
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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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.
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.
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