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 1

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,

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.