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Matrices and Determinants question

2010 · Shift 1 · Q51
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  5. /2010 · Shift 1 · Q51

Matrices and Determinants question

2010 · Shift 1 · Q51

JEE AdvancedMathematicsMatrices and DeterminantsMCQ+3 / −1
The number of 3×33 \times 33×3 matrices A whose entries are either 0 or 1 and for which the system A[xyz]=[100]\mathrm{A}\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]A​xyz​​=​100​​ has exactly two distinct solutions, is
  1. A
    0
  2. B
    29−12^9-129−1
  3. C
    168
  4. D
    2
View written solutionFree

Correct answer: A

  1. We need the system A[xyz]=[100]A\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\0\\0\end{bmatrix}A​xyz​​=​100​​ to have exactly two distinct solutions.

Let the unknown vector be u=[xyz]\mathbf{u}=\begin{bmatrix}x\\y\\z\end{bmatrix}u=​xyz​​ and the RHS be b=[100].\mathbf{b}=\begin{bmatrix}1\\0\\0\end{bmatrix}.b=​100​​. So we are studying the linear system Au=b.A\mathbf{u}=\mathbf{b}.Au=b.

  1. For any linear system over the real numbers, the set of solutions is of the form u=u0+N(A),\mathbf{u}=\mathbf{u}_0+\mathcal N(A),u=u0​+N(A), where N(A)\mathcal N(A)N(A) is the null space of AAA.

Hence there are only three possibilities:

  • no solution,
  • exactly one solution,
  • infinitely many solutions.

A linear system can never have exactly two distinct solutions.

Why? If u1\mathbf{u}_1u1​ and u2\mathbf{u}_2u2​ are two distinct solutions, then Au1=b,Au2=b.A\mathbf{u}_1=\mathbf{b},\qquad A\mathbf{u}_2=\mathbf{b}.Au1​=b,Au2​=b. Subtracting, A(u1−u2)=0.A(\mathbf{u}_1-\mathbf{u}_2)=0.A(u1​−u2​)=0. So u1−u2∈N(A)\mathbf{u}_1-\mathbf{u}_2\in \mathcal N(A)u1​−u2​∈N(A) and is nonzero. Then for every real number ttt,

= A\mathbf{u}_1 + tA(\mathbf{u}_1-\mathbf{u}_2) = \mathbf{b}+t\cdot 0 =\mathbf{b}.$$ Thus infinitely many solutions exist. So having exactly two distinct solutions is impossible for any matrix $A$. 3. Therefore, among all $3\times 3$ matrices with entries $0$ or $1$, the number satisfying the condition is $$0.$$ 4. Checking options: - A: $0$ ✓ - B: $2^9-1$ ✗ - C: $168$ ✗ - D: $2$ ✗ Hence the correct option is **A**.
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