Question

solve the determinant by using ementary operations​

Answer 1

Step-by-step explanation:

Given:

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

To find: Determinant using elementary operations

Solution:

Perform

$R_1->R_1-R_3$

$R_2->R_2-R_3$

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

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

Taking common 2 from first row and second row

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

Perform$R_3->R_3-R_1$

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

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

Expand the determinant along C1

$= 4(4 + 1) \\ \\ = 4 \times 5 \\ \\ = 20$

Final answer:

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

Hope it helps you.

To learn more on brainly:

if a=(3 0/5 1) and b=(-4 2/1 0) A^2-2ab+b^2

https://brainly.in/question/43921889