Question

How many \[3\times 3\] matrices M with entries from {0, 1, and 2} are there, for which the sum of the diagonal entries of \[{{M}^{T}}M\] is 5?

Answer 1

Hint:
To solve the answer to this question, we should have proper knowledge of finding the transpose of a matrix, and also we should have knowledge of multiplying the matrix. In this question, first we will take a matrix in the form of \[M=\left( \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right)\] and find the transpose of a matrix M. After that, we will multiply the matrix M with the transpose of the same matrix M. After multiplying the matrices, we will get the resultant matrix. After getting the resultant matrix, we will find the sum of diagonal elements of the resultant matrix. After that, we will find the combinations of getting the sum as 5 by putting the values of a, b, c, d, e, f, g, h, and i as 0, 1 and 2. And, one should also have a proper knowledge of permutation and combination such as \[{}^{n}{{C}_{r}}=\left( \begin{matrix}
   n \\
   r \\
\end{matrix} \right)=\dfrac{n!}{r!(n-r)!}\].

Complete step by step answer:
Let us suppose, we have a matrix of M order of \[3\times 3\],
\[M=\left( \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right)\]
Therefore, transpose of M will be
\[{{M}^{T}}=\left( \begin{matrix}
   a & d & g \\
   b & e & h \\
   c & f & i \\
\end{matrix} \right)\]
Then, \[{{M}^{T}}M=\left( \begin{matrix}
   a & d & g \\
   b & e & h \\
   c & f & i \\
\end{matrix} \right)\left( \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right)\]
\[\Rightarrow \] \[{{M}^{T}}M=\left( \begin{matrix}
   {{a}^{2}}+{{d}^{2}}+{{g}^{2}} & ab+de+gh & ac+df+gi \\
   ba+ed+hg & {{b}^{2}}+{{e}^{2}}+{{h}^{2}} & bc+ef+hi \\
   ca+fd+ig & cb+fe+ih & {{c}^{2}}+{{f}^{2}}+{{i}^{2}} \\
\end{matrix} \right)\]
Now, we will find the sum of the diagonal entries.
Let, S be the sum of sum of diagonal entries
\[S={{a}^{2}}+{{d}^{2}}+{{g}^{2}}+{{b}^{2}}+{{e}^{2}}+{{h}^{2}}+{{c}^{2}}+{{f}^{2}}+{{i}^{2}}\]
\[\Rightarrow \] \[S={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}+{{e}^{2}}+{{f}^{2}}+{{g}^{2}}+{{h}^{2}}+{{i}^{2}}\]
Now, we will have to find the combinations of S=5 by putting the values of a, b, c, d, e, f, g, h, and i as 0, 1 and 2.
For the above, we are having two possibilities,
First, \[5={{1}^{2}}+{{1}^{2}}+{{1}^{2}}+{{1}^{2}}+{{1}^{2}}+{{0}^{2}}+{{0}^{2}}+{{0}^{2}}+{{0}^{2}}\]
Second, \[~5={{2}^{2}}+{{1}^{2}}+{{0}^{2}}+{{0}^{2}}+{{0}^{2}}+{{0}^{2}}+{{0}^{2}}+{{0}^{2}}+{{0}^{2}}\]
In both the above cases, we are having 9 entries.
In the first case, we have 5 elements of 1 and 4 elements of 0.
Therefore, the total number of combinations we will have \[{}^{9}{{C}_{5}}\times {}^{4}{{C}_{4}}=\left( \begin{matrix}
   9 \\
   5 \\
\end{matrix} \right)\times \left( \begin{matrix}
   4 \\
   4 \\
\end{matrix} \right)=\dfrac{9!}{5!\left( 9-5 \right)!}\times \dfrac{4!}{4!\left( 4-4 \right)!}=\dfrac{9\times 8\times 7\times 6\times 5}{5\times 4\times 3\times 2\times 1}\times 1=126\]
In the second case, we have one element of 2, one element of 1 and 6 elements of 0.
Therefore, the total number of combinations we will have in the second case\[{}^{9}{{C}_{7}}\times {}^{2}{{C}_{1}}\times {}^{1}{{C}_{1}}=\left( \begin{matrix}
   9 \\
   7 \\
\end{matrix} \right)\times \left( \begin{matrix}
   2 \\
   1 \\
\end{matrix} \right)\times \left( \begin{matrix}
   1 \\
   1 \\
\end{matrix} \right)=\dfrac{9!}{7!(9-7)!}\times \dfrac{2!}{1!(2-1)!}\times 1=\dfrac{9\times 8}{2\times 1}\times 2=72\]
Hence, the total number of combinations ( i.e. total number of matrices with entries from {0, 1 and 2} for which the sum of diagonal entries of \[{{M}^{T}}M\]is 5) is 126+72 = 198.

Note:
One should always remember the formulas of permutation and combinations. And also have the knowledge in the matrix. For example, multiplying the matrices and finding the transpose of a matrix. Someone can make mistake in finding the transpose and multiplying the matrices. So, be aware of that thing.