If all elements of a third order determinant are equal to 1 or -1, then the value of the determinant is:
an even number but not necessarily 0
Understanding Third Order Determinants with Elements 1 or -1
Let's analyze the properties of a third order (3x3) determinant where every element can only be either 1 or -1. We want to determine what values such a determinant can possibly take.
Calculating a 3x3 Determinant
A general 3x3 determinant is given by:
\[ \Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} \]The value of this determinant can be calculated using the cofactor expansion method (along the first row, for example) or using Sarrus' rule.
Using cofactor expansion along the first row:
\[ \Delta = a_{11} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12} \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13} \begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix} \] \[ \Delta = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \]Using Sarrus' Rule:
\[ \Delta = (a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32}) - (a_{13}a_{22}a_{31} + a_{11}a_{23}a_{32} + a_{12}a_{21}a_{33}) \]In this problem, each element \(a_{ij}\) is either 1 or -1. Let's examine the terms in the expansion formula.
Analyzing the Terms in the Determinant Expansion
Look at the terms in the Sarrus' rule expansion:
- \(a_{11}a_{22}a_{33}\)
- \(a_{12}a_{23}a_{31}\)
- \(a_{13}a_{21}a_{32}\)
- \(a_{13}a_{22}a_{31}\)
- \(a_{11}a_{23}a_{32}\)
- \(a_{12}a_{21}a_{33}\)
Each term is a product of three elements, and each element is either 1 or -1. The product of three numbers, each being 1 or -1, will also be either 1 or -1.
- \(1 \times 1 \times 1 = 1\)
- \(1 \times 1 \times -1 = -1\)
- \(1 \times -1 \times 1 = -1\)
- \(-1 \times 1 \times 1 = -1\)
- \(1 \times -1 \times -1 = 1\)
- \(-1 \times 1 \times -1 = 1\)
- \(-1 \times -1 \times 1 = 1\)
- \(-1 \times -1 \times -1 = -1\)
So, each of the six product terms in the determinant expansion (\(a_{11}a_{22}a_{33}\), \(a_{12}a_{23}a_{31}\), etc.) is either 1 or -1.
The determinant is the sum of three of these terms and the difference of the other three terms. Let's write the determinant value as the sum of six terms, where each term is either 1 or -1 (after considering the minus signs in the formula):
\[ \Delta = T_1 + T_2 + T_3 + T_4 + T_5 + T_6 \]where each \(T_i \in \{1, -1\}\). For example, \(T_1 = a_{11}a_{22}a_{33}\), \(T_4 = -(a_{13}a_{22}a_{31})\). Since \(a_{13}a_{22}a_{31}\) is 1 or -1, \(T_4\) will also be -1 or 1 respectively.
So, the determinant is the sum of six numbers, each of which is either 1 or -1.
Let \(p\) be the number of terms that are +1, and \(q\) be the number of terms that are -1 among these six terms. The total number of terms is 6, so \(p + q = 6\). The value of the determinant is \(p \times (+1) + q \times (-1) = p - q\).
Since \(p + q = 6\), we have \(q = 6 - p\). Substituting this into the expression for the determinant:
\[ \Delta = p - (6 - p) = p - 6 + p = 2p - 6 \]Since \(p\) is the number of terms equal to +1, \(p\) can be any integer from 0 to 6 (\(p \in \{0, 1, 2, 3, 4, 5, 6\}\)).
Let's see the possible values for \(2p - 6\):
- If \(p=0\), \(\Delta = 2(0) - 6 = -6\)
- If \(p=1\), \(\Delta = 2(1) - 6 = -4\)
- If \(p=2\), \(\Delta = 2(2) - 6 = -2\)
- If \(p=3\), \(\Delta = 2(3) - 6 = 0\)
- If \(p=4\), \(\Delta = 2(4) - 6 = 2\)
- If \(p=5\), \(\Delta = 2(5) - 6 = 4\)
- If \(p=6\), \(\Delta = 2(6) - 6 = 6\)
The possible values for the determinant are -6, -4, -2, 0, 2, 4, 6. All these values are even numbers.
