Algebra 2 by Richard Wright
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Objectives:
SDA NAD Content Standards (2018): AII.4.1, AII.6.1
The building is the central building of Leuphana University of Lüneburg, Germany. A side is shaped like a triangle. If you wanted to find the area of that triangle, you could find the coordinates of the vertices by overlaying a grid. The area of that triangle could then easily be found using a determinant of a matrix.
This starts to apply matrices to various situations. The uses of matrices investigated today include solving systems of equations and finding the area of a triangle when the coordinates of its vertices are known. These are all done with a determinant.
The determinant of a matrix is a number associated with a square matrix. Each square matrix has one determinant, but different matrices can have the same determinant. It is symbolized by det A or |A|. When talking about matrices, the vertical lines mean determinant, not absolute value.
To find the determinant of a 2×2 matrix,
Evaluate \(\left|\begin{matrix}5&2\\-1&3\\\end{matrix}\right|\).
Solution
Multiply the down diagonal.
$$ \require{cancel} \left|\bcancel{\begin{matrix}5&2\\-1&3\\\end{matrix}}\right|=5(3) $$
Subtract the product of the up diagonal.
$$ \require{cancel} \left|\cancel{\begin{matrix}5&2\\-1&3\\\end{matrix}}\right|=5\left(3\right)-\mathbf{(-1)(2)} $$
Simplify.
$$ \left|\begin{matrix}5&2\\-1&3\\\end{matrix}\right|=5\left(3\right)-\left(-1\right)\left(2\right)=\mathbf{17} $$
To find the determinant of a 3×3 Matrix,
Evaluate \(\left|\begin{matrix}1&0&-2\\2&4&-3\\-1&3&5\\\end{matrix}\right| \).
Solution
Copy the first 2 columns behind the matrix.
$$ \left|\begin{matrix}1&0&-2\\2&4&-3\\-1&3&5\\\end{matrix}\right|\begin{matrix}1&0\\2&4\\-1&3\\\end{matrix} $$
Add the products of the down diagonals.
$$ \require{cancel} \left|\begin{matrix}\color{red}{\bcancel{\color{black}{1}}}&\color{blue}{\bcancel{\color{black}{0}}}&\color{green}{\bcancel{\color{black}{-2}}}\\2&\color{red}{\bcancel{\color{black}{4}}}&\color{blue}{\bcancel{\color{black}{-3}}}\\-1&3&\color{red}{\bcancel{\color{black}{5}}}\\\end{matrix}\right|\begin{matrix}1&0\\\color{green}{\bcancel{\color{black}{2}}}&4\\\color{blue}{\bcancel{\color{black}{-1}}}&\color{green}{\bcancel{\color{black}{3}}}\\\end{matrix}=\left(\color{red}{1·4·5}\right)+\left(\color{blue}{0·-3·-1}\right)+\left(\color{green}{-2·2·3}\right) $$
Subtract the products of the up diagonals.
$$ \require{cancel} \left|\begin{matrix}1&0&\color{red}{\cancel{\color{black}{-2}}}\\2&\color{red}{\cancel{\color{black}{4}}}&\color{blue}{\cancel{\color{black}{-3}}}\\\color{red}{\cancel{\color{black}{-1}}}&\color{blue}{\cancel{\color{black}{3}}}&\color{green}{\cancel{\color{black}{5}}}\\\end{matrix}\right|\begin{matrix}\color{blue}{\cancel{\color{black}{1}}}&\color{green}{\cancel{\color{black}{0}}}\\\color{green}{\cancel{\color{black}{2}}}&4\\-1&3\\\end{matrix}=\left(1·4·5\right)+\left(0·-3·-1\right)+\left(-2·2·3\right)-\left(\color{red}{-1·4·-2}\right)-\left(\color{blue}{3·-3·1}\right)-\left(\color{green}{5·2·0}\right) $$
Simplify.
= 9
The first application of a determinant is to find the area of a triangle. In geometry, when we were given the coordinates of the vertices of a triangle, we had to use the distance formula to find the length of the base. Then we had to use slopes and the distance formula to find the length of the height. Finally, we used \(A=\frac{1}{2}bh\) to find the area. Determinants make it much easier.
To find the area of a triangle from vertices, use
The ± indicates to make the final area positive.
Find the area of a triangle with vertices of (−1, 2), (4, 0), and (3, 5).
