Question

Let $M = \left[ {\begin{array}{*{20}{c}}
  a&{ - 360} \\
  b&c
\end{array}} \right]$, where $a,b$ and $c$ are integers. Then the smallest positive value of $b$ such that ${M^2} = O$, where $O$ denotes $2 \times 2$ null matrix is $l$ then sum of digits of $l$ is:

Answer 1

Hint: We will substitute the value of $M$ in the given equation. We will multiply $M$ with $M$ to find the value of ${M^2}$. We will then compare the entries of the matrices to form the equation. We will find the value of $b$ such that the given condition holds. Then, we will add the digits of $b$ to determine the value of $b$.

Complete step-by-step answer:
We are given that ${M^2} = O$.
We will substitute the value of $M$ in the given equation.
$
  \left[ {\begin{array}{*{20}{c}}
  a&{ - 360} \\
  b&c
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  a&{ - 360} \\
  b&c
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&0 \\
  0&0
\end{array}} \right] \\
   \Rightarrow \left[ {\begin{array}{*{20}{c}}
  {{a^2} - 360b}&{ - 360a - 360c} \\
  {ab + bc}&{ - 360b + {c^2}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&0 \\
  0&0
\end{array}} \right] \\
$
On comparing the corresponding terms, we will get,
The ${a_{11}}$ term is:
$
\Rightarrow {a^2} - 360b = 0 \\
   \Rightarrow b = \dfrac{{{a^2}}}{{360}} \\
$
Here,
 $ \Rightarrow$ $ {a^2}$ = 360b
 $ \Rightarrow$ a = $\sqrt {360b}$
But, we are given that $a$ and $b$ are integers.
This implies $360b$ has to be a perfect square.
Therefore, the smallest value of $b$ will be $10$
Then, $a = \sqrt {3600} = 60$
Thus, the value of \[l\] is 10.
The sum of digits of $l$ is $1 + 0 = 1$.

Note: Two matrices are equal if their corresponding terms are equal. We can also use other entries of the matrices to form equations. The null matrix is a matrix whose all entries are zero. Also, multiplication of the matrices should be correct.