Question

In the following exercises, find and then evaluate the indicated minors.

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

Find the minor $\left(a\right)a_1\left(b\right)b_2\left(c\right)c_3$

Short answer

Part (a) The minor of $a_1$is $2.$

Part (b) The minor of $b_2$is $31.$

Part (c) The minor of $c_3$is $-1.$

Step-by-step solution

Part (a)  Step 1. Given Information

The given matrix is

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

We have to find the minor of$a_1.$

Part (a)  Step 2. Evaluate the minor of a1

To evaluate the minor of $a_1$.

Let's eliminate the row and column that contains $a_1$.

Then write the remaining determinants as $2\times 2$matrix.

Part (a)  Step 3. Solve the matrix

So, the matrix is

$$\begin{gathered} \left|\begin{array}{cc}0& -2\\ 1& 5\end{array}\right| \\ =\left(0\right)\left(5\right)-(-2)\left(1\right) \\ =0+2 \\ =2 \end{gathered}$$

Thus, the minor of$a_1$is$2.$

Part (b) Step 1. Given Information

The given matrix is

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

We have to find the minor of$b_2$.

Part (b)  Step 2. Evaluate the minor of b2

To evaluate the minor of $b_2$

Let's eliminate the row and column that contains$b_2$

Then write the remaining determinants as $2\times 2$matrix.

Part (b)  Step 3. Solve the matrix 

So, the matrix is

$$\begin{gathered} \left|\begin{array}{cc}3& 4\\ -4& 5\end{array}\right| \\ =\left(3\right)\left(5\right)-\left(4\right)(-4) \\ =15+16 \\ =31 \end{gathered}$$

Thus, the minor of$b_2$is$31.$

Part (c)  Step 1. Given Information.

The given matrix is

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

We have to find the minor of$c_3.$

Part (c)  Step 2. Evaluate the minor of c3

To evaluate the minor of$c_3$

Let's eliminate the row and column that contains$c_3$

Then write the remaining determinants as $2\times 2$matrix.

Part (c)  Step 3. Solve the matrix

So, the matrix is,

$$\begin{gathered} \left|\begin{array}{cc}3& -1\\ -1& 0\end{array}\right| \\ =\left(3\right)\left(0\right)-(-1)(-1) \\ =0-1 \\ =-1 \end{gathered}$$

Thus, the minor of $c_3$is $-1.$