Calculation of determinants

Calculation of determinants of the 2nd and 3rd orders. Minors, cofactors.

A square table $A = (\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix})$ composed of four real or complex numbers is called a square matrix of the 2nd order.The determinant of the 2nd order corresponding to matrix $A$ (or simply the determinant of matrix $A$ ) is called the number $det A = | \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} | = a_{11} a_{22} - a_{12} a_{21} .$

Similarly, if $A = (\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix})$

- a square matrix of the 3rd order, then the corresponding determinant of the 3rd order is called the number

$det A = | \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix} | =$ $a_{11} a_{22} a_{33} + a_{21} a_{32} a_{13} + a_{12} a_{23} a_{31} - a_{13} a_{22} a_{31} - a_{12} a_{21} a_{33} - a_{23} a_{32} a_{11} .$

This formula is called the "triangle rule": one of the three terms in the right-hand side with a "+" sign is the product of the elements on the main diagonal of the matrix, each of the other two terms is the product of the elements lying on parallels to this diagonal and the element from the opposite corner of the matrix, and the terms with a "-" sign are constructed in the same way, but with respect to the second (side) diagonal.

Examples.

To calculate determinants of the second order:

1. $| \begin{matrix} - 1 & 4 \\ - 5 & 2 \end{matrix} |$

Solution.

$| \begin{matrix} - 1 & 4 \\ - 5 & 2 \end{matrix} | = - 1 \cdot 2 - (- 5) \cdot 4 = - 2 + 20 = 18.$

Answer: 18.

2. $| \begin{matrix} a + b & a - b \\ a - b & a + b \end{matrix} |$

Solution.

$| \begin{matrix} a + b & a - b \\ a - b & a + b \end{matrix} | = (a + b)^{2} - (a - b)^{2} = a^{2} + 2 a b + b^{2} - a^{2} + 2 a b - b^{2} = 4 a b .$

Answer: $4 a b .$

3. Solve the equation:

$| \begin{matrix} x & x + 1 \\ - 4 & x + 1 \end{matrix} | = 0.$

Solution.

$| \begin{matrix} x & x + 1 \\ - 4 & x + 1 \end{matrix} | = x (x + 1) - (- 4) (x + 1) = x^{2} + x + 4 x - 4 = x^{2} + 5 x + 4.$

The quadratic equation needs to be solved $x^{2} + 5 x + 4 = 0 :$

$D = 25 - 16 = 9$

$x_{1} = \frac{- 5 - 3}{2} = - 4; x_{2} = \frac{- 5 + 3}{2} = - 1.$

Answer: $x_{1} = - 4; x_{2} = - 1.$

4. $| \begin{matrix} 3 & 4 & - 5 \\ 8 & 7 & - 2 \\ 2 & - 1 & 8 \end{matrix} | .$

Solution.

$| \begin{matrix} 3 & 4 & - 5 \\ 8 & 7 & - 2 \\ 2 & - 1 & 8 \end{matrix} | = 3 \cdot 7 \cdot 8 + (- 5) \cdot 8 \cdot (- 1) + 4 \cdot (- 2) \cdot 2 - (- 5) \cdot 7 \cdot 2 -$

$- 4 \cdot 8 \cdot 8 - 3 \cdot (- 2) \cdot (- 1) = 168 + 40 - 16 + 70 - 256 - 6 = 0.$

Answer: $0.$

5. $| \begin{matrix} \sin α & \cos α & 1 \\ \sin β & \cos β & 1 \\ \sin γ & \cos γ & 1 \end{matrix} | .$

Solution.

$| \begin{matrix} \sin α & \cos α & 1 \\ \sin β & \cos β & 1 \\ \sin γ & \cos γ & 1 \end{matrix} | = \sin α \cos β + \sin β \cos γ + \cos α \sin γ -$

$- \cos β \sin γ - \sin α \cos γ - \cos α \sin β =$

$= \sin (α - β) + \sin (β - γ) + \sin (γ - α) .$

Answer: $\sin (α - β) + \sin (β - γ) + \sin (γ - α) .$

Properties of the determinant:

1) If the matrix is transposed, the determinant remains unchanged: $det A^{T} = det A .$

2) If all elements of a row (column) are multiplied by the same number, the determinant is multiplied by that number.

3) If two rows (columns) are interchanged, the determinant changes sign. In particular, if two rows (columns) are identical, the determinant equals zero.

