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Matrices and Determinants question

2022 · 30 Jun · Shift 1 · Q26
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  5. /2022 · 30 Jun · Shift 1 · Q26

Matrices and Determinants question

2022 · 30 Jun · Shift 1 · Q26

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let A=[1−2αα2−1]A = \left[ {\begin{matrix} 1 & { - 2} & \alpha \\ \alpha & 2 & { - 1} \\ \end{matrix} } \right]A=[1α​−22​α−1​] and B=[2α−124−5], α∈CB = \left[ {\begin{matrix} 2 & \alpha \\ { - 1} & 2 \\ 4 & { - 5} \\ \end{matrix} } \right],\,\alpha \in CB=​2−14​α2−5​​,α∈C. Then the absolute value of the sum of all values of α\alphaα for which det(AB) = 0 is :
  1. A
    3
  2. B
    4
  3. C
    2
  4. D
    5
View written solutionFree

Correct answer: A

  1. Write the matrices and dimensions

We have

B=\begin{bmatrix}2&\alpha\\ -1&2\\ 4&-5\end{bmatrix}.$$ Here, $A$ is $2\times 3$ and $B$ is $3\times 2$, so $AB$ is a $2\times 2$ matrix. --- 2. **Compute $AB$** Using row-column multiplication: $$AB= \begin{bmatrix} 1&-2&\alpha\\ \alpha&2&-1 \end{bmatrix} \begin{bmatrix} 2&\alpha\\ -1&2\\ 4&-5 \end{bmatrix}.

Now compute each entry:

  • First row, first column: 1⋅2+(−2)(−1)+α⋅4=2+2+4α=4+4α.1\cdot 2+(-2)(-1)+\alpha\cdot 4=2+2+4\alpha=4+4\alpha.1⋅2+(−2)(−1)+α⋅4=2+2+4α=4+4α.

  • First row, second column: 1⋅α+(−2)⋅2+α(−5)=α−4−5α=−4α−4.1\cdot \alpha+(-2)\cdot 2+\alpha(-5)=\alpha-4-5\alpha=-4\alpha-4.1⋅α+(−2)⋅2+α(−5)=α−4−5α=−4α−4.

  • Second row, first column: α⋅2+2(−1)+(−1)⋅4=2α−2−4=2α−6.\alpha\cdot 2+2(-1)+(-1)\cdot 4=2\alpha-2-4=2\alpha-6.α⋅2+2(−1)+(−1)⋅4=2α−2−4=2α−6.

  • Second row, second column: α⋅α+2⋅2+(−1)(−5)=α2+4+5=α2+9.\alpha\cdot \alpha+2\cdot 2+(-1)(-5)=\alpha^2+4+5=\alpha^2+9.α⋅α+2⋅2+(−1)(−5)=α2+4+5=α2+9.

So, AB=[4+4α−4α−42α−6α2+9].AB=\begin{bmatrix}4+4\alpha&-4\alpha-4\\ 2\alpha-6&\alpha^2+9\end{bmatrix}.AB=[4+4α2α−6​−4α−4α2+9​].

  1. Find det⁡(AB)\det(AB)det(AB)

det⁡(AB)=(4+4α)(α2+9)−(−4α−4)(2α−6).\det(AB)=(4+4\alpha)(\alpha^2+9)-(-4\alpha-4)(2\alpha-6).det(AB)=(4+4α)(α2+9)−(−4α−4)(2α−6).

Factor where possible: 4+4α=4(1+α),−4α−4=−4(α+1).4+4\alpha=4(1+\alpha),\qquad -4\alpha-4=-4(\alpha+1).4+4α=4(1+α),−4α−4=−4(α+1).

Thus, det⁡(AB)=4(α+1)(α2+9)−[−4(α+1)](2α−6).\det(AB)=4(\alpha+1)(\alpha^2+9)-[-4(\alpha+1)](2\alpha-6).det(AB)=4(α+1)(α2+9)−[−4(α+1)](2α−6).

=4(α+1)(α2+9)+4(α+1)(2α−6).=4(\alpha+1)(\alpha^2+9)+4(\alpha+1)(2\alpha-6).=4(α+1)(α2+9)+4(α+1)(2α−6).

Take common factor 4(α+1)4(\alpha+1)4(α+1): det⁡(AB)=4(α+1)[(α2+9)+(2α−6)].\det(AB)=4(\alpha+1)\big[(\alpha^2+9)+(2\alpha-6)\big].det(AB)=4(α+1)[(α2+9)+(2α−6)].

det⁡(AB)=4(α+1)(α2+2α+3).\det(AB)=4(\alpha+1)(\alpha^2+2\alpha+3).det(AB)=4(α+1)(α2+2α+3).

So we need 4(α+1)(α2+2α+3)=0.4(\alpha+1)(\alpha^2+2\alpha+3)=0.4(α+1)(α2+2α+3)=0.

Hence, α=−1\alpha=-1α=−1 or α2+2α+3=0.\alpha^2+2\alpha+3=0.α2+2α+3=0.

Solve the quadratic: α=−2±4−122=−1±i2.\alpha=\frac{-2\pm\sqrt{4-12}}{2}=-1\pm i\sqrt{2}.α=2−2±4−12​​=−1±i2​.

Therefore all values of α\alphaα are: −1, −1+i2, −1−i2.-1,\ -1+i\sqrt2,\ -1-i\sqrt2.−1, −1+i2​, −1−i2​.

  1. Find their sum

(−1)+(−1+i2)+(−1−i2)=−3.(-1)+(-1+i\sqrt2)+(-1-i\sqrt2)=-3.(−1)+(−1+i2​)+(−1−i2​)=−3.

Absolute value: ∣−3∣=3.|-3|=3.∣−3∣=3.

  1. Check with options

So the required value is 3.\boxed{3}.3​.

This matches Option A.

  1. Comparison with stored answer

Stored correct answer: A

Derived answer: A

So they agree.

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