Question & Answer
QUESTION

Let, A = $\left[ \begin{matrix}
   3{{x}^{2}} \\
   1 \\
   6x \\
\end{matrix} \right]$, B = \[\left[ \begin{matrix}
   a & b & c \\
\end{matrix} \right]\] and C = \[\left[ \begin{matrix}
   {{\left( x+2 \right)}^{2}} & 5{{x}^{2}} & 2x \\
   5{{x}^{2}} & 2x & {{\left( x+2 \right)}^{2}} \\
   2x & {{\left( x+2 \right)}^{2}} & 5{{x}^{2}} \\
\end{matrix} \right]\] be three given matrices, where a, b, c and x \[\in \] R, given that ‘tr. (AB) = tr. (c)’ \[\vee \] \[x\in R\], where tr. (A) denotes trace of A. Find the value of (a + b + c)
A.6
B.7
C.8
D.9

ANSWER Verified Verified
Hint: Use the formula \[\left[ \begin{matrix}
   3{{x}^{2}} \\
   1 \\
   6x \\
\end{matrix} \right]\times \left[ \begin{matrix}
   a & b & c \\
\end{matrix} \right]=\left[ \begin{matrix}
   3{{x}^{2}}\times a & 3{{x}^{2}}\times b & 3{{x}^{2}}\times c \\
   1\times a & 1\times b & 1\times c \\
   6x\times a & 6x\times b & 6x\times c \\
\end{matrix} \right]\] to find the value of (AB) and then find the trace of the matrix (AB) and matrix C by simply adding their diagonal elements and then equate them. By equating their coefficients you will get the values of a, b, and c. Add a, b, and c to get the final answer.

Complete step by step answer:
To solve the above problem we will write the given values first,
A = $\left[ \begin{matrix}
   3{{x}^{2}} \\
   1 \\
   6x \\
\end{matrix} \right]$, B = \[\left[ \begin{matrix}
   a & b & c \\
\end{matrix} \right]\], and C = \[\left[ \begin{matrix}
   {{\left( x+2 \right)}^{2}} & 5{{x}^{2}} & 2x \\
   5{{x}^{2}} & 2x & {{\left( x+2 \right)}^{2}} \\
   2x & {{\left( x+2 \right)}^{2}} & 5{{x}^{2}} \\
\end{matrix} \right]\] ……………………….. (1)
Also, tr. (AB) = tr. (c)
As we have given the condition that ‘tr. (AB) = tr. (c)’ and as matrix ‘C’ is given therefore we have to find the matrix (AB) and for that we should know the formula of multiplication of matrix given below,
Formula:
If P = \[\left[ \begin{matrix}
   p \\
   q \\
   r \\
\end{matrix} \right]\] and Q = \[\left[ \begin{matrix}
   s & t & u \\
\end{matrix} \right]\] the (PQ) will be given as,
\[PQ=\left[ \begin{matrix}
   p \\
   q \\
   r \\
\end{matrix} \right]\times \left[ \begin{matrix}
   s & t & u \\
\end{matrix} \right]=\left[ \begin{matrix}
   p\times s & p\times t & p\times u \\
   q\times s & q\times t & q\times u \\
   r\times s & r\times t & r\times u \\
\end{matrix} \right]\]
By using the above formula and the given value of matrix A and matrix B from (1) we can write (AB) as,
\[\therefore AB=\left[ \begin{matrix}
   3{{x}^{2}} \\
   1 \\
   6x \\
\end{matrix} \right]\times \left[ \begin{matrix}
   a & b & c \\
\end{matrix} \right]=\left[ \begin{matrix}
   3{{x}^{2}}\times a & 3{{x}^{2}}\times b & 3{{x}^{2}}\times c \\
   1\times a & 1\times b & 1\times c \\
   6x\times a & 6x\times b & 6x\times c \\
\end{matrix} \right]\]
\[\therefore AB=\left[ \begin{matrix}
   3a{{x}^{2}} & 3b{{x}^{2}} & 3c{{x}^{2}} \\
   a & b & c \\
   6ax & 6bx & 6cx \\
\end{matrix} \right]\] …………………………………………….. (2)
To find the trace of matrix (AB) and C we have to know the formula given below,
Formula:
If \[A=\left[ \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right]\] then the trace of A is given by the sum of its diagonal elements as shown below,
tr. A = a + e + i ……………………………………………. (3)
By using the equation (3) and equation (2) we can write the trace of (AB) as follows,
\[tr.\text{ }\left( AB \right)\text{ }=~3a{{x}^{2}}+b+6cx\]………………………………. (4)
Also by using equation (3) and the value of matrix C from (1) we will get,
\[tr.(C)={{\left( x+2 \right)}^{2}}+2x+5{{x}^{2}}\] …………………………….. (5)
\[\therefore tr.(C)={{x}^{2}}+2\times \left( 2x \right)+4+2x+5{{x}^{2}}\]
\[\therefore tr.(C)={{x}^{2}}+4x+4+2x+5{{x}^{2}}\]
\[\therefore tr.(C)=6{{x}^{2}}+6x+4\] …………………………………….. (6)
As we have given in the question,
 tr. (AB) = tr. (c)
If we put the value of equation (6) and equation (4) in above equation we will get,
\[\therefore ~3a{{x}^{2}}+b+6cx=6{{x}^{2}}+6x+4\]
By rearranging the above equation we will get,
\[\therefore ~3a{{x}^{2}}+6cx+b=6{{x}^{2}}+6x+4\]
Now by equating the coefficients of right hand side with left hand side of the above equations we will get,
3a = 6, 6c = 6, and b = 4
Therefore,
\[a=\dfrac{6}{3}\], \[c=\dfrac{6}{6}\] and b = 4
Therefore,
a = 2, c = 1, and b = 4
Therefore, we will get the values of a, b, and c as,
a = 2,
b = 4 and
c = 1
Now assume L = (a + b + c) ………………………………… (7)
If we put the values of a, b, and c in the above equation we will get,
Therefore, L = 2 + 4 + 1
Therefore, L = 7
From equation (7) and the above equation we will get,
a + b + c = 7
Therefore the value of (a + b + c) is 7.

Note: While solving the equation \[3a{{x}^{2}}+b+6cx=6{{x}^{2}}+6x+4\] solve directly by equating the coefficients otherwise you won’t get any answer.Don't
 try to solve by forming a quadratic equation.