Question

Solve for \(x\) \(\left|\begin{array}{rrr}x & 2 & 3 \\ 1 & x & 0 \\ 6 & 1 & -2\end{array}\right|=7\)

Short answer

x = 0 or x = -9

Step-by-step solution

- Write the Determinant Formula

For a 3x3 matrix, the determinant \(\text{det}(A)\) can be calculated using the formula:\[ \text{det}(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \]

- Identify Elements of the Matrix

For our given matrix \(\begin{pmatrix} x & 2 & 3 \ 1 & x & 0 \ 6 & 1 & -2 \end{pmatrix},\)Identify the elements:\[a_{11} = x, a_{12} = 2, a_{13} = 3 \]\[a_{21} = 1, a_{22} = x, a_{23} = 0 \]\[a_{31} = 6, a_{32} = 1, a_{33} = -2 \]

- Substitute Elements into the Determinant Formula

Substitute the elements into the determinant formula:\[ \text{det}(A) = x(x \times -2 - 0 \times 1) - 2(1 \times -2 - 0 \times 6) + 3(1 \times 1 - x \times 6) \]

- Simplify the Determinant Expression

Simplify the expression within the determinants:\[ \text{det}(A) = x(-2x) - 2(-2) + 3(1 - 6x) \] \[ \text{det}(A) = -2x^2 + 4 + 3 - 18x \] Combine like terms:\[ \text{det}(A) = -2x^2 - 18x + 7 \]

- Set Determinant Equal to Given Value

Set the determinant equal to the given value (7): \[ -2x^2 - 18x + 7 = 7 \]

- Solve the Quadratic Equation

Move all terms to one side of the equation: \[ -2x^2 - 18x + 7 - 7 = 0 \] \[ -2x^2 - 18x = 0 \]Factor out the common term: \[ -2x(x + 9) = 0 \]Use the zero-product property: \[ x = 0 \] or \[ x + 9 = 0 \] \[ x = -9 \]

Key concepts

3x3 Matrix

A 3x3 matrix is an arrangement of numbers in a square grid with three rows and three columns. Each element in the grid is denoted using two subscripts, the first indicating the row and the second indicating the column. For example, in matrix \(\begin{pmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \ \), the element in the second row and third column is 6 (denoted as \(a_{23}\)).
Recognizing the structure of a 3x3 matrix is crucial for solving determinants. Understanding the position of each element helps in accurately substituting values into the determinant formula. A matrix like \(\begin{pmatrix} x & 2 & 3 \ 1 & x & 0 \ 6 & 1 & -2 \ \) involves identifying elements such as \(a_{11} = x\), \(a_{12} = 2\), and so on.

Determinant

The determinant of a 3x3 matrix is a scalar value that provides significant information about the matrix, such as whether it is invertible. The determinant for a 3x3 matrix is calculated using a specific formula involving the elements of the matrix. For matrix A, it is given by:
\[ \text{det}(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \]
This method needs careful substitution and simplification. For example, in our solved problem, substituting matrix elements into the formula yields: \[ \text{det}(A) = x(x \times -2 - 0 \times 1) - 2(1 \times -2 - 0 \times 6) + 3(1 \times 1 - 6 \times x) \].

Zero-Product Property

The zero-product property is a fundamental principle in algebra which states that if the product of two factors is zero, at least one of the factors must be zero. This property is essential for solving quadratic equations and other polynomial equations.
In the given exercise, once we simplify the determinant expression to a quadratic form \( -2x(x + 9) = 0 \), we apply the zero-product property. This gives us two equations: \( x = 0\) and \(x + 9 = 0\). Solving these equations individually results in the solutions: \( x = 0\) and \(x = -9\).

Quadratic Equation

A quadratic equation is a second-order polynomial equation in a single variable x, with the general form \( ax^2 + bx + c = 0 \). In our problem, the simplified determinant equation \( -2x^2 - 18x + 7 = 7 \) transforms into a standard quadratic form after setting it to zero: \( -2x^2 - 18x = 0 \).
Solving quadratic equations often involves factoring, completing the square, or using the quadratic formula. In this case, we factored out the common term: \( -2x(x + 9) = 0 \). Utilizing the zero-product property, we solved for x, resulting in the solutions \( x = 0 \) and \(x = -9\).
Understanding these methods is crucial for solving quadratic equations efficiently.