4) If each element of a row (column) of the determinant is represented as the sum of two terms, the determinant equals the sum of two determinants, where all rows (columns) except the given one remain the same, and in the given row (column) of the first determinant are the first terms, and in the second determinant are the second terms.

5) If one row (column) is a linear combination of the other rows (columns), the determinant equals zero.

6) The determinant remains unchanged if a linear combination of its other rows (columns) is added to one of its rows (columns).

Examples:

1. Using the properties of the determinant, prove the following identity: $| \begin{matrix} a_{1} + b_{1} x & a_{1} - b_{1} x & c_{1} \\ a_{2} + b_{2} x & a_{2} - b_{2} x & c_{2} \\ a_{3} + b_{3} x & a_{3} - b_{3} x & c_{3} \end{matrix} | = - 2 x | \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} | .$

Proof.

$| \begin{matrix} a_{1} + b_{1} x & a_{1} - b_{1} x & c_{1} \\ a_{2} + b_{2} x & a_{2} - b_{2} x & c_{2} \\ a_{3} + b_{3} x & a_{3} - b_{3} x & c_{3} \end{matrix} | =$

$| \begin{matrix} a_{1} & a_{1} - b_{1} x & c_{1} \\ a_{2} & a_{2} - b_{2} x & c_{2} \\ a_{3} & a_{3} - b_{3} x & c_{3} \end{matrix} | +$ $| \begin{matrix} b_{1} x & a_{1} - b_{1} x & c_{1} \\ b_{2} x & a_{2} - b_{2} x & c_{2} \\ b_{3} x & a_{3} - b_{3} x & c_{3} \end{matrix} | =$

$= | \begin{matrix} a_{1} & a_{1} & c_{1} \\ a_{2} & a_{2} & c_{2} \\ a_{3} & a_{3} & c_{3} \end{matrix} | -$ $| \begin{matrix} a_{1} & b_{1} x & c_{1} \\ a_{2} & b_{2} x & c_{2} \\ a_{3} & b_{3} x & c_{3} \end{matrix} | +$ $| \begin{matrix} b_{1} x & a_{1} & c_{1} \\ b_{2} x & a_{2} & c_{2} \\ b_{3} x & a_{3} & c_{3} \end{matrix} | -$ $| \begin{matrix} b_{1} x & b_{1} x & c_{1} \\ b_{2} x & b_{2} x & c_{2} \\ b_{3} x & b_{3} x & c_{3} \end{matrix} | =$

$- | \begin{matrix} a_{1} & b_{1} x & c_{1} \\ a_{2} & b_{2} x & c_{2} \\ a_{3} & b_{3} x & c_{3} \end{matrix} | +$ $| \begin{matrix} b_{1} x & a_{1} & c_{1} \\ b_{2} x & a_{2} & c_{2} \\ b_{3} x & a_{3} & c_{3} \end{matrix} | =$ $- | \begin{matrix} a_{1} & b_{1} x & c_{1} \\ a_{2} & b_{2} x & c_{2} \\ a_{3} & b_{3} x & c_{3} \end{matrix} | -$ $| \begin{matrix} a_{1} & b_{1} x & c_{1} \\ a_{2} & b_{2} x & c_{2} \\ a_{3} & b_{3} x & c_{3} \end{matrix} | =$

$- 2 | \begin{matrix} a_{1} & b_{1} x & c_{1} \\ a_{2} & b_{2} x & c_{2} \\ a_{3} & b_{3} x & c_{3} \end{matrix} | =$ $- 2 x | \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} | .$

Q.E.D.

2. Check that the determinant $| \begin{matrix} 1 & 1 & 1 \\ x & y & z \\ x^{2} & y^{2} & z^{2} \end{matrix} |$ is divisible by $x - y$ , $y - z$ , and $z - x$ .

Checking.

1) Using the 6th property of determinants, we subtract the second column from the first, then we use the 2nd property and factor out the common factor from the determinant.

$| \begin{matrix} 1 & 1 & 1 \\ x & y & z \\ x^{2} & y^{2} & z^{2} \end{matrix} | =$ $| \begin{matrix} 0 & 1 & 1 \\ x - y & y & z \\ x^{2} - y^{2} & y^{2} & z^{2} \end{matrix} | =$ $| \begin{matrix} 0 & 1 & 1 \\ x - y & y & z \\ (x - y) (x + y) & y^{2} & z^{2} \end{matrix} | =$

$(x - y) | \begin{matrix} 0 & 1 & 1 \\ 1 & y & z \\ x + y & y^{2} & z^{2} \end{matrix} | .$

Thus, we have shown that the determinant is divisible bа $x - y .$ The other two cases can be proved in exactly the same way.

