Solving the Determinant Equation to Find Values of x

We are given a determinant equation involving a 3x3 matrix, and we need to find the values of \(x\) that satisfy this equation. The equation is:

\(\left| {\;\begin{array}{} {x + 2}&2&2\\ 2&{x + 2}&2\\ 2&2&{x + 2} \end{array}} \right| = 0\)

To solve this determinant equation, we can evaluate the determinant. There are several ways to evaluate a 3x3 determinant. A helpful approach for this specific determinant is to use row or column operations to simplify it before expansion.

Simplifying the Matrix Determinant

Let the given determinant be \(D\). We can perform a column operation \(C_1 \to C_1 + C_2 + C_3\). This means we replace the first column with the sum of the first, second, and third columns. This operation does not change the value of the determinant.

\(D = \left| {\;\begin{array}{} {x + 2 + 2 + 2}&2&2\\ {2 + x + 2 + 2}&{x + 2}&2\\ {2 + 2 + x + 2}&2&{x + 2} \end{array}} \right| = \left| {\;\begin{array}{} {x + 6}&2&2\\ {x + 6}&{x + 2}&2\\ {x + 6}&2&{x + 2} \end{array}} \right|\)

Now, we can factor out the common term \((x + 6)\) from the first column.

\(D = (x + 6)\left| {\;\begin{array}{} 1&2&2\\ 1&{x + 2}&2\\ 1&2&{x + 2} \end{array}} \right|\)

So the determinant equation becomes \((x + 6)\left| {\;\begin{array}{} 1&2&2\\ 1&{x + 2}&2\\ 1&2&{x + 2} \end{array}} \right| = 0\).

This equation is satisfied if either \((x + 6) = 0\) or the remaining 3x3 determinant is equal to 0. From \((x + 6) = 0\), we get one value of \(x\): \(x = -6\).

Evaluating the Remaining Determinant

Now we need to evaluate the second determinant: \(\left| {\;\begin{array}{} 1&2&2\\ 1&{x + 2}&2\\ 1&2&{x + 2} \end{array}} \right|\). We can use row operations to create zeros in the first column, which simplifies expansion.

Perform operations \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - R_1\).

• For \(R_2 \to R_2 - R_1\): \(1 - 1 = 0\), \((x + 2) - 2 = x\), \(2 - 2 = 0\). The new \(R_2\) is \([\begin{matrix} 0 & x & 0 \end{matrix}]\).
• For \(R_3 \to R_3 - R_1\): \(1 - 1 = 0\), \(2 - 2 = 0\), \((x + 2) - 2 = x\). The new \(R_3\) is \([\begin{matrix} 0 & 0 & x \end{matrix}]\).

The determinant becomes:

\(\left| {\;\begin{array}{} 1&2&2\\ 0&x&0\\ 0&0&x \end{array}} \right|\)

This is the determinant of an upper triangular matrix. The determinant of an upper triangular matrix is the product of its diagonal elements.

Determinant value = \(1 \times x \times x = x^2\).

Finding All Values of x Satisfying the Determinant Equation

Substituting this back into our main determinant equation, we get:

\((x + 6)(x^2) = 0\)

This cubic equation yields the values of \(x\) that make the equation true. The equation is satisfied if either factor is zero:

• \(x + 6 = 0 \implies x = -6\)
• \(x^2 = 0 \implies x = 0\) (This root has a multiplicity of 2)

Thus, the values of \(x\) satisfying the determinant equation are \(0\), \(0\), and \(-6\). These are the roots of the cubic equation obtained by evaluating the matrix determinant.

The values of \(x\) that solve for \(x\) in this determinant equation are \(0, 0, -6\).

These values can be listed as \(0, 0, -6\), which correctly identifies the solutions to the determinant equation.