Question

If A is a square matrix of order 5 and $9{{A}^{-1}}=4{{A}^{T}}$ . Then $\left| adj(adj\left( adj\text{ }A \right) \right|$ contains how many digits.
Here ${{A}^{-1}},{{A}^{T}}\text{ and }adj(A)$ means Inverse of A, Transpose of A and adjugate matrix of A respectively. ($\log 3=0.477,\log 2=0.303$)
$\begin{aligned} & \text{a) 56 digits} \\\ & \text{b) 60 digits} \\\ & \text{c) 58 digits} \\\ & \text{d) 53 digits} \\\ \end{aligned}$

Answer 1

First we will try to solve the given equation by taking the determinant. We can use the properties $|{{A}^{T}}|=|A|$ and $|{{A}^{-1}}|=\dfrac{1}{|A|}$ to find the value of determinant of A.

Complete step by step answer:
Then we know the relation between determinant of adjugate A and determinant of A which is $|adj(A)|=|A{{|}^{n-1}}$ using this relation 3 times successively we get $\left| adj(adj\left( adj\text{ }A \right) \right|$ and hence we can find the number of digits in $\left| adj(adj\left( adj\text{ }A \right) \right|$ is greatest integer of $\left| adj(adj\left( adj\text{ }A \right) \right|$+ 1
Now we are given with the equation $9{{A}^{-1}}=4{{A}^{T}}$ and the order of Matrix A is 5.
Taking determinant on both sides we get $|9{{A}^{-1}}|=|4{{A}^{T}}|$
Now we will use the property of determinant which says if A is a matrix of order n then
$|pA|={{p}^{n}}|A|$. Hence we get
${{9}^{5}}|{{A}^{-1}}|={{4}^{5}}|{{A}^{T}}|$.
Now we also know that $|{{A}^{T}}|=|A|$ and $|{{A}^{-1}}|=\dfrac{1}{|A|}$ .
Using this properties we get ${{9}^{5}}\dfrac{1}{|A|}={{4}^{5}}|A|$
Now we will rearrange the terms by taking ${{4}^{5}}$ to LHS and $\dfrac{1}{|A|}$ to RHS. So we get
$\begin{aligned} & \dfrac{{{9}^{5}}}{{{4}^{5}}}=|A{{|}^{2}} \\\ & \Rightarrow |A{{|}^{2}}={{\left( \dfrac{9}{4} \right)}^{5}} \\\ & \Rightarrow |A|={{\left( \dfrac{9}{4} \right)}^{\dfrac{5}{2}}} \\\ & \Rightarrow |A|={{\left( \dfrac{3}{2} \right)}^{5}} \\\ \end{aligned}$
Now we have $|A|={{\left( \dfrac{3}{2} \right)}^{5}}.....................(1)$
Now we have a property of determinant which says $|adj(A)|=|A{{|}^{n-1}}$
Let us successively use this property

& |adjA|=|A{{|}^{5-1}}=|A{{|}^{4}} \\\ & \Rightarrow |adj(adjA)|={{(|A{{|}^{4}})}^{5-1}} \\\ & \Rightarrow |adj[adj(adjA)]|={{({{(|A{{|}^{4}})}^{4}})}^{5-1}} \\\ & \Rightarrow |adj[adj(adjA)]|={{({{(|A{{|}^{4}})}^{4}})}^{4}}=|A{{|}^{{{4}^{3}}}} \\\ \end{aligned}$$ Now we substitute the value of |A| from equation (1) so we get $$|adj[adj(adjA)]|={{\left( {{\left( \dfrac{3}{2} \right)}^{5}} \right)}^{{{4}^{3}}}}$$ Now we know that ${{({{a}^{m}})}^{n}}={{a}^{mn}}$ using this property we get $$\Rightarrow |adj[adj(adjA)]|={{\left( \dfrac{3}{2} \right)}^{5\times {{4}^{3}}}}={{\left( \dfrac{3}{2} \right)}^{320}}$$ Now number of digits in $$\left| adj(adj\left( adj\text{ }A \right) \right|$$ $$=\left[ \log {{\left( \dfrac{3}{2} \right)}^{320}} \right]+1$$ Where [] is greatest integer function $$=\left[ \log \dfrac{{{3}^{320}}}{{{2}^{320}}} \right]+1$$ Now we apply the property of log which is $\log \dfrac{a}{b}=\log a-\log b$ $$=\left[ \log {{3}^{320}}-\log {{2}^{320}} \right]+1$$ Now we also know that $\log {{a}^{b}}=b\log a$ $$=\left[ 320\log 3-320\log 2 \right]+1$$ We are given the values of log3 and log2, we will substitute those values in the equation to get $$=\left[ 320(0.477)-320(0.303) \right]+1$$ Now 320 × 0.477 = 152.64 and 320 × 0.303 = 96.96. $$\begin{aligned} & =\left[ (152.64)-(96.96) \right]+1 \\\ & =[55.6]+1 \\\ & =55+1 \\\ & =56 \\\ \end{aligned}$$ **Hence the number of digits in $$\left| adj(adj\left( adj\text{ }A \right) \right|$$ is equal to 56** **Note:** The characteristic of the logarithm of a positive number is positive and it is one less than the number of digits in the number. Hence we use find number of digits in $$\left| adj(adj\left( adj\text{ }A \right) \right|$$ by taking greatest integer of log of $$\left| adj(adj\left( adj\text{ }A \right) \right|$$ and adding 1.