2) $| \begin{matrix} 1 & 1 & 1 \\ x & y & z \\ x^{2} & y^{2} & z^{2} \end{matrix} | =$ $| \begin{matrix} 1 & 0 & 1 \\ x & y - z & z \\ x^{2} & y^{2} - z^{2} & z^{2} \end{matrix} | =$ $| \begin{matrix} 1 & 0 & 1 \\ x & y - z & z \\ x^{2} & (y - z) (y + z) & z^{2} \end{matrix} | =$

$(y - z) | \begin{matrix} 1 & 0 & 1 \\ x & 1 & z \\ x^{2} & y + z & z^{2} \end{matrix} | .$

3) $| \begin{matrix} 1 & 1 & 1 \\ x & y & z \\ x^{2} & y^{2} & z^{2} \end{matrix} | =$ $| \begin{matrix} 1 & 1 & 0 \\ x & y & z - x \\ x^{2} & y^{2} & z^{2} - x^{2} \end{matrix} | =$ $| \begin{matrix} 1 & 1 & 0 \\ x & y & z - x \\ x^{2} & y^{2}) & (z - x) (z + x) \end{matrix} | =$

$(z - x) | \begin{matrix} 1 & 1 & 0 \\ x & y & 1 \\ x^{2} & y^{2} & z + x \end{matrix} | .$

The minor $M_{i j}$ of the element $a_{i j}$ of the $n$ -th order matrix $A$ is defined as the determinant of the $(n - 1)$ -th order obtained from the original determinant by removing the $i$ -th row and $j$ -th column.:

$M_{i j} =$ $| \begin{matrix} a_{11} & \dots & a_{1, j - 1} & a_{1, j + 1} & \dots & a_{1 n} \\ \dots & \dots & \dots & \dots & \dots & \dots \\ a_{i - 1, 1} & \dots & a_{i - 1, j - 1} & a_{i - 1, j + 1} & \dots & a_{i - 1, n} \\ a_{i + 1, 1} & \dots & a_{i + 1, j - 1} & a_{i + 1, j + 1} & \dots & a_{i + 1, n} \\ \dots & \dots & \dots & \dots & \dots & \dots \\ a_{n 1} & \dots & a_{n, j - 1} & a_{n, j + 1} & \dots & a_{n n} \end{matrix} |$

The algebraic complement $A_{i j}$ of the element $a_{i j}$ of the matrix $A$ of order $n$ is defined as the number equal to the product of the minor $M_{i j}$ by $(- 1)^{i + j}$ : $A_{i j} = (- 1)^{i + j} M_{i j}$ .

Determinants of order $n$ are computed using the method of lowering the order - according to the formula $det A = \sum_{j = 1}^{n} a_{i j} A_{i j}$ (with $i$ fixed) --- expansion along the $i$ -th row.

From this formula and the second property of determinants - $det A^{T} = det A,$ it follows that the expansion formula along the $j$ -th column is also valid: $det A = \sum_{i = 1}^{n} a_{i j} A_{i j}$ (with $j$ fixed).

The method of reducing to triangular form consists of transforming the determinant such that all elements lying on one side of one of its diagonals become zero. In this case, the determinant equals the product of the diagonal elements.

Examples.

Calculate determinants, using the appropriate expansion along a row or column.

3. $| \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 0 \\ 2 & 0 & 3 \end{matrix} | .$

Solution.

We will calculate this determinant using expansion along the second row:

$| \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 0 \\ 2 & 0 & 3 \end{matrix} | =$ $0 \cdot (- 1)^{2 + 1} | \begin{matrix} 0 & 2 \\ 0 & 3 \end{matrix} | +$ $2 \cdot (- 1)^{2 + 2} | \begin{matrix} 1 & 2 \\ 2 & 3 \end{matrix} | +$ $0 \cdot (- 1)^{2 + 3} | \begin{matrix} 1 & 0 \\ 2 & 0 \end{matrix} |$ $= 2 (3 - 4) = - 2.$

4. $| \begin{matrix} 2 & - 3 & 4 & 1 \\ 4 & - 2 & 3 & 2 \\ a & b & c & d \\ 3 & - 1 & 4 & 3 \end{matrix} | .$

Solution.

