Question

Let three matrices A = \left( {\begin{array}{*{20}{c}} 2&1 \\\ 4&1 \end{array}} \right); B = \left( {\begin{array}{*{20}{c}} 3&4 \\\ 2&3 \end{array}} \right) and C = \left( {\begin{array}{*{20}{c}} 3&{ - 4} \\\ { - 2}&3 \end{array}} \right), then what is the value of ${t_r}\left( A \right) + {t_r}\left( {\dfrac{{ABC}}{2}} \right) + {t_r}\left( {\dfrac{{A{{\left( {BC} \right)}^2}}}{4}} \right) + {t_r}\left( {\dfrac{{A{{\left( {BC} \right)}^3}}}{8}} \right) + ... + \infty$ ?
A) $6$
B) $9$
C) $12$
D) None of these.

Answer 1

In this question, we are given three matrices and an equation whose value is to be found using the three given matrices. First, find the value of matrix BC. Then, find the value of ${t_r}\left( A \right)$. Put all the values in the given equation. You will notice that we have been asked to find out the sum of infinite terms of the G.P. Using the formula, find the answer.

Formula used: ${S_\infty } = \dfrac{a}{{\left( {1 - r} \right)}}$

Complete step-by-step answer:
We are given three matrices and we have been asked to find the value of the given equation using the three matrices.
Now, if you look closely at the given equation, you will notice that we have been asked the sum of a geometric progression with an infinite number of terms. The first term of this G.P is ${t_r}\left( A \right)$ and the common ratio is $\dfrac{{BC}}{2}$. But before that we have to find $BC$.
On multiplying the matrix B with C, we will get -
BC = \left( {\begin{array}{*{20}{c}} 3&4 \\\ 2&3 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 3&{ - 4} \\\ { - 2}&3 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&0 \\\ 0&1 \end{array}} \right)
Hence, the value of $BC$ will be 1.
Therefore, our equation has now been simplified to –
${t_r}\left( A \right) + {t_r}\left( {\dfrac{A}{2}} \right) + {t_r}\left( {\dfrac{A}{4}} \right) + {t_r}\left( {\dfrac{A}{8}} \right) + ... + \infty$
Let us first find the value of ${t_r}\left( A \right)$. Since trace of A is the sum of the diagonal elements,
$\Rightarrow$ ${t_r}\left( A \right)$= $2 + 1 = 3$
Now, let us put the values in the equation ${t_r}\left( A \right) + {t_r}\left( {\dfrac{A}{2}} \right) + {t_r}\left( {\dfrac{A}{4}} \right) + {t_r}\left( {\dfrac{A}{8}} \right) + ... + \infty$,
$\Rightarrow 3 + \dfrac{3}{2} + \dfrac{3}{4} + \dfrac{3}{8} + ... + \infty$
To find the sum, we will use the formula of sum of infinite G.P $\Rightarrow {S_\infty } = \dfrac{a}{{\left( {1 - r} \right)}}$
We know $a = 3$ and $r = \dfrac{1}{2}$. Putting the values in the formula –
$\Rightarrow {S_\infty } = \left( {\dfrac{{\dfrac{3}{1}}}{{1 - \dfrac{1}{2}}}} \right)$
On solving we will get,
$\Rightarrow {S_\infty } = \left( {\dfrac{{\dfrac{3}{1}}}{{\dfrac{1}{2}}}} \right)$
$\Rightarrow {S_\infty } = 6$
$\therefore$ ${t_r}\left( A \right) + {t_r}\left( {\dfrac{{ABC}}{2}} \right) + {t_r}\left( {\dfrac{{A{{\left( {BC} \right)}^2}}}{4}} \right) + {t_r}\left( {\dfrac{{A{{\left( {BC} \right)}^3}}}{8}} \right) + ... + \infty$= 6

Option A is the correct answer.

Note: We have to know that the sign $'{t_r}'$stands for “trace”. Trace is the sum of the main diagonal elements, starting from upper left element to bottom right element.
For example, in the given question,
A = \left( {\begin{array}{*{20}{c}} 2&1 \\\ 4&1 \end{array}} \right); ${t_r}\left( A \right)$ $= 2 + 1 = 3$
B = \left( {\begin{array}{*{20}{c}} 3&4 \\\ 2&3 \end{array}} \right); ${t_r}\left( B \right) = 3 + 3 = 6$
C = \left( {\begin{array}{*{20}{c}} 3&{ - 4} \\\ { - 2}&3 \end{array}} \right); ${t_r}\left( C \right) = 3 + 3 = 6$