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. ($\log 3=0.477,\log 2=0.303$)
$\begin{align}
  & \text{a) 56 digits} \\
 & \text{b) 60 digits} \\
 & \text{c) 58 digits} \\
 & \text{d) 53 digits} \\
\end{align}$

Answer 1

Hint: 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
$\begin{align}
  & \dfrac{{{9}^{5}}}{{{4}^{5}}}=|A{{|}^{2}} \\
 & \Rightarrow |A{{|}^{2}}={{\left( \dfrac{9}{4} \right)}^{5}} \\
 & \Rightarrow |A|={{\left( \dfrac{9}{4} \right)}^{\dfrac{5}{2}}} \\
 & \Rightarrow |A|={{\left( \dfrac{3}{2} \right)}^{5}} \\
\end{align}$
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
\[\begin{align}
  & |adjA|=|A{{|}^{5-1}}=|A{{|}^{4}} \\
 & \Rightarrow |adj(adjA)|={{(|A{{|}^{4}})}^{5-1}} \\
 & \Rightarrow |adj[adj(adjA)]|={{({{(|A{{|}^{4}})}^{4}})}^{5-1}} \\
 & \Rightarrow |adj[adj(adjA)]|={{({{(|A{{|}^{4}})}^{4}})}^{4}}=|A{{|}^{{{4}^{3}}}} \\
\end{align}\]
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.
\[\begin{align}
  & =\left[ (152.64)-(96.96) \right]+1 \\
 & =[55.6]+1 \\
 & =55+1 \\
 & =56 \\
\end{align}\]

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.