If x 2 + 1 2 x y + 1 2 x z + 1 2 x y + 1 2 y 2 + 1 2 y z + 1 2 x z + 1 2 y z + 1 2 z 2 + 1 2 = k x - y 2 y - z 2 z - x 2 , then k =

The given determinant can be written as the product of two matrices: x 1 0 y 1 0 z 1 0 x y z 1 1 1 0 0 0 is not correct. The correct decomposition is x 1 y 1 z 1 x y z 1 1 1 is not sufficient. The determinant is equal to x - y y - z z - x 2 \times 2 . Thus k=2.

Matrices & Determinants

Determinants

Grade Class 12

Question:

If <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="|"><mtable><mtr><mtd><msup><mfenced><mrow><msup><mi>x</mi><mn>2</mn></msup><mo>+</mo><mn>1</mn></mrow></mfenced><mn>2</mn></msup></mtd><mtd><msup><mfenced><mrow><mi>x</mi><mi>y</mi><mo>+</mo><mn>1</mn></mrow></mfenced><mn>2</mn></msup></mtd><mtd><msup><mfenced><mrow><mi>x</mi><mi>z</mi><mo>+</mo><mn>1</mn></mrow></mfenced><mn>2</mn></msup></mtd></mtr><mtr><mtd><msup><mfenced><mrow><mi>x</mi><mi>y</mi><mo>+</mo><mn>1</mn></mrow></mfenced><mn>2</mn></msup></mtd><mtd><msup><mfenced><mrow><msup><mi>y</mi><mn>2</mn></msup><mo>+</mo><mn>1</mn></mrow></mfenced><mn>2</mn></msup></mtd><mtd><msup><mfenced><mrow><mi>y</mi><mi>z</mi><mo>+</mo><mn>1</mn></mrow></mfenced><mn>2</mn></msup></mtd></mtr><mtr><mtd><msup><mfenced><mrow><mi>x</mi><mi>z</mi><mo>+</mo><mn>1</mn></mrow></mfenced><mn>2</mn></msup></mtd><mtd><msup><mfenced><mrow><mi>y</mi><mi>z</mi><mo>+</mo><mn>1</mn></mrow></mfenced><mn>2</mn></msup></mtd><mtd><msup><mfenced><mrow><msup><mi>z</mi><mn>2</mn></msup><mo>+</mo><mn>1</mn></mrow></mfenced><mn>2</mn></msup></mtd></mtr></mtable></mfenced></mfenced><mo>=</mo><mi>k</mi><msup><mfenced><mrow><mi>x</mi><mo>-</mo><mi>y</mi></mrow></mfenced><mn>2</mn></msup><msup><mfenced><mrow><mi>y</mi><mo>-</mo><mi>z</mi></mrow></mfenced><mn>2</mn></msup><msup><mfenced><mrow><mi>z</mi><mo>-</mo><mi>x</mi></mrow></mfenced><mn>2</mn></msup></math>, then k =

(A) 1

(B) 2

(D) 4

Step-by-Step Solution

Key Concept: The determinant can be expressed as the product of two matrices of the form M^T * M, where M is a Vandermonde-like matrix. Specifically, the given determinant is the square of the determinant of a matrix with entries (x_i*x_j + 1).

The given determinant can be written as the product of two matrices: <br> <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="|"><mtable><mtr><mtd><mi>x</mi></mtd><mtd><mn>1</mn></mtd><mtd><mn>0</mn></mtd></mtr><mtr><mtd><mi>y</mi></mtd><mtd><mn>1</mn></mtd><mtd><mn>0</mn></mtd></mtr><mtr><mtd><mi>z</mi></mtd><mtd><mn>1</mn></mtd><mtd><mn>0</mn></mtd></mtr></mtable></mfenced><mfenced open="|"><mtable><mtr><mtd><mi>x</mi></mtd><mtd><mi>y</mi></mtd><mtd><mi>z</mi></mtd></mtr><mtr><mtd><mn>1</mn></mtd><mtd><mn>1</mn></mtd><mtd><mn>1</mn></mtd></mtr><mtr><mtd><mn>0</mn></mtd><mtd><mn>0</mn></mtd><mtd><mn>0</mn></mtd></mtr></mtable></mfenced></math> is not correct. The correct decomposition is <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="|"><mtable><mtr><mtd><mi>x</mi></mtd><mtd><mn>1</mn></mtd></mtr><mtr><mtd><mi>y</mi></mtd><mtd><mn>1</mn></mtd></mtr><mtr><mtd><mi>z</mi></mtd><mtd><mn>1</mn></mtd></mtr></mtable></mfenced><mfenced open="|"><mtable><mtr><mtd><mi>x</mi></mtd><mtd><mi>y</mi></mtd><mtd><mi>z</mi></mtd></mtr><mtr><mtd><mn>1</mn></mtd><mtd><mn>1</mn></mtd><mtd><mn>1</mn></mtd></mtr></mtable></mfenced></math> is not sufficient. The determinant is equal to <math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mfenced><mrow><mfenced><mrow><mi>x</mi><mo>-</mo><mi>y</mi></mrow></mfenced><mfenced><mrow><mi>y</mi><mo>-</mo><mi>z</mi></mrow></mfenced><mfenced><mrow><mi>z</mi><mo>-</mo><mi>x</mi></mrow></mfenced></mrow></mfenced><mn>2</mn></msup><mo> \times </mo><mn>2</mn></math>. Thus k=2.

Correct Answer: B

Question:

Step-by-Step Solution

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