Example 7 from Algebra and Trigonometry, 11.8 Solving Systems with Cramer's Rule

Illustrating Properties of Determinants

Illustrate each of the properties of determinants.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
Property 1 states that if the matrix is in upper triangular form, the determinant is the product of the entries down the main diagonal.

$A = [\begin{array}{r} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 0 & 0 & - 1 \end{array}]$
Augment $A$ with the first two columns.

$A = [\begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 0 & 0 & - 1 \end{matrix} | \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \begin{matrix} 2 \\ 2 \\ 0 \end{matrix}]$

Then

$\begin{array}{l} \det (A) = 1 (2) (−1) + 2 (1) (0) + 3 (0) (0) - 0 (2) (3) - 0 (1) (1) + 1 (0) (2) \\ = −2 \end{array}$
Property 2 states that interchanging rows changes the sign. Given

$\begin{array}{l} \begin{array}{l} A = [\begin{matrix} −1 & 5 \\ 4 & −3 \end{matrix}], \det (A) = (−1) (−3) - (4) (5) = 3 - 20 = −17 \end{array} \\ B = [\begin{matrix} 4 & - 3 \\ - 1 & 5 \end{matrix}], \det (B) = (4) (5) - (−1) (−3) = 20 - 3 = 17 \end{array}$
Property 3 states that if two rows or two columns are identical, the determinant equals zero.

$\begin{array}{l} A = [\begin{matrix} 1 & 2 & 2 \\ 2 & 2 & 2 \\ −1 & 2 & 2 \end{matrix} | \begin{matrix} 1 \\ 2 \\ −1 \end{matrix} \begin{matrix} 2 \\ 2 \\ 2 \end{matrix}] \\ \det (A) = 1 (2) (2) + 2 (2) (−1) + 2 (2) (2) + 1 (2) (2) - 2 (2) (1) - 2 (2) (2) \\ = 4 - 4 + 8 + 4 - 4 - 8 = 0 \end{array}$
Property 4 states that if a row or column equals zero, the determinant equals zero. Thus,

$A = [\begin{matrix} 1 & 2 \\ 0 & 0 \end{matrix}], \det (A) = 1 (0) - 2 (0) = 0$

Property 5 states that the determinant of an inverse matrix $A^{- 1}$ is the reciprocal of the determinant $A .$ Thus,
$\begin{array}{l} A = [\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}], \det (A) = 1 (4) - 3 (2) = −2 \\ A^{- 1} = [\begin{matrix} - 2 & 1 \\ \frac{3}{2} & - \frac{1}{2} \end{matrix}], \det (A^{- 1}) = - 2 (- \frac{1}{2}) - (\frac{3}{2}) (1) = - \frac{1}{2} \end{array}$
Property 6 states that if any row or column of a matrix is multiplied by a constant, the determinant is multiplied by the same factor. Thus,

$\begin{array}{l} A = [\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}], \det (A) = 1 (4) - 2 (3) = −2 \\ B = [\begin{matrix} 2 (1) & 2 (2) \\ 3 & 4 \end{matrix}], \det (B) = 2 (4) - 3 (4) = −4 \end{array}$

How did it go?

Solution

Solution