Question

In the following exercise, evaluate each determinant by expanding by minors.

$\left|\begin{array}{rrr}3& 5& 4\\ -1& 3& 0\\ -2& 6& 1\end{array}\right|$

Short answer

The value of the given determinant is$14$

Step-by-step solution

Step 1. Given information

The given determinant is:

$\left|\begin{array}{rrr}3& 5& 4\\ -1& 3& 0\\ -2& 6& 1\end{array}\right|$

Step 2. Expanding the determinant along the first row by minors 

The expansion of general determinant along the first row is given as :

$\left|\begin{array}{lll}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|=a_1\left|\begin{array}{ll}{b}_2& c_2\\ b_3& c_3\end{array}\right|-b_1\left|\begin{array}{ll}{a}_2& c_2\\ a_3& c_3\end{array}\right|+c_1\left|\begin{array}{ll}{a}_2& b_2\\ a_3& b_3\end{array}\right|$

Similarly, for the given determinant, we have:

$\left|\begin{array}{ccc}3& 5& 4\\ -1& 3& 0\\ -2& 6& 1\end{array}\right|=1\left|\begin{array}{ll}5& 4\\ 6& 1\end{array}\right|+3\left|\begin{array}{cc}3& 4\\ -2& 1\end{array}\right|+0\left|\begin{array}{cc}3& 5\\ -2& 6\end{array}\right|$

Step 3. Evaluating the determinant

Evaluating the expanded determinant, we get:

$$\begin{gathered} =1\left|\begin{array}{ll}5& 4\\ 6& 1\end{array}\right|+3\left|\begin{array}{cc}3& 4\\ -2& 1\end{array}\right|+0\left|\begin{array}{cc}3& 5\\ -2& 6\end{array}\right| \\ =1[5\times 1-(4\times 6\left)\right]+3[3\times 1-(4\times -2\left)\right]+0 \\ =(5-24)+3(3+8) \\ =-19+33 \\ =14 \end{gathered}$$