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.6B.7C.8D.9

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.