Question

Let matrix $A=\left[ \begin{matrix}
   x & y & -z \\
   1 & 2 & 3 \\
   1 & 1 & 2 \\
\end{matrix} \right]$ , where $x,y,z\in N$ . If $\det \left( adj\left( adjA \right) \right)={{2}^{8}}{{3}^{4}}$ , where $adjA$ denotes the adjoint of square matrix A, then the number of such matrices A is:
(a) 220
(b) 45
(c) 55
(d) 110

Answer 1

Hint: To find the number of matrices like A, we have to use the property \[\left| adj\left( adjA \right) \right|={{\left| A \right|}^{{{\left( n-1 \right)}^{2}}}}\] (where n is the order of the matrix A) to find the value of determinant A. Then, we have to find the determinant of A from the given matrix and equate the value of $\left| A \right|$ from the previous step. The result of this step will be an equation of the form ${{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}=m$ whose solution is given by $^{m-1}{{C}_{n-1}}$ . The value of this combination will be the required answer.

Complete step-by-step answer:
We have to find the number of matrices like A. We are given that $\det \left( adj\left( adjA \right) \right)={{2}^{8}}{{3}^{4}}$ .
$\left| adj\left( adjA \right) \right|={{2}^{8}}{{3}^{4}}$
We know that \[\left| adj\left( adjA \right) \right|={{\left| A \right|}^{{{\left( n-1 \right)}^{2}}}}\] , where n is the order of the matrix A. Here, we can see that the order of the matrix, $n=3$ .Therefore, we can write the given condition as
$\begin{align}
  & \Rightarrow \left| adj\left( adjA \right) \right|={{2}^{8}}{{3}^{4}} \\
 & \Rightarrow {{\left| A \right|}^{{{\left( 3-1 \right)}^{2}}}}={{2}^{8}}{{3}^{4}} \\
\end{align}$
Let us simplify the LHS of the above equation.
$\begin{align}
  & \Rightarrow {{\left| A \right|}^{{{2}^{2}}}}={{2}^{8}}{{3}^{4}} \\
 & \Rightarrow {{\left| A \right|}^{4}}={{2}^{8}}{{3}^{4}} \\
\end{align}$
We know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ . Let us rewrite the exponents in the RHS using this rule.
$\Rightarrow {{\left| A \right|}^{4}}={{\left( {{2}^{2}}\times 3 \right)}^{4}}$
From the above equation, we can see that the exponents on both the sides are equal. Hence, we can equate their bases.
$\Rightarrow \left| A \right|=\left( {{2}^{2}}\times 3 \right)$
Now, let us simplify the RHS of the above equation.
$\begin{align}
  & \Rightarrow \left| A \right|=4\times 3 \\
 & \Rightarrow \left| A \right|=12...\left( i \right) \\
\end{align}$
Now, we are given with $A=\left[ \begin{matrix}
   x & y & -z \\
   1 & 2 & 3 \\
   1 & 1 & 2 \\
\end{matrix} \right]$ . Let us find the determinant of the matrix A.
$\begin{align}
  & \Rightarrow \left| A \right|=\left| \begin{matrix}
   x & y & -z \\
   1 & 2 & 3 \\
   1 & 1 & 2 \\
\end{matrix} \right| \\
 & \Rightarrow \left| A \right|=x\left( 4-3 \right)-y\left( 2-3 \right)-z\left( 1-2 \right) \\
\end{align}$
Let us simplify the above equation.
$\Rightarrow \left| A \right|=x+y+z$
We have to substitute (i) in the above equation.
$\Rightarrow x+y+z=12...\left( ii \right)$
We know that for equation of the form ${{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}=m$ , the solution is given as $^{m-1}{{C}_{n-1}}$ . Here, from the above equation, we will get $m=12$ and $n=3$ . Hence, we can write the solution of the equation (ii) as
$\begin{align}
  & ^{12-1}{{C}_{3-1}} \\
 & {{\Rightarrow }^{11}}{{C}_{2}} \\
\end{align}$
We know that $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ . Let us expand the above combination.
$\begin{align}
  & {{\Rightarrow }^{11}}{{C}_{2}}=\dfrac{11!}{2!\left( 11-2 \right)!} \\
 & {{\Rightarrow }^{11}}{{C}_{2}}=\dfrac{11!}{2!9!} \\
\end{align}$
We know that $n!=n\times \left( n-2 \right)\times \left( n-3 \right)!$ . Hence, we can write the above equation as
${{\Rightarrow }^{11}}{{C}_{2}}=\dfrac{11\times 10\times 9!}{\left( 2\times 1 \right)9!}$
Let us cancel 9! From the numerator and denominator of the RHS and simplify.
$\begin{align}
  & {{\Rightarrow }^{11}}{{C}_{2}}=\dfrac{11\times 10}{\left( 2\times 1 \right)} \\
 & {{\Rightarrow }^{11}}{{C}_{2}}=\dfrac{110}{2} \\
 & {{\Rightarrow }^{11}}{{C}_{2}}=55 \\
\end{align}$
Therefore, 55 matrices like A are possible.

So, the correct answer is “Option (c)”.

Note: Students must know the properties and rules associated with matrices, determinants and adjoints. They must know to find the determinant of a matrix. They must be thorough with the basic solutions like the solution of the equation of the form ${{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}=m$ . Students must know the formula of permutations and combinations.