Q 239 - Intermediate Algebra

Question

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

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

Find the minor $\left(a\right) a_3 \left(b\right) b_3 \left(c\right) c_3$

Step-by-Step Solution

Answer

Part a. The minor of $a_3$ is $-9.$

Part b. The minor of $b_3$ is $-3.$

Part c. The minor of $c_3$ is $8.$

Part a Step 1. Given Information

The given matrix is

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

We have to find the minor of $a_3.$

Part a Step 2. Evaluate the minor of a 3

To evaluate the minor of $a_3$

Let's eliminate the row and column that contains $a_3$

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

Part a Step 3. Solve the matrix

So, the matrix is

$$\begin{gathered} \left|\begin{array}{cc}-2& 3\\ -3& 0\end{array}\right| \\ =(-2)\left(0\right)-\left(3\right)\left(3\right) \\ =0-9 \\ =-9 \end{gathered}$$

Thus, the minor of $a_3$ is $-9.$

Part b Step 1. Given Information

The given matrix is

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

We have to find the minor of $b_3.$

Part b Step 2. Evaluate the minor of b 3

To evaluate the minor of $b_3$

Let's eliminate the row and column that contains $b_3$

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

Part b Step 3. Solve the matrix

So, the matrix is

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

Thus, the minor of $b_3$ is $-3.$

Part c Step 1. Given Information

The given matrix is

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

We have to find the minor of $c_3.$

Part c Step 2. Evaluate the minor of c 3

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

$$\begin{gathered} \left|\begin{array}{cc}-2& -2\\ 1& -3\end{array}\right| \\ =(-2)(-3)-\left(1\right)(-2) \\ =6+2 \\ =8 \end{gathered}$$

Thus, the minor of $c_3$ is $8.$