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Matrices and Determinants question

2002 · Shift 0 · Q78
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  5. /2002 · Shift 0 · Q78

Matrices and Determinants question

2002 · Shift 0 · Q78

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If a>0a\gt 0a>0 and discriminant of  ax2+2bx+c\,a{x^2} + 2bx + cax2+2bx+c is −ve-ve−ve, then ∣abax+bbcbx+cax+bbx+c0∣\left| {\begin{matrix} a & b & {ax + b} \\ b & c & {bx + c} \\ {ax + b} & {bx + c} & 0 \\ \end{matrix} } \right|​abax+b​bcbx+c​ax+bbx+c0​​ is equal to
  1. A
    +ve+ve+ve
  2. B
    (ac−b2)(ax2+2bx+c)\left( {ac - {b^2}} \right)\left( {a{x^2} + 2bx + c} \right)(ac−b2)(ax2+2bx+c)
  3. C
    −ve-ve−ve
  4. D
    000
View written solutionFree

Correct answer: C

  1. Let
∣abax+bbcbx+cax+bbx+c0∣.\begin{vmatrix} a & b & ax+b\\ b & c & bx+c\\ ax+b & bx+c & 0 \end{vmatrix}.​abax+b​bcbx+c​ax+bbx+c0​​.

We need to determine its sign.

  1. Use a column operation: C3→C3−xC1.C_3 \to C_3 - xC_1.C3​→C3​−xC1​. Then
  • third entry in row 1 becomes ax+b−xa=b,ax+b-xa=b,ax+b−xa=b,
  • third entry in row 2 becomes bx+c−xb=c,bx+c-xb=c,bx+c−xb=c,
  • third entry in row 3 becomes 0−x(ax+b)=−(ax+b)x=−ax2−bx.0-x(ax+b)=-(ax+b)x=-ax^2-bx.0−x(ax+b)=−(ax+b)x=−ax2−bx.

So,

∣abbbccax+bbx+c−ax2−bx∣.\begin{vmatrix} a & b & b\\ b & c & c\\ ax+b & bx+c & -ax^2-bx \end{vmatrix}.​abax+b​bcbx+c​bc−ax2−bx​​.
  1. Now use another column operation: C3→C3−C2.C_3 \to C_3 - C_2.C3​→C3​−C2​. Then
∣ab0bc0ax+bbx+c−(ax2+2bx+c)∣.\begin{vmatrix} a & b & 0\\ b & c & 0\\ ax+b & bx+c & -(ax^2+2bx+c) \end{vmatrix}.​abax+b​bcbx+c​00−(ax2+2bx+c)​​.
  1. Expand along the third column:
∣abbc∣.\begin{vmatrix} a & b\\ b & c \end{vmatrix}.​ab​bc​​.

Hence, D=−(ax2+2bx+c)(ac−b2).D=-(ax^2+2bx+c)(ac-b^2).D=−(ax2+2bx+c)(ac−b2).

Equivalently, D=(b2−ac)(ax2+2bx+c).D=(b^2-ac)(ax^2+2bx+c).D=(b2−ac)(ax2+2bx+c).

5. Now use the given condition: the discriminant of $$ax^2+2bx+c$$ is negative. Its discriminant is $$\Delta=(2b)^2-4ac=4(b^2-ac)<0.$$ So, $$b^2-ac<0 \quad \Rightarrow \quad ac-b^2>0.$$ Also, since $a>0$ and the quadratic has negative discriminant, it is always positive for all real $x$: $$ax^2+2bx+c>0.$$ Therefore, $$D=-(ac-b^2)(ax^2+2bx+c).$$ Since both factors are positive, $$D<0.$$ 6. So the determinant is negative. Therefore the correct option is: $$\boxed{\text{C: } -ve}.$$ 7. Comparison with stored answer: Stored correct answer = C, which matches our result.
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