Solution
Substitute the vertices into the area determinant formula. The order of the points does not matter.
$$ Area=±\frac{1}{2}\left|\begin{matrix}-1&2&1\\4&0&1\\3&5&1\\\end{matrix}\right| $$
Find the determinant.
$$ \require{cancel} Area=\pm\frac{1}{2}\left|\begin{matrix} \color{red}{\bcancel{\color{black}{-1}}}&\color{blue}{\bcancel{\color{black}{2}}}&\color{green}{\bcancel{\color{orange}{\cancel{\color{black}{1}}}}} \\ 4&\color{red}{\bcancel{\color{orange}{\cancel{\color{black}{0}}}}}&\color{blue}{\bcancel{\color{purple}{\cancel{\color{black}{1}}}}} \\ \color{orange}{\cancel{\color{black}{3}}}&\color{purple}{\cancel{\color{black}{5}}}&\color{red}{\bcancel{\color{gray}{\cancel{\color{black}{1}}}}} \end{matrix}\right|\begin{matrix} \color{purple}{\cancel{\color{black}{-1}}}&\color{gray}{\cancel{\color{black}{2}}} \\ \color{green}{\bcancel{\color{gray}{\cancel{\color{black}{4}}}}}&0 \\ \color{blue}{\bcancel{\color{black}{3}}}&\color{green}{\bcancel{\color{black}{5}}} \end{matrix} $$
$$ =\pm\frac{1}{2}\left(\left(\color{red}{-1·0·1}\right)+\left(\color{blue}{2·1·3}\right)+\left(\color{green}{-1·4·5}\right)-\left(\color{orange}{3·0·1}\right)-\left(\color{purple}{5·1·-1}\right)-\left(\color{gray}{1·4·2}\right)\right) $$
$$ =\pm\frac{1}{2}\left(23\right) $$
$$ =\frac{\mathbf{23}}{\mathbf{2}}=\mathbf{11.5} $$
The area of the triangle is \(\frac{23}{2}\). If the determinant was negative, then make the final area positive. For this purpose, area is always positive.
The second application of determinants is Cramer's Rule. This is a way to solve a system of equations. The advantage of this is that it will solve for one variable without needing to solve for any of the other variables. You can just find y, for example without doing anything with x.
To solve a two-dimensional system using Cramer's Rule,
If \(\begin{matrix}ax+by=\\cx+dy=\\\end{matrix}\fbox{$\begin{matrix}e\\f\\\end{matrix}$}\), then \(x=\frac{\left|\begin{matrix}\fbox{$\begin{matrix}e\\f\\\end{matrix}$}&\begin{matrix}b\\d\\\end{matrix}\\\end{matrix}\right|}{\left|\begin{matrix}a&b\\c&d\\\end{matrix}\right|},y=\frac{\left|\begin{matrix}\begin{matrix}a\\b\\\end{matrix}&\fbox{$\begin{matrix}e\\f\\\end{matrix}$}\\\end{matrix}\right|}{\left|\begin{matrix}a&b\\c&d\\\end{matrix}\right|}\)
Notice that the numerator and denominator are the same except for the columns containing the coefficients of the variable you are solving for is replaced with the numbers from the constants column.
Solve using Cramer's Rule: \(\left\{\begin{alignat}{3} x&+&3y&=&6 \\ 3x&-&2y&=&7 \end{alignat}\right.\).
Solution
Fill out the formula for x. The determinant of the coefficient matrix goes in the denominator. The numerator is the same thing, but replace the x column (1st column) with the constants.
$$ x=\frac{\left|\begin{matrix}\fbox{$\begin{matrix}6\\7\\\end{matrix}$}&\begin{matrix}3\\-2\\\end{matrix}\\\end{matrix}\right|}{\left|\begin{matrix}1&3\\3&-2\\\end{matrix}\right|} $$
Evaluate the determinants.
$$ \require{cancel} x=\frac{\left|\color{red}{\bcancel{\color{blue}{\cancel{\color{black}{\begin{matrix} 6&3 \\ 7&-2 \end{matrix}}}}}}\right|}{\left|\color{orange}{\bcancel{\color{green}{\cancel{\color{black}{\begin{matrix}1&3\\3&-2\\\end{matrix}}}}}}\right|}=\frac{\color{red}{6\left(-2\right)}-\color{blue}{7\left(3\right)}}{1\left(-2\right)-3\left(3\right)}=\frac{-33}{-11}=\mathbf{3} $$
Repeat for the y. In the numerator, replace the y column (2nd column) with the constants. The denominator is the same for all variables, so it does not need to be calculated again.
