Answer
0
Work Step by Step
Let \[A=\left|\begin{array}{ccc} 12&13&14\\
15&16&17\\
18&19&20 \end{array}\right|\]
Applying column operations on A
\[C_2\rightarrow C_2-C_1\]
and \[C_3\rightarrow C_3-C_1\]
\[\left|\begin{array}{ccc} 12&1&2\\
15&1&2\\
18&1&2 \end{array}\right|\]
We will use the property of determinant
\[\left|\begin{array}{ccc} a&b&kc\\
d&e&kf\\
g&h&ki\end{array}\right|=k\left|\begin{array}{ccc} a&b&c\\
d&e&f\\
g&h&i\end{array}\right|\;\;\;...(1)\]
By using (1)
\[\left|\begin{array}{ccc} 12&1&2\\
15&1&2\\
18&1&2 \end{array}\right|=2\left|\begin{array}{ccc} 12&1&1\\
15&1&1\\
18&1&1 \end{array}\right|\]
Now $C_2 $ and $C_3$ are identical
\[\Rightarrow 2\left|\begin{array}{ccc} 12&1&1\\
15&1&1\\
18&1&1 \end{array}\right|=2(0)=0\]
Hence, $A=0$.
