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Matrices and Determinants question

2023 · 13 Apr · Shift 1 · Q32
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Matrices and Determinants question

2023 · 13 Apr · Shift 1 · Q32

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let B=[13α123αα4],α>2B=\left[\begin{array}{lll}1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4\end{array}\right], \alpha \gt 2B=​11α​32α​α34​​,α>2 be the adjoint of a matrix AAA and ∣A∣=2|A|=2∣A∣=2. Then [α−2αα]B[α−2αα]\left[\begin{array}{ccc}\alpha & -2 \alpha & \alpha\end{array}\right] B\left[\begin{array}{c}\alpha \\ -2 \alpha \\ \alpha\end{array}\right][α​−2α​α​]B​α−2αα​​ is equal to :
  1. A
    32
  2. B
    −-− 16
  3. C
    0
  4. D
    16
View written solutionFree

Correct answer: B

  1. Use the property of adjoint

Since B=adj⁡(A)B=\operatorname{adj}(A)B=adj(A) for a 3×33\times 33×3 matrix AAA and ∣A∣=2|A|=2∣A∣=2, we use

∣adj⁡(A)∣=∣A∣n−1|\operatorname{adj}(A)| = |A|^{n-1}∣adj(A)∣=∣A∣n−1

for an n×nn\times nn×n matrix. Here n=3n=3n=3, so

∣B∣=∣adj⁡(A)∣=∣A∣2=22=4.|B|=|\operatorname{adj}(A)|=|A|^2=2^2=4.∣B∣=∣adj(A)∣=∣A∣2=22=4.

So the determinant of the given matrix BBB must be 444.

  1. Compute ∣B∣|B|∣B∣ in terms of α\alphaα

Given

B=[13α123αα4]B=\begin{bmatrix} 1 & 3 & \alpha\\ 1 & 2 & 3\\ \alpha & \alpha & 4 \end{bmatrix}B=​11α​32α​α34​​

Now,

∣B∣=1∣23α4∣−3∣13α4∣+α∣12αα∣|B|= 1\begin{vmatrix}2&3\\ \alpha&4\end{vmatrix} -3\begin{vmatrix}1&3\\ \alpha&4\end{vmatrix} +\alpha\begin{vmatrix}1&2\\ \alpha&\alpha\end{vmatrix}∣B∣=1​2α​34​​−3​1α​34​​+α​1α​2α​​ =1(8−3α)−3(4−3α)+α(α−2α)=1(8-3\alpha)-3(4-3\alpha)+\alpha(\alpha-2\alpha)=1(8−3α)−3(4−3α)+α(α−2α) =8−3α−12+9α−α2=8-3\alpha-12+9\alpha-\alpha^2=8−3α−12+9α−α2 =−α2+6α−4=-\alpha^2+6\alpha-4=−α2+6α−4

Since ∣B∣=4|B|=4∣B∣=4,

−α2+6α−4=4-\alpha^2+6\alpha-4=4−α2+6α−4=4 −α2+6α−8=0-\alpha^2+6\alpha-8=0−α2+6α−8=0 α2−6α+8=0\alpha^2-6\alpha+8=0α2−6α+8=0 (α−2)(α−4)=0(\alpha-2)(\alpha-4)=0(α−2)(α−4)=0

Given α>2\alpha>2α>2, we get

α=4.\alpha=4.α=4.
  1. Evaluate the quadratic form

We need to find

[α−2αα]B[α−2αα].\begin{bmatrix}\alpha & -2\alpha & \alpha\end{bmatrix} B \begin{bmatrix}\alpha\\-2\alpha\\\alpha\end{bmatrix}.[α​−2α​α​]B​α−2αα​​.

Let

v=[α−2αα].v=\begin{bmatrix}\alpha\\-2\alpha\\\alpha\end{bmatrix}.v=​α−2αα​​.

With α=4\alpha=4α=4,

v=[4−84]=4[1−21].v=\begin{bmatrix}4\\-8\\4\end{bmatrix} =4\begin{bmatrix}1\\-2\\1\end{bmatrix}.v=​4−84​​=4​1−21​​.

Also,

B=[134123444].B=\begin{bmatrix} 1&3&4\\ 1&2&3\\ 4&4&4 \end{bmatrix}.B=​114​324​434​​.

Now compute

B[1−21]=[1(1)+3(−2)+4(1)1(1)+2(−2)+3(1)4(1)+4(−2)+4(1)]=[−100].B\begin{bmatrix}1\\-2\\1\end{bmatrix} = \begin{bmatrix} 1(1)+3(-2)+4(1)\\ 1(1)+2(-2)+3(1)\\ 4(1)+4(-2)+4(1) \end{bmatrix} = \begin{bmatrix} -1\\ 0\\ 0 \end{bmatrix}.B​1−21​​=​1(1)+3(−2)+4(1)1(1)+2(−2)+3(1)4(1)+4(−2)+4(1)​​=​−100​​.

Hence,

Bv=4[−100]=[−400].Bv = 4\begin{bmatrix}-1\\0\\0\end{bmatrix} =\begin{bmatrix}-4\\0\\0\end{bmatrix}.Bv=4​−100​​=​−400​​.

Therefore,

v^T B v =egin{bmatrix}4&-8&4\end{bmatrix} \begin{bmatrix}-4\\0\\0\end{bmatrix} =4(-4)+(-8)(0)+4(0) =-16.
  1. Final answer
−16\boxed{-16}−16​

So the correct option is B.

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