$$ y=\frac{\left|\begin{matrix}\begin{matrix}1\\3\\\end{matrix}&\fbox{$\begin{matrix}6\\7\\\end{matrix}$}\\\end{matrix}\right|}{\left|\begin{matrix}1&3\\3&-2\\\end{matrix}\right|} $$
$$ y=\frac{\left|\color{red}{\bcancel{\color{blue}{\cancel{\color{black}{\begin{matrix} 1&6 \\ 3&7\end{matrix}}}}}}\right|}{\left|\begin{matrix}1&3\\3&-2\\\end{matrix}\right|}=\frac{\color{red}{1\left(7\right)}-\color{blue}{3\left(6\right)}}{-11}=\frac{-11}{-11}=\mathbf{1} $$
The solution is (3, 1).
Solve \(\left\{\begin{alignat}{4} 2x&-&y&+&z&=&4 \\ 2x&+&y&-&2z&=&0 \\ -x&-&2y&+&3z&=&3 \end{alignat}\right.\).
Solution
Fill out the formula for x. The determinant of the coefficient matrix goes in the denominator. The numerator is the same thing, but replace the x column (1st column) with the constants.
$$ x=\frac{\left|\begin{matrix}\fbox{$\begin{matrix}4\\0\\3\\\end{matrix}$}&\begin{matrix}-1\\1\\-2\\\end{matrix}&\begin{matrix}1\\-2\\3\\\end{matrix}\\\end{matrix}\right|}{\left|\begin{matrix}2&-1&1\\2&1&-2\\-1&-2&3\\\end{matrix}\right|} $$
Evaluate the determinants.
$$ x=\frac{\left|\begin{matrix} \color{red}{\bcancel{\color{black}{4}}}&\color{blue}{\bcancel{\color{black}{-1}}}&\color{green}{\bcancel{\color{orange}{\cancel{\color{black}{1}}}}} \\ 0&\color{red}{\bcancel{\color{orange}{\cancel{\color{black}{1}}}}}&\color{blue}{\bcancel{\color{purple}{\cancel{\color{black}{-2}}}}} \\ \color{orange}{\cancel{\color{black}{3}}}&\color{purple}{\cancel{\color{black}{-2}}}&\color{red}{\bcancel{\color{gray}{\cancel{\color{black}{3}}}}} \end{matrix}\right|\begin{matrix} \color{purple}{\cancel{\color{black}{4}}}&\color{gray}{\cancel{\color{black}{-1}}} \\ \color{green}{\bcancel{\color{gray}{\cancel{\color{black}{0}}}}}&1 \\ \color{blue}{\bcancel{\color{black}{3}}}&\color{green}{\bcancel{\color{black}{-2}}} \end{matrix}} {\left|\begin{matrix} \color{red}{\bcancel{\color{black}{2}}}&\color{blue}{\bcancel{\color{black}{-1}}}&\color{green}{\bcancel{\color{orange}{\cancel{\color{black}{1}}}}} \\ 2&\color{red}{\bcancel{\color{orange}{\cancel{\color{black}{1}}}}}&\color{blue}{\bcancel{\color{purple}{\cancel{\color{black}{-2}}}}} \\ \color{orange}{\cancel{\color{black}{-1}}}&\color{purple}{\cancel{\color{black}{-2}}}&\color{red}{\bcancel{\color{gray}{\cancel{\color{black}{3}}}}} \end{matrix}\right|\begin{matrix} \color{purple}{\cancel{\color{black}{2}}}&\color{gray}{\cancel{\color{black}{-1}}} \\ \color{green}{\bcancel{\color{gray}{\cancel{\color{black}{2}}}}}&1 \\ \color{blue}{\bcancel{\color{black}{-1}}}&\color{green}{\bcancel{\color{black}{-2}}} \end{matrix}} $$
$$ =\frac{\left(\color{red}{4·1·3}\right)+\left(\color{blue}{-1·-2·3}\right)+\left(\color{green}{1·0·-2}\right)-\left(\color{orange}{3·1·1}\right)-\left(\color{purple}{-2·-2·4}\right)-\left(\color{grey}{3·0·-1}\right)}{\left(\color{red}{2·1·3}\right)+\left(\color{blue}{-1·-2·-1}\right)+\left(\color{green}{1·2·-2}\right)-\left(\color{orange}{-1·1·1}\right)-\left(\color{purple}{-2·-2·2}\right)-\left(\color{grey}{3·2·-1}\right)} $$
$$ =\frac{-1}{-1} = \mathbf{1} $$
Repeat for the y. In the numerator, replace the y column (2nd column) with the constants. The denominator is the same for all variables, so it does not need to be calculated again.