Thus, the value of the determinant must be an even number.
Examples of Determinants with Elements 1 or -1
Let's look at some examples to confirm that 0 is a possible value and that non-zero even numbers are also possible values.
Example 1: A determinant with value 0
Consider the determinant where all elements are 1:
\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} \]Since the first two rows are identical, the determinant value is 0. 0 is an even number.
Example 2: A determinant with a non-zero even value
Consider the determinant:
\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} \]Calculating the value:
\[ \Delta = 1((-1)(-1) - 1 \times 1) - 1(1(-1) - 1 \times 1) + 1(1 \times 1 - (-1) \times 1) \] \[ \Delta = 1(1 - 1) - 1(-1 - 1) + 1(1 - (-1)) \] \[ \Delta = 1(0) - 1(-2) + 1(2) \] \[ \Delta = 0 + 2 + 2 = 4 \]The value is 4, which is an even number and is not 0.
From the analysis and examples, we see that the determinant value must be an even number, and it can be 0 or a non-zero even number.
Conclusion on Determinant Value
Based on the structure of the 3x3 determinant expansion and the fact that each element is 1 or -1, each of the six terms in the expansion is either 1 or -1. The determinant value is the sum/difference of these six terms, which simplifies to the sum of six values, each being 1 or -1. The sum of an even number of integers (in this case, six integers), each being 1 or -1, is always an even number. This is because the sum can be written as \(p - q\), where \(p+q=6\), leading to \(p - q = 2p - 6\), which is always even.
Revision Table: Third Order Determinant
| Concept | Description | Key Property (for elements ∈ {1, -1}) |
|---|---|---|
| Third Order Determinant | A 3x3 matrix determinant. | Calculated as sum/difference of 6 terms. |
| Determinant Term | Product of 3 elements in expansion. | Value is 1 or -1. |
| Determinant Value | Sum/Difference of 6 terms. | Sum of 6 values, each 1 or -1. Result is \(2p-6\). |
| Parity of Value | Even or Odd. | Always Even. |
Additional Information: Determinant Properties and Applications
Determinants are fundamental in linear algebra and have various applications. For a matrix with integer entries, the determinant is always an integer.
- Invertibility: A square matrix is invertible if and only if its determinant is non-zero.
- Linear Systems: Determinants are used in Cramer's rule to solve systems of linear equations.
- Geometric Interpretation: The absolute value of a determinant of a 3x3 matrix represents the volume of the parallelepiped formed by the column (or row) vectors.
- Eigenvalues: Determinants are used in finding eigenvalues of a matrix by solving the characteristic equation \(det(A - \lambda I) = 0\).
This specific problem highlights how constraints on matrix elements affect the possible values of the determinant, revealing properties like parity (evenness or oddness).
Important Questions from Determinants
If \(A=\left[\begin{array}{rrr} 2 & -1 & 0 \\ -1 & 3 & 0 \\ 1 & 0 & 1 \end{array}\right]\), then what is the value of det[adj(adjA)] ?
If A, B and C are square matrices of order 3 and det(BC) = 2 det(A), then what is the value of det(2A-1BC)?
If \(A=\left[\begin{array}{rrr} 0 & 3 & 4 \\ -3 & 0 & 5 \\ -4 & -5 & 0 \end{array}\right]\), then which one of the following statements is correct?
If \(\left|\begin{array}{ccc} x^2+3 x & x-1 & x+3 \\ x+1 & -2 x & x-4 \\ x-3 & x+4 & 3 x \end{array}\right|\) = ax4 + bx3 + cx2 + dx + e, then what is the value of e?"
If a, b, c are in AP, then what is \(\left|\begin{array}{lll} x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4 \\ x+a & x+b & x+c \end{array}\right|\) equal to ?
