Let a determinant is given by A = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="|" close="|"><mtable><mtr><mtd><mi>a</mi></mtd><mtd><mi>b</mi></mtd><mtd><mi>c</mi></mtd></mtr><mtr><mtd><mi>p</mi></mtd><mtd><mi>q</mi></mtd><mtd><mi>r</mi></mtd></mtr><mtr><mtd><mi>x</mi></mtd><mtd><mi>y</mi></mtd><mtd><mi>z</mi></mtd></mtr></mtable></mfenced></math> and suppose A = 6. If B = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="|" close="|"><mtable><mtr><mtd><mi>p</mi><mo>+</mo><mi>x</mi></mtd><mtd><mi>q</mi><mo>+</mo><mi>y</mi></mtd><mtd><mi>r</mi><mo>+</mo><mi>z</mi></mtd></mtr><mtr><mtd><mi>a</mi><mo>+</mo><mi>x</mi></mtd><mtd><mi>b</mi><mo>+</mo><mi>y</mi></mtd><mtd><mi>c</mi><mo>+</mo><mi>z</mi></mtd></mtr><mtr><mtd><mi>a</mi><mo>+</mo><mi>p</mi></mtd><mtd><mi>b</mi><mo>+</mo><mi>q</mi></mtd><mtd><mi>c</mi><mo>+</mo><mi>r</mi></mtd></mtr></mtable></mfenced></math> then
Step-by-Step Solution
Key Concept: The determinant B can be expressed as 2 times the determinant A using row operations and properties of determinants.
Given A = |a b c; p q r; x y z| = 6. Let R1, R2, R3 be the rows of A. B = |p+x q+y r+z; a+x b+y c+z; a+p b+q c+r|. Applying R1 -> R1+R2+R3, we get B = |2(a+p+x) 2(b+q+y) 2(c+r+z); a+x b+y c+z; a+p b+q c+r| = 2 |a+p+x b+q+y c+r+z; a+x b+y c+z; a+p b+q c+r|. Applying R1 -> R1-R2-R3, we get B = 2 |p q r; a b c; x y z|. By swapping rows twice to match the order of A, we get B = 2 * (-1)^2 * |a b c; p q r; x y z| = 2 * A = 2 * 6 = 12.
Correct Answer: 3