$$ y=\frac{\left|\begin{matrix}\begin{matrix}2\\2\\-1\\\end{matrix}&\fbox{$\begin{matrix}4\\0\\3\\\end{matrix}$}&\begin{matrix}1\\-2\\3\\\end{matrix}\\\end{matrix}\right|}{\left|\begin{matrix}2&-1&1\\2&1&-2\\-1&-2&3\\\end{matrix}\right|} $$
Evaluate the determinants.
$$ y=\frac{\left|\begin{matrix} \color{red}{\bcancel{\color{black}{2}}}&\color{blue}{\bcancel{\color{black}{4}}}&\color{green}{\bcancel{\color{orange}{\cancel{\color{black}{1}}}}} \\ 2&\color{red}{\bcancel{\color{orange}{\cancel{\color{black}{0}}}}}&\color{blue}{\bcancel{\color{purple}{\cancel{\color{black}{-2}}}}} \\ \color{orange}{\cancel{\color{black}{-1}}}&\color{purple}{\cancel{\color{black}{3}}}&\color{red}{\bcancel{\color{gray}{\cancel{\color{black}{3}}}}} \end{matrix}\right|\begin{matrix} \color{purple}{\cancel{\color{black}{2}}}&\color{gray}{\cancel{\color{black}{4}}} \\ \color{green}{\bcancel{\color{gray}{\cancel{\color{black}{2}}}}}&0 \\ \color{blue}{\bcancel{\color{black}{-1}}}&\color{green}{\bcancel{\color{black}{3}}} \end{matrix}} {-1} $$
$$ =\frac{\left(\color{red}{2·0·3}\right)+\left(\color{blue}{4·-2·-1}\right)+\left(\color{green}{1·2·3}\right)-\left(\color{orange}{-1·0·1}\right)-\left(\color{purple}{3·-2·2}\right)-\left(\color{grey}{3·2·4}\right)}{-1} $$
$$ =\frac{2}{-1 }= -\mathbf{2} $$
Repeat for the z. In the numerator, replace the z column (3rd column) with the constants. The denominator is the same for all variables, so it does not need to be calculated again.
$$ z=\frac{\left|\begin{matrix}\begin{matrix}2\\2\\-1\\\end{matrix}&\begin{matrix}-1\\1\\-2\\\end{matrix}&\fbox{$\begin{matrix}4\\0\\3\\\end{matrix}$}\\\end{matrix}\right|}{\left|\begin{matrix}2&-1&1\\2&1&-2\\-1&-2&3\\\end{matrix}\right|} $$
Evaluate the determinants.
$$ z=\frac{\left|\begin{matrix} \color{red}{\bcancel{\color{black}{2}}}&\color{blue}{\bcancel{\color{black}{-1}}}&\color{green}{\bcancel{\color{orange}{\cancel{\color{black}{4}}}}} \\ 2&\color{red}{\bcancel{\color{orange}{\cancel{\color{black}{1}}}}}&\color{blue}{\bcancel{\color{purple}{\cancel{\color{black}{0}}}}} \\ \color{orange}{\cancel{\color{black}{-1}}}&\color{purple}{\cancel{\color{black}{-2}}}&\color{red}{\bcancel{\color{gray}{\cancel{\color{black}{3}}}}} \end{matrix}\right|\begin{matrix} \color{purple}{\cancel{\color{black}{2}}}&\color{gray}{\cancel{\color{black}{-1}}} \\ \color{green}{\bcancel{\color{gray}{\cancel{\color{black}{2}}}}}&1 \\ \color{blue}{\bcancel{\color{black}{-1}}}&\color{green}{\bcancel{\color{black}{-2}}} \end{matrix}} {-1} $$
$$ =\frac{\left(\color{red}{2·1·3}\right)+\left(\color{blue}{-1·0·-1}\right)+\left(\color{green}{4·2·-2}\right)-\left(\color{orange}{-1·1·4}\right)-\left(\color{purple}{-2·0·2}\right)-\left(\color{grey}{3·2·-1}\right)}{-1} $$
$$ =\frac{0}{-1}=\mathbf{0} $$
The solution is (1, −2, 0).
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