Elementary Matrices
More on Determinants

Example 1.

We would like to find the determinant of

(1)   \begin{equation*}A =\det \left (\begin{array}{cc}1 & -1 \\ 3 & 6\end{array}\right ) \end{equation*}

We write

(2)   \begin{equation*}\det A =\left (1\right )\left (6\right ) -( -1)\left (3\right ) =9 \end{equation*}

 

Example 2.

We would like to find the determinant of

(3)   \begin{equation*}A =\left (\begin{array}{ccc}4 & -2 & -1 \\ -5 & 6 & 1 \\ 1 & 5 & 3\end{array}\right ) \end{equation*}

We start by using the first row and definition of determinant

(4)   \begin{align*}\det A =4\det \left (\begin{array}{cc}6 & 1 \\ 5 & 3\end{array}\right ) -\left ( -2\right )\det \left (\begin{array}{cc} -5 & 1 \\ 1 & 3\end{array}\right ) +\left ( -1\right )\det \left (\begin{array}{cc} -5 & 6 \\ 1 & 5\end{array}\right ) \\ =4\left [\left (6\right )\left (3\right ) -\left (1\right )\left (5\right )\right ] +2\left [\left ( -5\right )\left (3\right ) -\left (1\right )\left (1\right )\right ] -\left [\left ( -5\right )\left (5\right ) -\left (6\right )\left (1\right )\right ] \\ =51\end{align*}

 

Example 3.

Suppose we are given

(5)   \begin{equation*}A =\left (\begin{array}{cc}\lambda -1 & -1 \\ 1 & \lambda -1\end{array}\right ) \end{equation*}

We want to find the value(s) of \lambda such that \det A =0. We know determinant of A is computed by

(6)   \begin{equation*}(\lambda -1)\left (\lambda -1\right ) -( -1)\left (1\right ) =\left (\lambda -1\right )^{2} +1 \end{equation*}

For this to be equal to zero, we must have

(7)   \begin{equation*}\left (\lambda -1\right )^{2} = -1 \end{equation*}

Which means

(8)   \begin{align*}\lambda -1 = \pm i \\ \lambda _{1} =1 +i \\ \lambda _{2} =1 -i\end{align*}

 

Example 4.

Suppose we are given

(9)   \begin{equation*}A =\left (\begin{array}{ccc}4 & -8 & 6 \\ 1 & 2 & 3 \\ 5 & 4 & 12\end{array}\right ) \end{equation*}

 

We would like to use elementary row operations to find the determinant of A.

 

1. Interchange the first and second row. We know this will result in

(10)   \begin{equation*}\det A = -\det \left (\begin{array}{ccc}1 & 2 & 3 \\ 4 & -8 & 6 \\ 5 & 4 & 12\end{array}\right ) \end{equation*}

 

2. Multiply the first row by -4 and add the first row to the second. This will result in no change in determinant

(11)   \begin{equation*}\det A = -\det \left (\begin{array}{ccc}1 & 2 & 3 \\ 0 & -16 & -6 \\ 5 & 4 & 12\end{array}\right ) \end{equation*}

 

3. Next multiply the second row by -\frac{1}{16}

(12)   \begin{equation*}\det A = -\left ( -16\right )\det \left (\begin{array}{ccc}1 & 2 & 3 \\ 0 & 1 & \frac{3}{8} \\ 5 & 4 & 12\end{array}\right ) \end{equation*}

 

4. Next, multiply the first row by -5 and add to the third row. No change in determinant

(13)   \begin{equation*}\det A = -\left ( -16\right )\det \left (\begin{array}{ccc}1 & 2 & 3 \\ 0 & 1 & \frac{3}{8} \\ 0 & -6 & -3\end{array}\right ) \end{equation*}

 

5. Now multiply the second row by 6 and add it to the third row (again no change in determinant)

(14)   \begin{equation*}\det A = -\left ( -16\right )\det \left (\begin{array}{ccc}1 & 2 & 3 \\ 0 & 1 & \frac{3}{8} \\ 0 & 0 & -\frac{3}{4}\end{array}\right ) \end{equation*}

 

We can see now that the determinant of the last obtained matrix is \frac{ -3}{4} and so

(15)   \begin{equation*}\det A =16\left ( -\frac{3}{4}\right ) = -12 \end{equation*}

 

Example 5.

Suppose we are given the matrix

(16)   \begin{equation*}A =\left (\begin{array}{ccc} -1 & 2 & -2 \\ 3 & -6 & -4 \\ 5 & -10 & 3\end{array}\right ) \end{equation*}

We would like to use elementary row operations to find determinant of A . We start by noticing the following steps:

 

1. If we multiply the first row of A by -1 and obtain

(17)   \begin{equation*}\det \left (\begin{array}{ccc} -1 & 2 & -2 \\ 3 & -6 & -4 \\ 5 & -10 & 3\end{array}\right ) =\left ( -1\right )\det \left (\begin{array}{ccc}1 & -2 & 2 \\ 3 & -6 & -4 \\ 5 & -10 & 3\end{array}\right ) \end{equation*}

 

2. Now multiply the first row by -3 and add the first row to the second row. In this case no change happens in determinant

(18)   \begin{equation*}\det \left (\begin{array}{ccc} -1 & 2 & -2 \\ 3 & -6 & -4 \\ 5 & -10 & 3\end{array}\right ) =\left ( -1\right )\det \left (\begin{array}{ccc}1 & -2 & 2 \\ 0 & 0 & -10 \\ 5 & -10 & 3\end{array}\right ) \end{equation*}

 

3. Next multiply the first row by -5 and add the first row to the last row. Here again no change in determinant

(19)   \begin{equation*}\det \left (\begin{array}{ccc} -1 & 2 & -2 \\ 3 & -6 & -4 \\ 5 & -10 & 3\end{array}\right ) =\left ( -1\right )\det \left (\begin{array}{ccc}1 & -2 & 2 \\ 0 & 0 & -10 \\ 0 & 0 & -7\end{array}\right ) \end{equation*}

As you see the matrix obtained which is row equivalent to A is an upper triangular matrix and determinant of an upper or lower triangular matrix is the multiplication of the diagonal elements

(20)   \begin{equation*}\det A =\left ( -1\right )\left (1\right )\left (0\right )\left ( -7\right ) =0 \end{equation*}

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