Question

A determinant is given as follows

Answer 1

The determinant is divisible by x².

Answer 2

The problem statement "A determinant is given as follows" is incomplete as it does not provide the determinant. Based on the provided similar_question, which itself contains a typo, the intended determinant is assumed to be the corrected one from the similar_question's explanation:

$\Delta = \begin{vmatrix} a^2+x & ab & ac \\ ab & b^2+x & bc \\ ac & bc & c^2+x \end{vmatrix}$

The question implicitly asks to analyze this determinant, likely to find its factors or express its value.

Solution:

The determinant can be evaluated using properties of matrices or by direct expansion with strategic row/column operations.

Method 1: Using Matrix Eigenvalues (More advanced but elegant)

Let the given determinant be $\Delta = \det(M+xI)$, where $I$ is the identity matrix and $M$ is the matrix: $M = \begin{pmatrix} a^2 & ab & ac \\ ab & b^2 & bc \\ ac & bc & c^2 \end{pmatrix}$ This matrix $M$ can be expressed as the outer product of a vector $\mathbf{v} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$ with itself, i.e., $M = \mathbf{v}\mathbf{v}^T$. A matrix of the form $\mathbf{v}\mathbf{v}^T$ (a rank-1 matrix) has only one non-zero eigenvalue, which is $\mathbf{v}^T\mathbf{v} = a^2+b^2+c^2$. The other eigenvalues are zero. For a 3x3 matrix, the eigenvalues of $M$ are $\lambda_1 = a^2+b^2+c^2$, $\lambda_2 = 0$, and $\lambda_3 = 0$.

The eigenvalues of $M+xI$ are $\lambda_1+x, \lambda_2+x, \lambda_3+x$. So, the eigenvalues of the matrix corresponding to $\Delta$ are $(a^2+b^2+c^2+x)$, $x$, and $x$. The determinant of a matrix is the product of its eigenvalues. Therefore, $\Delta = (a^2+b^2+c^2+x) \cdot x \cdot x$ $\Delta = x^2(a^2+b^2+c^2+x)$

Method 2: Using Row/Column Operations

1. Multiply the first row by $a$, the second row by $b$, and the third row by $c$. To keep the determinant value unchanged, divide the entire determinant by $abc$: $\Delta = \frac{1}{abc} \begin{vmatrix} a(a^2+x) & a^2b & a^2c \\ b(ab) & b(b^2+x) & b^2c \\ c(ac) & c(bc) & c(c^2+x) \end{vmatrix}$
2. Take out common factors $a$ from the first column, $b$ from the second column, and $c$ from the third column. The $abc$ in the denominator cancels out: $\Delta = \frac{abc}{abc} \begin{vmatrix} a^2+x & a^2 & a^2 \\ b^2 & b^2+x & b^2 \\ c^2 & c^2 & c^2+x \end{vmatrix}$ Let $A=a^2$, $B=b^2$, $C=c^2$. The determinant becomes: $\Delta = \begin{vmatrix} A+x & A & A \\ B & B+x & B \\ C & C & C+x \end{vmatrix}$
3. Perform column operations $C_1 \to C_1 - C_2$ and $C_2 \to C_2 - C_3$: $\Delta = \begin{vmatrix} (A+x)-A & A-A & A \\ B-(B+x) & (B+x)-B & B \\ C-C & C-(C+x) & C+x \end{vmatrix}$ $\Delta = \begin{vmatrix} x & 0 & A \\ -x & x & B \\ 0 & -x & C+x \end{vmatrix}$
4. Expand the determinant along the first column ($C_1$): $\Delta = x \cdot \begin{vmatrix} x & B \\ -x & C+x \end{vmatrix} - (-x) \cdot \begin{vmatrix} 0 & A \\ -x & C+x \end{vmatrix} + 0 \cdot (\dots)$ $\Delta = x[x(C+x) - B(-x)] + x[0(C+x) - A(-x)]$ $\Delta = x[xC + x^2 + Bx] + x[Ax]$ $\Delta = x^2C + x^3 + x^2B + x^2A$ Factor out $x^2$: $\Delta = x^2(A+B+C+x)$
5. Substitute back $A=a^2, B=b^2, C=c^2$: $\Delta = x^2(a^2+b^2+c^2+x)$

Both methods yield the same result. From the expression $\Delta = x^2(a^2+b^2+c^2+x)$, it is clear that the determinant is divisible by $x^2$.

Explanation of the solution:

The determinant is assumed to be $\Delta = \begin{vmatrix} a^2+x & ab & ac \\ ab & b^2+x & bc \\ ac & bc & c^2+x \end{vmatrix}$. This determinant can be simplified by first transforming it to a known form. Multiply $R_1$ by $a$, $R_2$ by $b$, $R_3$ by $c$, and divide the determinant by $abc$. Then, take out common factors $a$ from $C_1$, $b$ from $C_2$, $c$ from $C_3$. This results in: $\Delta = \begin{vmatrix} a^2+x & a^2 & a^2 \\ b^2 & b^2+x & b^2 \\ c^2 & c^2 & c^2+x \end{vmatrix}$ Let $A=a^2, B=b^2, C=c^2$. So, $\Delta = \begin{vmatrix} A+x & A & A \\ B & B+x & B \\ C & C & C+x \end{vmatrix}$. Apply column operations $C_1 \to C_1 - C_2$ and $C_2 \to C_2 - C_3$: $\Delta = \begin{vmatrix} x & 0 & A \\ -x & x & B \\ 0 & -x & C+x \end{vmatrix}$ Expand along $C_1$: $\Delta = x(x(C+x) - B(-x)) - (-x)(0(C+x) - A(-x))$ $\Delta = x(xC+x^2+Bx) + x(Ax)$ $\Delta = x^2C+x^3+x^2B+x^2A = x^2(A+B+C+x)$ Substitute back $A=a^2, B=b^2, C=c^2$: $\Delta = x^2(a^2+b^2+c^2+x)$ From this expression, it is evident that $\Delta$ is divisible by $x^2$.