The determinant of a matrix. Sarrus' rule
To every square matrix we can associate a single number that summarises very valuable information about it: whether it has an inverse, whether its rows are "independent", and what area or volume its vectors span. That number is the determinant. In this section you will learn how to calculate it for orders 2 and 3; in the following ones you will see its properties and how to calculate it for higher orders.
Determinant of order 2
The product of the leading diagonal is added; the product of the secondary diagonal is subtracted.
In graphic design software, a scaling transformation is defined by a matrix of order 2. For the image not to be completely distorted, it is necessary to calculate the…
In graphic design software, a scaling transformation is defined by a matrix of order 2. For the image not to be completely distorted, it is necessary to calculate the determinant of that matrix. Calculate the value of the determinant of the following matrix $A$:
$$A = \begin{pmatrix} 5 & -3 \\ 2 & 4 \end{pmatrix}$$For a temperature control system to be stable, the determinant of its coefficient matrix must be non-zero. Given the matrix $B$, determine for which value of the…
For a temperature control system to be stable, the determinant of its coefficient matrix must be non-zero. Given the matrix $B$, determine for which value of the parameter $x$ the determinant of the matrix is equal to $10$:
$$B = \begin{pmatrix} x & 2 \\ 3 & x-1 \end{pmatrix}$$Determinant of order 3: Sarrus' rule
With a $3\times3$ matrix there is no longer a single diagonal of each type: you have to complete three products "to the right" (positive) and three "to the left" (negative), mentally repeating the first two rows or columns after the matrix.
Calculate, applying Sarrus' rule, the value of the determinant of the matrix $M$. Use the visual diagonal diagram to organise your products: $$|M| = \begin{vmatrix} 1 &…
Calculate, applying Sarrus' rule, the value of the determinant of the matrix $M$. Use the visual diagonal diagram to organise your products:
$$|M| = \begin{vmatrix} 1 & 2 & 0 \\ 3 & -1 & 4 \\ 5 & 0 & 2 \end{vmatrix}$$Solve the following equation by setting the determinant of the matrix $A$ equal to zero, applying Sarrus' rule to obtain the resulting polynomial: $$A = \begin{pmatrix}…
Solve the following equation by setting the determinant of the matrix $A$ equal to zero, applying Sarrus' rule to obtain the resulting polynomial:
$$A = \begin{pmatrix} x & 1 & 0 \\ 1 & x & 1 \\ 1 & 1 & x \end{pmatrix}$$Find all real solutions to the equation $|A| = 0$.
Singular matrix and regular matrix
Notice that when the determinant is $0$, the parallelogram (or the space) spanned by the rows is "flattened" into a line or a plane; it ends up with no area or no volume. That is why a singular matrix has no inverse: there is no way to "undo" a transformation that collapses space.
A square matrix is said to be singular if its determinant is zero (which implies it has no inverse). Check whether the following matrix $C$ is singular or non-singular…
A square matrix is said to be singular if its determinant is zero (which implies it has no inverse). Check whether the following matrix $C$ is singular or non-singular by calculating its determinant:
$$C = \begin{pmatrix} 2 & 1 & 3 \\ 0 & 4 & -1 \\ 2 & 5 & 2 \end{pmatrix}$$Find the value of $k$ so that the following equality holds, evaluating the determinant on the left using Sarrus' rule: $$\begin{vmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1…
Find the value of $k$ so that the following equality holds, evaluating the determinant on the left using Sarrus' rule:
$$\begin{vmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{vmatrix} = 0$$Given the matrix $M$ which depends on the real parameter $k$: $$M = \begin{pmatrix} 1 & k & 1 \\ k & 1 & 1 \\ 2 & 1 & k \end{pmatrix}$$ Determine, using Sarrus' rule,…
Given the matrix $M$ which depends on the real parameter $k$:
$$M = \begin{pmatrix} 1 & k & 1 \\ k & 1 & 1 \\ 2 & 1 & k \end{pmatrix}$$Determine, using Sarrus' rule, for which values of $k$ the determinant of the matrix is zero. Justify whether for $k = 0$ the matrix has an inverse based on the result obtained.
Area and volume with determinants
The same idea extends to three dimensions: the volume of a parallelepiped defined by three vectors $\vec{u}, \vec{v}, \vec{w}$ starting from the same vertex is the absolute value of the determinant of the matrix formed by their components as rows. If that determinant is $0$, the three vectors are coplanar (they do not "fill" a volume): they lie in the same plane.
The area of a triangle in the plane, whose vertices are $P_1(x_1, y_1)$, $P_2(x_2, y_2)$ and $P_3(x_3, y_3)$, can be calculated using the formula: $$Área = \frac{1}{2}…
The area of a triangle in the plane, whose vertices are $P_1(x_1, y_1)$, $P_2(x_2, y_2)$ and $P_3(x_3, y_3)$, can be calculated using the formula:
$$Área = \frac{1}{2} \cdot | \det(A) |$$where $A$ is the matrix formed by the coordinates of the points and a column of ones. Calculate the area of the triangle in the figure whose vertices are $A(1, 2)$, $B(3, 4)$ and $C(0, 5)$:
$$A = \begin{pmatrix} 1 & 2 & 1 \\ 3 & 4 & 1 \\ 0 & 5 & 1 \end{pmatrix}$$In a strategy video game, the area of a triangular region is essential for calculating the range of a defence tower. If the vertices of the triangle on the coordinate…
In a strategy video game, the area of a triangular region is essential for calculating the range of a defence tower. If the vertices of the triangle on the coordinate map are $P(2, 1)$, $Q(6, 2)$ and $R(3, 5)$, calculate the area of this region using the value of the determinant of the matrix formed by their coordinates:
$$Área = \frac{1}{2} \cdot \left| \det \begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{pmatrix} \right|$$It is known that the volume of a parallelepiped defined by three vectors $\vec{u}, \vec{v}$ and $\vec{w}$ originating from the same vertex is given by the absolute value…
It is known that the volume of a parallelepiped defined by three vectors $\vec{u}, \vec{v}$ and $\vec{w}$ originating from the same vertex is given by the absolute value of the determinant of the matrix formed by their components.
If the vectors are $\vec{u} = (k, 1, 1)$, $\vec{v} = (1, k, 1)$ and $\vec{w} = (1, 1, k)$, calculate using Sarrus' rule the value of $k$ so that the parallelepiped "flattens" (its volume is zero) and justify the connection between this result and the linear dependence of the vectors.