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

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.