a & 2b & 2c \\\ 3 & b & c \\\ 4 & a & b \\\ \end{matrix} \right|=0.
We have to find the value of abcabc.
Now, let us first solve the given matrix. We know that we can solve the given matrix by multiplying the element by 2×22\times 2 determinant. The determinant of a 3×33\times 3 matrix is calculated for a matrix having 3 rows and 3 columns. Then we will get
⇒a(b×b−a×c)−2b(3×b−4×c)+2c(3×a−4×b)\Rightarrow a\left( b\times b-a\times c \right)-2b\left( 3\times b-4\times c \right)+2c\left( 3\times a-4\times b \right)
Now, simplifying the above obtained equation we will get
⇒a(b2−ac)−2b(3b−4c)+2c(3a−4b) ⇒ab2−a2c−6b2+8bc+6ac−8bc ⇒ab2−a2c−6b2+6ac \begin{aligned} & \Rightarrow a\left( {{b}^{2}}-ac \right)-2b\left( 3b-4c \right)+2c\left( 3a-4b \right) \\\ & \Rightarrow a{{b}^{2}}-{{a}^{2}}c-6{{b}^{2}}+8bc+6ac-8bc \\\ & \Rightarrow a{{b}^{2}}-{{a}^{2}}c-6{{b}^{2}}+6ac \\\ \end{aligned}
We have given that a≠6,b,ca\ne 6,b,c satisfy matrix value equal to zero.
Then we will get
⇒ab2−a2c−6b2+6ac=0 ⇒ab2−6b2−a2c+6ac=0 ⇒b2(a−6)−ac(a−6)=0 ⇒(a−6)(b2−ac)=0 \begin{aligned} & \Rightarrow a{{b}^{2}}-{{a}^{2}}c-6{{b}^{2}}+6ac=0 \\\ & \Rightarrow a{{b}^{2}}-6{{b}^{2}}-{{a}^{2}}c+6ac=0 \\\ & \Rightarrow {{b}^{2}}\left( a-6 \right)-ac\left( a-6 \right)=0 \\\ & \Rightarrow \left( a-6 \right)\left( {{b}^{2}}-ac \right)=0 \\\ \end{aligned}
If a≠6a\ne 6 then (b2−ac)=0\left( {{b}^{2}}-ac \right)=0 then simplifying the obtained equation we will get
⇒b2−ac=0 ⇒b2=ac ⇒abc=b3 \begin{aligned} & \Rightarrow {{b}^{2}}-ac=0 \\\ & \Rightarrow {{b}^{2}}=ac \\\ & \Rightarrow abc={{b}^{3}} \\\ \end{aligned}
Hence we get the value of abcabc as b3{{b}^{3}}.

Option C is the correct answer.

Note:
Students must remember the condition given in the question that a≠6a\ne 6. If we consider the factor a−6=0a-6=0 then we will get the value a=6a=6. So we need to ignore the factor. In matrices, determinants are the special numbers calculated from the square matrix. Square matrix should have an equal number of rows and columns.