We will calculate this determinant using expansion along the third row:

$| \begin{matrix} 2 & - 3 & 4 & 1 \\ 4 & - 2 & 3 & 2 \\ a & b & c & d \\ 3 & - 1 & 4 & 3 \end{matrix} | =$ $a \cdot (- 1)^{3 + 1}$ $| \begin{matrix} - 3 & 4 & 1 \\ - 2 & 3 & 2 \\ - 1 & 4 & 3 \end{matrix} | +$ $b \cdot (- 1)^{3 + 2} | \begin{matrix} 2 & 4 & 1 \\ 4 & 3 & 2 \\ 3 & 4 & 3 \end{matrix} | +$ $+ c \cdot (- 1)^{3 + 3} | \begin{matrix} 2 & - 3 & 1 \\ 4 & - 2 & 2 \\ 3 & - 1 & 3 \end{matrix} | +$ $d \cdot (- 1)^{3 + 4} | \begin{matrix} 2 & - 3 & 4 \\ 4 & - 2 & 3 \\ 3 & - 1 & 4 \end{matrix} | =$

$= a (- 27 - 8 - 8 + 3 + 24 + 24) - b (18 + 16 + 24 - 9 - 16 - 48) +$

$+ c (- 12 - 4 - 18 + 6 + 4 + 36) - d (- 16 - 16 - 27 + 24 + 6 + 48) =$

$= 8 a + 15 b + 12 c - 19 d .$

Answer: $8 a + 15 b + 12 c - 19 d .$

5. Calculate the determinant: $| \begin{matrix} 2 & 1 & 1 & 1 & 1 \\ 1 & 3 & 1 & 1 & 1 \\ 1 & 1 & 4 & 1 & 1 \\ 1 & 1 & 1 & 5 & 1 \\ 1 & 1 & 1 & 1 & 6 \end{matrix} | .$

Solution.

Let's compute this determinant by transforming it into triangular form:

$| \begin{matrix} 2 & 1 & 1 & 1 & 1 \\ 1 & 3 & 1 & 1 & 1 \\ 1 & 1 & 4 & 1 & 1 \\ 1 & 1 & 1 & 5 & 1 \\ 1 & 1 & 1 & 1 & 6 \end{matrix} | =$

Subtract the fifth row from each of the first four rows

$= | \begin{matrix} 1 & 0 & 0 & 0 & - 5 \\ 0 & 2 & 0 & 0 & - 5 \\ 0 & 0 & 3 & 0 & - 5 \\ 0 & 0 & 0 & 4 & - 5 \\ 1 & 1 & 1 & 1 & 6 \end{matrix} | =$

Subtract the first row from the fifth row, then multiply the fifth row by two

$= | \begin{matrix} 1 & 0 & 0 & 0 & - 5 \\ 0 & 2 & 0 & 0 & - 5 \\ 0 & 0 & 3 & 0 & - 5 \\ 0 & 0 & 0 & 4 & - 5 \\ 0 & 1 & 1 & 1 & 11 \end{matrix} | =$ $\frac{1}{2} | \begin{matrix} 1 & 0 & 0 & 0 & - 5 \\ 0 & 2 & 0 & 0 & - 5 \\ 0 & 0 & 3 & 0 & - 5 \\ 0 & 0 & 0 & 4 & - 5 \\ 0 & 2 & 2 & 2 & 22 \end{matrix} | =$

Next, subtract the second row from the fifth row, then multiply the fifth row by $\frac{3}{2} :$

$= \frac{1}{2} | \begin{matrix} 1 & 0 & 0 & 0 & - 5 \\ 0 & 2 & 0 & 0 & - 5 \\ 0 & 0 & 3 & 0 & - 5 \\ 0 & 0 & 0 & 4 & - 5 \\ 0 & 0 & 2 & 2 & 27 \end{matrix} | =$ $\frac{1}{2} \frac{2}{3} | \begin{matrix} 1 & 0 & 0 & 0 & - 5 \\ 0 & 2 & 0 & 0 & - 5 \\ 0 & 0 & 3 & 0 & - 5 \\ 0 & 0 & 0 & 4 & - 5 \\ 0 & 0 & 3 & 3 & 40, 5 \end{matrix} | =$

Now, subtract the third row from the fifth row, then multiply the fifth row by $\frac{4}{3} :$

