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Matrices and Determinants question

2019 · 12 Apr · Shift 1 · Q30
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  5. /2019 · 12 Apr · Shift 1 · Q30

Matrices and Determinants question

2019 · 12 Apr · Shift 1 · Q30

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If B=[52α1021α3−1]B = \left[ {\begin{matrix} 5 & {2\alpha } & 1 \\ 0 & 2 & 1 \\ \alpha & 3 & { - 1} \\ \end{matrix} } \right]B=​50α​2α23​11−1​​ is the inverse of a 3 × 3 matrix A, then the sum of all values of α\alphaα for which det(A) + 1 = 0, is :
  1. A
    2
  2. B
    - 1
  3. C
    0
  4. D
    1
View written solutionFree

Correct answer: D

  1. Since BBB is the inverse of matrix AAA, we have A−1=B ⇒ det⁡(A)det⁡(B)=1.A^{-1}=B \,\Rightarrow\, \det(A)\det(B)=1.A−1=B⇒det(A)det(B)=1. Hence, det⁡(A)=1det⁡(B).\det(A)=\frac{1}{\det(B)}.det(A)=det(B)1​.

  2. The condition given is det⁡(A)+1=0⇒det⁡(A)=−1.\det(A)+1=0 \Rightarrow \det(A)=-1.det(A)+1=0⇒det(A)=−1. So, 1det⁡(B)=−1⇒det⁡(B)=−1.\frac{1}{\det(B)}=-1 \Rightarrow \det(B)=-1.det(B)1​=−1⇒det(B)=−1.

  3. Now compute det⁡(B)\det(B)det(B) for B=[52α1021α3−1].B=\begin{bmatrix}5&2\alpha&1\\0&2&1\\\alpha&3&-1\end{bmatrix}.B=​50α​2α23​11−1​​. Expand along the first row: det⁡(B)=5∣213−1∣−2α∣01α−1∣+1∣02α3∣.\det(B)=5\begin{vmatrix}2&1\\3&-1\end{vmatrix}-2\alpha\begin{vmatrix}0&1\\\alpha&-1\end{vmatrix}+1\begin{vmatrix}0&2\\\alpha&3\end{vmatrix}.det(B)=5​23​1−1​​−2α​0α​1−1​​+1​0α​23​​.

Now evaluate each minor: ∣213−1∣=2(−1)−1(3)=−2−3=−5,\begin{vmatrix}2&1\\3&-1\end{vmatrix}=2(-1)-1(3)=-2-3=-5,​23​1−1​​=2(−1)−1(3)=−2−3=−5, ∣01α−1∣=0(−1)−1(α)=−α,\begin{vmatrix}0&1\\\alpha&-1\end{vmatrix}=0(-1)-1(\alpha)=-\alpha,​0α​1−1​​=0(−1)−1(α)=−α, ∣02α3∣=0⋅3−2α=−2α.\begin{vmatrix}0&2\\\alpha&3\end{vmatrix}=0\cdot 3-2\alpha=-2\alpha.​0α​23​​=0⋅3−2α=−2α.

Therefore, det⁡(B)=5(−5)−2α(−α)+1(−2α)\det(B)=5(-5)-2\alpha(-\alpha)+1(-2\alpha)det(B)=5(−5)−2α(−α)+1(−2α) =−25+2α2−2α.=-25+2\alpha^2-2\alpha.=−25+2α2−2α.

  1. Use the condition det⁡(B)=−1\det(B)=-1det(B)=−1: −25+2α2−2α=−1.-25+2\alpha^2-2\alpha=-1.−25+2α2−2α=−1. 2α2−2α−24=02\alpha^2-2\alpha-24=02α2−2α−24=0 α2−α−12=0\alpha^2-\alpha-12=0α2−α−12=0 (α−4)(α+3)=0.(\alpha-4)(\alpha+3)=0.(α−4)(α+3)=0. So, α=4orα=−3.\alpha=4 \quad \text{or} \quad \alpha=-3.α=4orα=−3.

  2. Sum of all values of α\alphaα: 4+(−3)=1.4+(-3)=1.4+(−3)=1.

  3. Therefore, the correct option is D: 1.\boxed{\text{D: }1}.D: 1​.

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