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Matrices and Determinants question

2024 · 8 Apr · Shift 1 · Q31
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Matrices and Determinants question

2024 · 8 Apr · Shift 1 · Q31

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let A=[2a013105b]A=\left[\begin{array}{lll}2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b\end{array}\right]A=​210​a35​01b​​. If A3=4A2−A−21IA^3=4 A^2-A-21 IA3=4A2−A−21I, where III is the identity matrix of order 3×33 \times 33×3, then 2a+3b2 a+3 b2a+3b is equal to
  1. A
    −10-10−10
  2. B
    −12-12−12
  3. C
    −13-13−13
  4. D
    −9-9−9
View written solutionFree

Correct answer: C

We are given A=[2a013105b]A=\begin{bmatrix}2&a&0\\1&3&1\\0&5&b\end{bmatrix}A=​210​a35​01b​​ and A3=4A2−A−21I.A^3=4A^2-A-21I.A3=4A2−A−21I.

We need to find 2a+3b2a+3b2a+3b.

1. Use the given matrix equation

Rearrange: A3−4A2+A+21I=0.A^3-4A^2+A+21I=0.A3−4A2+A+21I=0. So the matrix AAA satisfies the polynomial p(x)=x3−4x2+x+21.p(x)=x^3-4x^2+x+21.p(x)=x3−4x2+x+21.

Factor this polynomial: x3−4x2+x+21=(x−3)(x2−x−7).x^3-4x^2+x+21=(x-3)(x^2-x-7).x3−4x2+x+21=(x−3)(x2−x−7). Hence the eigenvalues of AAA must be among the roots of this polynomial.

2. Use trace and determinant

For matrix A=[2a013105b],A=\begin{bmatrix}2&a&0\\1&3&1\\0&5&b\end{bmatrix},A=​210​a35​01b​​, its trace is tr⁡(A)=2+3+b=5+b.\operatorname{tr}(A)=2+3+b=5+b.tr(A)=2+3+b=5+b.

Since AAA satisfies the cubic polynomial above, it is natural to compare with its characteristic polynomial.

Let us compute the characteristic polynomial of AAA: det⁡(A−λI)=∣2−λa013−λ105b−λ∣.\det(A-\lambda I)=\begin{vmatrix}2-\lambda & a & 0\\1&3-\lambda&1\\0&5&b-\lambda\end{vmatrix}.det(A−λI)=​2−λ10​a3−λ5​01b−λ​​. Expand along the first row:

det⁡(A−λI)=(2−λ)∣3−λ15b−λ∣−a∣110b−λ∣.\det(A-\lambda I) =(2-\lambda)\begin{vmatrix}3-\lambda&1\\5&b-\lambda\end{vmatrix} -a\begin{vmatrix}1&1\\0&b-\lambda\end{vmatrix}.det(A−λI)=(2−λ)​3−λ5​1b−λ​​−a​10​1b−λ​​.

So

det⁡(A−λI)=(2−λ)((3−λ)(b−λ)−5)−a(b−λ).\det(A-\lambda I)=(2-\lambda)\big((3-\lambda)(b-\lambda)-5\big)-a(b-\lambda).det(A−λI)=(2−λ)((3−λ)(b−λ)−5)−a(b−λ).

Now expand:

(3−λ)(b−λ)=3b−(3+b)λ+λ2.(3-\lambda)(b-\lambda)=3b-(3+b)\lambda+\lambda^2.(3−λ)(b−λ)=3b−(3+b)λ+λ2.

Thus

(3−λ)(b−λ)−5=λ2−(b+3)λ+(3b−5).(3-\lambda)(b-\lambda)-5= \lambda^2-(b+3)\lambda+(3b-5).(3−λ)(b−λ)−5=λ2−(b+3)λ+(3b−5).

Hence

det⁡(A−λI)=(2−λ)[λ2−(b+3)λ+(3b−5)]−a(b−λ).\det(A-\lambda I)=(2-\lambda)\big[\lambda^2-(b+3)\lambda+(3b-5)\big]-a(b-\lambda).det(A−λI)=(2−λ)[λ2−(b+3)λ+(3b−5)]−a(b−λ).

Expanding,

det⁡(A−λI)=−λ3+(b+5)λ2+(a−5b−1)λ+b(6−a)−10.\det(A-\lambda I) =-\lambda^3+(b+5)\lambda^2+(a-5b-1)\lambda+b(6-a)-10.det(A−λI)=−λ3+(b+5)λ2+(a−5b−1)λ+b(6−a)−10.

Therefore the characteristic polynomial is

χ(λ)=det⁡(λI−A)=λ3−(b+5)λ2−(a−5b−1)λ−[b(6−a)−10].\chi(\lambda)=\det(\lambda I-A) =\lambda^3-(b+5)\lambda^2-(a-5b-1)\lambda-[b(6-a)-10].χ(λ)=det(λI−A)=λ3−(b+5)λ2−(a−5b−1)λ−[b(6−a)−10].

That is,

χ(λ)=λ3−(b+5)λ2+(5b−a+1)λ+(ab−6b+10).\chi(\lambda)=\lambda^3-(b+5)\lambda^2+(5b-a+1)\lambda+(ab-6b+10).χ(λ)=λ3−(b+5)λ2+(5b−a+1)λ+(ab−6b+10).

3. Apply Cayley–Hamilton

Since AAA satisfies A3−4A2+A+21I=0,A^3-4A^2+A+21I=0,A3−4A2+A+21I=0, and the characteristic polynomial is monic of degree 333, for a 3×33\times 33×3 matrix this strongly suggests χ(λ)=λ3−4λ2+λ+21.\chi(\lambda)=\lambda^3-4\lambda^2+\lambda+21.χ(λ)=λ3−4λ2+λ+21. So compare coefficients:

Coefficient of λ2\lambda^2λ2

−(b+5)=−4-(b+5)=-4−(b+5)=−4 b+5=4b+5=4b+5=4 b=−1.b=-1.b=−1.

Coefficient of λ\lambdaλ

5b−a+1=15b-a+1=15b−a+1=1 5b−a=05b-a=05b−a=0 a=5b=−5.a=5b=-5.a=5b=−5.

Check constant term: ab−6b+10=(−5)(−1)−6(−1)+10=5+6+10=21,ab-6b+10=(-5)(-1)-6(-1)+10=5+6+10=21,ab−6b+10=(−5)(−1)−6(−1)+10=5+6+10=21, which matches.

So a=−5,b=−1.a=-5,\qquad b=-1.a=−5,b=−1.

4. Compute the required value

2a+3b=2(−5)+3(−1)=−10−3=−13.2a+3b=2(-5)+3(-1)=-10-3=-13.2a+3b=2(−5)+3(−1)=−10−3=−13.

5. Option matching

The correct option is C: −13.\boxed{\text{C: }-13}.C: −13​.

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