Question

Solve for \(x\) \(\left|\begin{array}{lll}x & 1 & 2 \\ 1 & x & 3 \\ 0 & 1 & 2\end{array}\right|=-4 x\)

Short answer

The solutions for \(x\) are \(0\) and \(-\frac{1}{2}\).

Step-by-step solution

- Understand the Problem

The problem requires solving for the variable \(x\) given that the determinant of a 3x3 matrix is equal to \(-4x\).

- Write the Determinant

Write down the determinant of the given matrix: \(\left|\begin{array}{lll} x & 1 & 2 \ 1 & x & 3 \ 0 & 1 & 2 \end{array}\right|\)

- Calculate the Determinant Using Expansion

Expand the determinant along the first row: \(x \left|\begin{array}{ll} x & 3 \ 1 & 2 \end{array}\right| - 1 \left|\begin{array}{ll} 1 & 3 \ 0 & 2 \end{array}\right| + 2 \left|\begin{array}{ll} 1 & x \ 0 & 1 \end{array}\right|\)

- Simplify the Determinants of 2x2 Matrices

Simplify each 2x2 determinant: \(x(x \cdot 2 - 1 \cdot 3) - 1(1 \cdot 2 - 3 \cdot 0) + 2(1 \cdot 1 - 0 \cdot x)\)

- Perform Multiplications

Calculate the values inside the parentheses: \(x(2x - 3) - 1(2) + 2(1)\) becomes \(2x^2 - 3x - 2 + 2\).

- Combine Like Terms

Combine the terms to get the simplified determinant: \(2x^2 - 3x\).

- Set the Determinant to \(-4x\)

Set the simplified determinant equal to \(-4x\) as given in the problem: \(2x^2 - 3x = -4x\).

- Rearrange the Equation

Move all terms to one side of the equation: \(2x^2 - 3x + 4x = 0\) simplifies to \(2x^2 + x = 0\).

- Factor the Equation

Factor out an \(x\): \(x(2x + 1) = 0\).

- Solve for x

Set each factor to zero and solve: \(x = 0\) or \(2x + 1 = 0\) which gives \(x = 0\) or \(x = -\frac{1}{2}\).

Key concepts

3x3 matrix determinants

A determinant of a matrix is a special number that can be calculated from its elements. For a 3x3 matrix, the process involves calculating the determinant using minor matrices. This is a little more complex compared to a 2x2 matrix.
To find the determinant of a 3x3 matrix like \(\begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix}\), you use an expansion method, also known as cofactor expansion. This requires choosing any row or column to expand along. Typically, we often use the first row for simplicity.
So, for our matrix \(\begin{bmatrix} x & 1 & 2 \ 1 & x & 3 \ 0 & 1 & 2 \end{bmatrix}\), we expand along the first row: \(\text{Det} = x(\text{Det of } \begin{bmatrix} x & 3 \ 1 & 2 \end{bmatrix}) - 1(\text{Det of } \begin{bmatrix} 1 & 3 \ 0 & 2 \end{bmatrix}) + 2(\text{Det of } \begin{bmatrix} 1 & x \ 0 & 1 \end{bmatrix})\).
Each small \(\begin{bmatrix}2x2\text{ matrix}\end{bmatrix}\) is what we call a 'minor' of the original matrix. To solve, we find determinants of these 2x2 minors which will be explained in a later section.

Factoring quadratic equations

Once we simplify the determinant, we often end up with a quadratic equation like \(2x^2 + x = 0\). Solving a quadratic equation involves factoring it into simpler expressions that can be set to zero individually.
A quadratic equation typically has the form \(ax^2 + bx + c = 0\). In our problem, the equation simplifies to \(2x^2 + x = 0\). To factor it, we look for common factors. Here, \(x\) is a common factor in both terms. So, we factor \(x\) out: \(x(2x + 1) = 0\).
Setting each factor to zero gives us the possible solutions for \(x\). So, \(x = 0\) or \(2x + 1 = 0\). Solving \(2x + 1 = 0\) further, we get \(2x = -1\), thus \(x = -\frac{1}{2}\).
Factoring is a useful method in solving quadratic equations as it breaks down a complex equation into simpler, manageable parts.

Expansion of determinants

The determinant expansion method, also known as the cofactor expansion, is a method of breaking down the determinant calculation of a larger matrix into smaller, simpler determinants of \(\begin{bmatrix}2 \times 2 \end{bmatrix}\) matrices.
When expanding along the first row of our 3x3 matrix, each element in the row is multiplied by the determinant of the 2x2 minor matrix that is left after removing the row and column of that element. In our problem, for the matrix \(\begin{bmatrix} x & 1 & 2 \ 1 & x & 3 \ 0 & 1 & 2 \end{bmatrix}\), expanding along the first row gives us three 2x2 minors. We calculate the determinants of these minors as follows:

• For \(x(2x - 3)\), the minor matrix is \(\begin{bmatrix} x & 3 \ 1 & 2 \end{bmatrix} = x \times 2 - 1 \times 3 = 2x - 3\).
• For \(1(1 \times 2 - 0 \times 3) = 1 \times 2 = 2\), the minor matrix is \(\begin{bmatrix} 1 & 3 \ 0 & 2 \end{bmatrix} = 1 \times 2 - 3 \times 0\).
• For \(2(1 \times 1 - 0 \times x = 2)\), the minor matrix is \(\begin{bmatrix} 1 & x \ 0 & 1 \end{bmatrix} = 1 \times 1 - 0 \times x= 1 \times 1 = 1\).

Then we combine these results following their respective coefficients from the expansion: \(x(2x - 3) - 1(2) + 2(1) = 2x^2 - 3x - 2 + 2 \). The simplified determinant is \(2x^2 - 3x\), which we set to \(-4x\) and solve as shown in the above factoring section.