[amp_mcq option1=”\\[\\left| {\\begin{array}{*{20}{c}} 1&{{\\text{x}}\\left( {{\\text{x}} + 1} \\right)}&{{\\text{x}} + 1} \\\\ 1&{{\\text{y}}\\left( {{\\text{y}} + 1} \\right)}&{{\\text{y}} + 1} \\\\ 1&{{\\text{z}}\\left( {{\\text{z}} + 1} \\right)}&{{\\text{z}} + 1} \\end{array}} \\right|\\]” option2=”\\[\\left| {\\begin{array}{*{20}{c}} 1&{{\\text{x}} + 1}&{{{\\text{x}}^2} + 1} \\\\ 1&{{\\text{y}} + 1}&{{{\\text{y}}^2} + 1} \\\\ 1&{{\\text{z}} + 1}&{{{\\text{z}}^2} + 1} \\end{array}} \\right|\\]” option3=”\\[\\left| {\\begin{array}{*{20}{c}} 0&{{\\text{x}} – {\\text{y}}}&{{{\\text{x}}^2} – {{\\text{y}}^2}} \\\\ 0&{{\\text{y}} – {\\text{z}}}&{{{\\text{y}}^2} – {{\\text{z}}^2}} \\\\ 1&{\\text{z}}&{{{\\text{z}}^2}} \\end{array}} \\right|\\]” option4=”\\[\\left| {\\begin{array}{*{20}{c}} 2&{{\\text{x}} + {\\text{y}}}&{{{\\text{x}}^2} + {{\\text{y}}^2}} \\\\ 2&{{\\text{y}} + {\\text{z}}}&{{{\\text{y}}^2} + {{\\text{z}}^2}} \\\\ 1&{\\text{z}}&{{{\\text{z}}^2}} \\end{array}} \\right|\\]” correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\text{(B)}}$.<\/p>\n
The determinant of a 3×3 matrix can be computed using the following formula:<\/p>\n
$$\\det \\begin{bmatrix}
\na & b & c \\\\
\nd & e & f \\\\
\ng & h & i
\n\\end{bmatrix} = (aei – bg – ch) + (bf – ce)d + (cd – af)g$$<\/p>\n
In this case, we have the matrix:<\/p>\n
$$\\begin{bmatrix}
\n1 & x & x^2 \\\\
\n1 & y & y^2 \\\\
\n1 & z & z^2
\n\\end{bmatrix}$$<\/p>\n
Substituting into the formula, we get:<\/p>\n
$$\\det \\begin{bmatrix}
\n1 & x & x^2 \\\\
\n1 & y & y^2 \\\\
\n1 & z & z^2
\n\\end{bmatrix} = (1)(z^2 – y^2) – (x)(y^2 – z^2) + (x)(z^2 – x^2)$$<\/p>\n
Simplifying, we get:<\/p>\n
$$\\det \\begin{bmatrix}
\n1 & x & x^2 \\\\
\n1 & y & y^2 \\\\
\n1 & z & z^2
\n\\end{bmatrix} = x^2 – y^2 – z^2$$<\/p>\n
This is the determinant of the matrix in option $\\boxed{\\text{(B)}}$.<\/p>\n
The determinants of the matrices in options $\\boxed{\\text{(A)}}$, $\\boxed{\\text{(C)}}$, and $\\boxed{\\text{(D)}}$ are all equal to $0$. This is because they all have a row or column of identical elements.<\/p>\n
In option $\\boxed{\\text{(A)}}$, the first row is all $1$s. This means that the determinant of the matrix is equal to $1$ times the determinant of the matrix that is left after we remove the first row and column. The determinant of the remaining matrix is equal to $0$, since it is a 2×2 matrix with a row of identical elements. Therefore, the determinant of the matrix in option $\\boxed{\\text{(A)}}$ is also equal to $0$.<\/p>\n
In option $\\boxed{\\text{(C)}}$, the second column is all $0$s. This means that the determinant of the matrix is equal to $0$ times the determinant of the matrix that is left after we remove the second column and row. The determinant of the remaining matrix is equal to $0$, since it is a 2×2 matrix with a column of identical elements. Therefore, the determinant of the matrix in option $\\boxed{\\text{(C)}}$ is also equal to $0$.<\/p>\n
In option $\\boxed{\\text{(D)}}$, the third row is all $1$s. This means that the determinant of the matrix is equal to $1$ times the determinant of the matrix that is left after we remove the third row and column. The determinant of the remaining matrix is equal to $0$, since it is a 2×2 matrix with a row of identical elements. Therefore, the determinant of the matrix in option $\\boxed{\\text{(D)}}$ is also equal to $0$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”\\[\\left| {\\begin{array}{*{20}{c}} 1&{{\\text{x}}\\left( {{\\text{x}} + 1} \\right)}&{{\\text{x}} + 1} \\\\ 1&{{\\text{y}}\\left( {{\\text{y}} + 1} \\right)}&{{\\text{y}} + 1} \\\\ 1&{{\\text{z}}\\left( {{\\text{z}} + 1} \\right)}&{{\\text{z}} + 1} \\end{array}} \\right|\\]” option2=”\\[\\left| {\\begin{array}{*{20}{c}} 1&{{\\text{x}} + 1}&{{{\\text{x}}^2} + 1} \\\\ 1&{{\\text{y}} + 1}&{{{\\text{y}}^2} + 1} \\\\ 1&{{\\text{z}} + 1}&{{{\\text{z}}^2} + 1} \\end{array}} \\right|\\]” option3=”\\[\\left| {\\begin{array}{*{20}{c}} 0&{{\\text{x}} – {\\text{y}}}&{{{\\text{x}}^2} – … <\/p>\n