Question
If $\left|\begin{array}{lll}<\text{br}> &1 &\text{amp; } 0 &\text{amp; } 0\\ <\text{br}>&2 &\text{amp; } 3 &\text{amp; } 4\\ <\text{br}>&5 &\text{amp; } -6 &\text{amp; x}<\text{br}> \end{array}\right|=45$ then $\text{x}=$
  1. 4
  2. 7
  3. -5
  4. -7

Answer

  1. 7
Solution:
Given, $\left|\begin{array}{lll}<\text{br}> &1 &\text{amp; } 0 &\text{amp; } 0\\ <\text{br}>&2 &\text{amp; } 3 &\text{amp; } 4\\ <\text{br}>&5 &\text{amp; } -6 &\text{amp; x}<\text{br}> \end{array}\right|=45$
By operation of matrix (5),
$1(3\text{x}+24)=45$
$3\text{x}=21$
$\Rightarrow \text{x}=7$

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