Chapter 16, Problem 12CT

To determine

To find: The value of given determinant by using minor about third row.

Answer to Problem 12CT

The value of given determinant by using minor about third row is $\left|\begin{array}{ccc}3& 0& -1\\ 2& -1& 1\\ 1& -1& 0\end{array}\right|=10$ .

Explanation of Solution

Given information:

The givendeterminant is $\left|\begin{array}{ccc}3& 0& -1\\ 2& -1& 1\\ 1& -1& 0\end{array}\right|$ .

Calculation:

Multiply each element of third row by its minor. Also multiply the product for the element in the third row second column by $-1$ . Since the sum $1+2$ or $3$ .

Add the product.

  $\begin{array}{l}\left|\begin{array}{ccc}3& 0& -1\\ 2& -1& 1\\ 1& -1& 0\end{array}\right|=1\left(3\cdot 1-\left(-1\right)\cdot 2\right)-\left(-1\right)\left(3\cdot 1-\left(-1\right)\cdot 2\right)+0\left(3\left(-1\right)-20\right)\\ \left|\begin{array}{ccc}3& 0& -1\\ 2& -1& 1\\ 1& -1& 0\end{array}\right|=1\left(3+2\right)+\left(3+2\right)+0\\ \left|\begin{array}{ccc}3& 0& -1\\ 2& -1& 1\\ 1& -1& 0\end{array}\right|=5+5\\ \left|\begin{array}{ccc}3& 0& -1\\ 2& -1& 1\\ 1& -1& 0\end{array}\right|=10\end{array}$

Therefore,the value of given determinant by using minor about third rowis $\left|\begin{array}{ccc}3& 0& -1\\ 2& -1& 1\\ 1& -1& 0\end{array}\right|=10$ .