Q 237 - Intermediate Algebra

Question

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

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

Find the minor $\left(a\right) a_{1 }\left(b\right) b_1 \left(c\right) c_2$

Step-by-Step Solution

Answer

Part a. The minor of $a_1$ is $6.$

Part b. The minor of $b_1$ is $-14.$

Part c. The minor of $c_2$ is $-6.$

Part a Step 1. Given Information

The given matrix is

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

We have to find the minor of $a_1$.

Part a Step 2. Evaluate the minor of a 1

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

$$\begin{gathered} \left|\begin{array}{cc}-2& -1\\ 0& -3\end{array}\right| \\ =(-2)(-3)-\left(0\right)(-1) \\ =6-0 \\ =6 \end{gathered}$$

Thus, the minor of $a_1$ is $6.$

Part b Step 1. Given Information

The given matrix is

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

We have to find the minor of  $b_1.$

Part b Step 2. Evaluate the minor of b 1

To evaluate the minor of $b_1$

Let's eliminate the row and column that contains $b_1$

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

Part b Step 3. Solve the matrix

So, the matrix is

$$\begin{gathered} \left|\begin{array}{cc}4& -1\\ -2& -3\end{array}\right| \\ =\left(4\right)(-3)-(-1)(-2) \\ =(-12)-\left(2\right) \\ =-14 \end{gathered}$$

Thus, the minor of $b_1$ is $-14.$

Part c Step 1. Given Information

The given matrix is

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

We have to find the minor of  $c_2.$

Part c Step 2. Evaluate the minor of c 2

To evaluate the minor of $c_2$

Let's eliminate the row and column that contains $c_2$

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

Part c Step 3. Solve the matrix

So, the matrix is

$$\begin{gathered} \left|\begin{array}{cc}-1& -3\\ -2& 0\end{array}\right| \\ =(-1)\left(0\right)-(-2)(-3) \\ =\left(0\right)-\left(6\right) \\ =-6 \end{gathered}$$

Thus, the minor of $c_2$ is $-6.$