$= \frac{1}{3} | \begin{matrix} 1 & 0 & 0 & 0 & - 5 \\ 0 & 2 & 0 & 0 & - 5 \\ 0 & 0 & 3 & 0 & - 5 \\ 0 & 0 & 0 & 4 & - 5 \\ 0 & 0 & 0 & 3 & 45, 5 \end{matrix} | =$ $\frac{1}{3} \frac{3}{4} | \begin{matrix} 1 & 0 & 0 & 0 & - 5 \\ 0 & 2 & 0 & 0 & - 5 \\ 0 & 0 & 3 & 0 & - 5 \\ 0 & 0 & 0 & 4 & - 5 \\ 0 & 0 & 0 & 4 & \frac{182}{3} \end{matrix} | =$

Subtract the fourth row from the fifth row and multiply the diagonal elements, we get the answer:

$= \frac{1}{4} | \begin{matrix} 1 & 0 & 0 & 0 & - 5 \\ 0 & 2 & 0 & 0 & - 5 \\ 0 & 0 & 3 & 0 & - 5 \\ 0 & 0 & 0 & 4 & - 5 \\ 0 & 0 & 0 & 0 & \frac{197}{3} \end{matrix} | =$

$= \frac{1}{4} \cdot 2 \cdot 3 \cdot 4 \cdot \frac{197}{3} = 394.$

Answer: $394.$

Homework:

Calculate determinants of the second order:

1. $| \begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix} | .$

Answer: $1.$

2. $| \begin{matrix} \frac{1 - t^{2}}{1 + t^{2}} & \frac{2 t}{1 + t^{2}} \\ - \frac{2 t}{1 + t^{2}} & \frac{1 - t^{2}}{1 + t^{2}} \end{matrix} | .$

Answer: $1.$

Solve the equation:

3. $| \begin{matrix} \cos 8 x & - \sin 5 x \\ \sin 8 x & \cos 5 x \end{matrix} | = 0.$

Answer: $x = \frac{π}{6} + \frac{π k}{3},$ $k \in Z .$

Calculate determinants of the third order:

4. $| \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix} | .$

Answer: $0.$

5. $| \begin{matrix} α^{2} + 1 & α β & α γ \\ α β & β^{2} + 1 & β γ \\ α γ & β γ & γ^{2} + 1 \end{matrix} | .$

Answer: $α^{2} + β^{2} + γ^{2} + 1.$

6. Using the properties of the determinant, prove the following identity:

$| \begin{matrix} a_{1} + b_{1} x & a_{1} x + b_{1} & c_{1} \\ a_{2} + b_{2} x & a_{2} x + b_{2} & c_{2} \\ a_{3} + b_{3} x & a_{3} x + b_{3} & c_{3} \end{matrix} | =$ $(1 - x^{2}) | \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} | .$

7. Check that the determinant $| \begin{matrix} x & y & x + y \\ y & x + y & x \\ x + y & x & y \end{matrix} |$ is divisible by $x + y$ and $x^{2} - x y + y^{2}$ .

Compute the determinants using the appropriate expansion along a row or column.

8. $| \begin{matrix} - 1 & 5 & 2 \\ 0 & 7 & 0 \\ 1 & 2 & 0 \end{matrix} | .$

Answer: $- 14.$

9. $| \begin{matrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{matrix} | .$

Answer: $4.$

10. $| \begin{matrix} 5 & a & 2 & - 1 \\ 4 & b & 4 & - 3 \\ 2 & c & 3 & - 2 \\ 4 & d & 5 & - 4 \end{matrix} | .$

Answer: $2 a - 8 b + c + 5 d .$

11. Compute the determinant: $| \begin{matrix} 5 & 6 & 0 & 0 & 0 \\ 1 & 5 & 6 & 0 & 0 \\ 0 & 1 & 5 & 6 & 0 \\ 0 & 0 & 1 & 5 & 6 \\ 0 & 0 & 0 & 1 & 5 \end{matrix} | .$

Answer: $665.$

Tags: cofactors, determinant, linear algebra, matrix, calculation of determinants, minors

Tables and Formulas

Derivative Table.

Table of Derivatives of Composite Functions.

Table of Higher Order Derivatives.

Table of Integrals

Taylor Series

Comparison of functions O(f) and o(f).

Calculation of determinants

Tables and Formulas

Popular posts

Reducing the quadratic form to canonical form.

Calculation of determinants

Logical symbolism. Necessary and sufficient conditions.