Question

If every element of third order determinant of $\Delta$ is multiplied by 5 then value of new determinant equals to,

• A. $\Delta$
• B. $5\Delta$
• C. $25\Delta$
• D. $125\Delta$

Answer 1

The correct option is D $125\Delta$

$|kA|=k^n|A|$ where n is order of matrix
Let $\Delta =|A|$
So, ${\Delta }^{\prime }=|5A|=5^3|A|$
${\Delta }^{\prime }=125\Delta$