Let A and B be two square matrices of order 2.

Question

Let A and B be two square matrices of order 2. If $$det\,(A) = 2$$, $$det\,(B) = 3$$ and $$\det \left( {(\det \,5(det\,A)B){A^2}} \right) = {2^a}{3^b}{5^c}$$ for some a, b, c, $$\in$$ N, then a + b + c is equal to :

Accepted Answer

Given,

$$\det (A) = 2$$,

$$\det (B) = 3$$

and $$\det \left( {\left( {\det \left( {5\left( {\det A} \right)B} \right)} \right){A^2}} \right) = {2^a}{3^b}{5^c}$$

$$ \Rightarrow \left| {\det \left( {5\left( {\det A} \right)B} \right){A^2}} \right| = {2^a}{3^b}{5^c}$$

$$ \Rightarrow \left| {\left| {5\left( {\det A} \right)\left. B \right|{A^2}} \right.} \right| = {2^a}{3^b}{5^c}$$

$$ \Rightarrow \left| {\left| {5\left| {A\left| B \right.\left| {{A^2}} \right.} \right.} \right.} \right| = {2^a}{3^b}{5^c}$$

$$ \Rightarrow \left| {\left| {5\,.\,2\,.\,B} \right|{A^2}} \right| = {2^a}\,.\,{3^b}\,.\,{5^c}$$

$$\Rightarrow \left| {\left| {10B} \right|{A^2}} \right| = {2^a}\,.\,{3^b}\,.\,{5^c}$$

$$ \Rightarrow \left| {{{10}^2}\,.\,\left| B \right|{A^2}} \right| = {2^a}\,.\,{3^b}\,.\,{5^c}$$

As $$\left| {k\,.\,A} \right| = {k^n}|A|$$

$$ \Rightarrow \left| {100 \times 3{A^2}} \right| = {2^a}\,.\,{3^b}\,.\,{5^c}$$

$$ \Rightarrow {(300)^2}\,.\,|{A^2}| = {2^a}\,.\,{3^b}\,.\,{5^c}$$

$$ \Rightarrow {(300)^2}\,.\,|A{|^2} = {2^a}\,.\,{3^b}\,.\,{5^c}$$

$$ \Rightarrow {(300)^2}\,.\,{2^2} = {2^a}\,.\,{3^b}\,.\,{5^c}$$

$$ \Rightarrow 9 \times 100 \times 100 \times {2^2} = {2^a}\,.\,{3^b}\,.\,{5^c}$$

$$ \Rightarrow {3^2} \times {2^2} \times {5^2} \times {2^2} \times {5^2} \times {2^2} = {2^a}\,.\,{3^b}\,.\,{5^c}$$

$$ \Rightarrow {2^6}\,.\,{3^2}\,.\,{5^4} = {2^a}\,.\,{3^b}\,.\,{5^c}$$

Comparing both sides, we get

$a = 6$, $b = 2$, $c = 4$

$$\therefore$$ $a + b + c = 6 + 2 + 4 = 12$

Solutions · 1