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Using the properties of determinants, prove that
 \[\left| \begin{gathered}
  1 \\
  1 + 3y \\
  1 \\
\end{gathered} \right.\] $\begin{gathered}
  1 \\
  1 \\
  1 + 3z \\
\end{gathered} $ \[\left. \begin{gathered}
  1 + 3x \\
  1 \\
  1 \\
\end{gathered} \right| = 9(3xyz + xy + yz + xz)\]

Answer
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534.9k+ views
Hint: In these types of questions remember to use the row -column transformation method for example ${R_3} \to {R_3} - {R_1}$, ${R_2} \to {R_2} - {R_1}$ to solve the question. After doing this row-column transformation, then we just need to find the determinants. We will get the required answer.

Complete step-by-step answer:
According to the given information the given determinant is \[\left| \begin{gathered}
  1 \\
  1 + 3y \\
  1 \\
\end{gathered} \right.\]$\begin{gathered}
  1 \\
  1 \\
  1 + 3z \\
\end{gathered} $\[\left. \begin{gathered}
  1 + 3x \\
  1 \\
  1 \\
\end{gathered} \right|\]

Applying ${R_2} \to {R_2} - {R_1}$
So the determinant becomes \[\left| \begin{gathered}
  1 \\
  3y \\
  1 \\
\end{gathered} \right.\] $\begin{gathered}
  1 \\
  0 \\
  1 + 3z \\
\end{gathered} $ \[\left. \begin{gathered}
  1 + 3x \\
   - 3x \\
  1 \\
\end{gathered} \right|\]

Now applying ${R_3} \to {R_3} - {R_1}$
\[\left| \begin{gathered}
  1 \\
  3y \\
  0 \\
\end{gathered} \right.\] $\begin{gathered}
  1 \\
  0 \\
  3z \\
\end{gathered} $ \[\left. \begin{gathered}
  1 + 3x \\
   - 3x \\
   - 3x \\
\end{gathered} \right|\]

Now, solving the determinant form along ${C_1}$
\[ \Rightarrow \]1(3x) (3z) – 3y (-3x – 3z – 9xz) = 27xyz+9xy+9yz+9xz
\[ \Rightarrow \]9(3xyz+xy+yz+zx)
since LHS = RHS
Hence proved.

Note: In these types of questions first use the row-column transformation method for example ${R_2} \to {R_2} - {R_1}$, ${R_3} \to {R_3} - {R_1}$ to change the matrix such that when we solve the determinant of matrix simplify the result until the LHS becomes equal to RHS and we will get the result we required.