Question

Q1.solve the determinant by using ementary operations​
.

Pls

Answer 1

Step-by-step explanation:

Given:

$\begin{gathered}\left|\begin{array}{ccc}3&1&1\\1&3&1\\1&1&3\end{array}\right|\\\end{gathered}$

To find:

• Determinant using elementary operations

Solution:

Perform

$\begin{gathered}R_1- > R_1-R_3\\\end{gathered}$

$\begin{gathered}R_2- > R_2-R_3\\\end{gathered}$

$\begin{gathered}=\left|\begin{array}{ccc}3-1&1-1&1-3\\1-1&3-1&1-3\\1&1&3\end{array}\right|\\\end{gathered}$

$\begin{gathered}=\left|\begin{array}{ccc}2&0&-2\\0&2&-2\\1&1&3\end{array}\right|\\\end{gathered}$

Taking common 2 from first row and second row

$\begin{gathered}=4\left|\begin{array}{ccc}1&0&-1\\0&1&-1\\1&1&3\end{array}\right|\\\end{gathered}$

Perform

$\begin{gathered}R_3- > R_3-R_1\\\end{gathered}$

$\begin{gathered}=4\left|\begin{array}{ccc}1&0&-1\\0&1&-1\\1-1&1-0&3+1\end{array}\right|\\\end{gathered}$

$\begin{gathered}=4\left|\begin{array}{ccc}1&0&-1\\0&1&-1\\0&1&4\end{array}\right|\\\end{gathered}$

Expand the determinant along C1

$\begin{gathered} = 4(4 + 1) \\ \\ = 4 \times 5 \\ \\ = 20 \\ \end{gathered}$

Final answer:

$\begin{gathered}\left|\begin{array}{ccc}3&1&1\\1&3&1\\1&1&3\end{array}\right|=20\\\end{gathered}$