Related questions with answers
Use properties of determinants to evaluate the given determinant by inspection. Explain your reasoning.
$\left \begin{array} { r r r } { 2 } & { 3 } & {  4 } \\ { 1 } & {  3 } & {  2 } \\ {  1 } & { 5 } & { 2 } \end{array} \right$
Solutions
VerifiedRecall the following results.

If $C$ is a matrix obtained from a matrix $B$ by adding a multiple of $i^\text{th}$ column of $B$ to its $j^\text{th}$ column, then $\det(C)=\det(B)$.

If a column of a matrix is a zero column, then the determinant of the matrix is zero.
By, ispection, we can see that if we add $2$ times second row to first, first row will be :
$\begin{matrix} 0& 3& 0 \end{matrix}$
Also, if we add 1 times second row to third it will be:
$\begin{matrix} 0&2 &0 \end{matrix}$
Now, it is easy to see that if we add $\dfrac{2}{3}$ times first row to thrid, we would obtain zeros in third row, so by $\textbf{Theorem 4.3 a)}$, value of determinant is $0$.
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