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Matrices and Determinants question

2017 · Shift 1 · Q26
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  5. /2017 · Shift 1 · Q26

Matrices and Determinants question

2017 · Shift 1 · Q26

JEE AdvancedMathematicsMatrices and DeterminantsNumerical+3 / −1
For a real number α\alphaα, if the system [1αα2α1αα2α1][xyz]=[1−11]\left[ {\begin{matrix} 1 & \alpha & {{\alpha ^2}} \\ \alpha & 1 & \alpha \\ {{\alpha ^2}} & \alpha & 1 \\ \end{matrix} } \right]\left[ {\begin{matrix} x \\ y \\ z \\ \end{matrix} } \right] = \left[ {\begin{matrix} 1 \\ { - 1} \\ 1 \\ \end{matrix} } \right]​1αα2​α1α​α2α1​​​xyz​​=​1−11​​ of linear equations, has infinitely many solutions, then 1 +α\alphaα+α\alphaα 2 =
Numerical answer
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Correct answer: 1

  1. Let the coefficient matrix be A=[1αα2α1αα2α1],b=[1−11].A=\begin{bmatrix}1&\alpha&\alpha^2\\ \alpha&1&\alpha\\ \alpha^2&\alpha&1\end{bmatrix},\qquad b=\begin{bmatrix}1\\-1\\1\end{bmatrix}.A=​1αα2​α1α​α2α1​​,b=​1−11​​.

We need the system A[xyz]=bA\begin{bmatrix}x\\y\\z\end{bmatrix}=bA​xyz​​=b to have infinitely many solutions.

  1. For a system of 3 linear equations to have infinitely many solutions, we need:
  • det⁡(A)=0\det(A)=0det(A)=0 so that the matrix is singular,
  • and the system must be consistent, i.e. rank⁡(A)=rank⁡(A∣b)<3\operatorname{rank}(A)=\operatorname{rank}(A|b)<3rank(A)=rank(A∣b)<3.
  1. First compute det⁡(A)\det(A)det(A).

Expanding along the first row,

det⁡(A)=1∣1αα1∣−α∣ααα21∣+α2∣α1α2α∣.\det(A)=1\begin{vmatrix}1&\alpha\\ \alpha&1\end{vmatrix} -\alpha\begin{vmatrix}\alpha&\alpha\\ \alpha^2&1\end{vmatrix} +\alpha^2\begin{vmatrix}\alpha&1\\ \alpha^2&\alpha\end{vmatrix}.det(A)=1​1α​α1​​−α​αα2​α1​​+α2​αα2​1α​​.

Now, ∣1αα1∣=1−α2,\begin{vmatrix}1&\alpha\\ \alpha&1\end{vmatrix}=1-\alpha^2,​1α​α1​​=1−α2, ∣ααα21∣=α−α3=α(1−α2),\begin{vmatrix}\alpha&\alpha\\ \alpha^2&1\end{vmatrix}=\alpha-\alpha^3=\alpha(1-\alpha^2),​αα2​α1​​=α−α3=α(1−α2), ∣α1α2α∣=α2−α2=0.\begin{vmatrix}\alpha&1\\ \alpha^2&\alpha\end{vmatrix}=\alpha^2-\alpha^2=0.​αα2​1α​​=α2−α2=0.

Hence,

det⁡(A)=(1−α2)−α⋅α(1−α2)+α2⋅0=(1−α2)−α2(1−α2).\det(A)=(1-\alpha^2)-\alpha\cdot \alpha(1-\alpha^2)+\alpha^2\cdot 0 =(1-\alpha^2)-\alpha^2(1-\alpha^2).det(A)=(1−α2)−α⋅α(1−α2)+α2⋅0=(1−α2)−α2(1−α2).

So,

det⁡(A)=(1−α2)2.\det(A)=(1-\alpha^2)^2.det(A)=(1−α2)2.

Therefore, det⁡(A)=0  ⟺  α2=1  ⟺  α=±1.\det(A)=0 \iff \alpha^2=1 \iff \alpha=\pm 1.det(A)=0⟺α2=1⟺α=±1.

  1. Now check consistency for these values.

Case 1: α=1\alpha=1α=1

Then A=[111111111].A=\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}.A=​111​111​111​​. The system becomes x+y+z=1,x+y+z=1,x+y+z=1, x+y+z=−1,x+y+z=-1,x+y+z=−1, x+y+z=1.x+y+z=1.x+y+z=1. This is inconsistent, so there are no solutions.

Thus α=1\alpha=1α=1 is rejected.

Case 2: α=−1\alpha=-1α=−1

Then A=[1−11−11−11−11].A=\begin{bmatrix}1&-1&1\\-1&1&-1\\1&-1&1\end{bmatrix}.A=​1−11​−11−1​1−11​​. The system becomes x−y+z=1,x-y+z=1,x−y+z=1, −(x−y+z)=−1  ⟺  x−y+z=1,-(x-y+z)=-1 \iff x-y+z=1,−(x−y+z)=−1⟺x−y+z=1, x−y+z=1.x-y+z=1.x−y+z=1. All three equations are the same, so the system is consistent and has infinitely many solutions.

Thus, α=−1.\alpha=-1.α=−1.

  1. We need to find 1+α+α2.1+\alpha+\alpha^2.1+α+α2. Substitute α=−1\alpha=-1α=−1: 1+(−1)+(−1)2=1−1+1=1.1+(-1)+(-1)^2=1-1+1=1.1+(−1)+(−1)2=1−1+1=1.

Therefore, the required value is 1.\boxed{1}.1​.

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