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NCERT Class 12 · Ch 4 Exercise 4.5 · Q9

NCERT Class 12 Maths · Chapter 4 Determinants · Exercise 4.5 Question 9

Use the matrix method for AX=B, where A = \(\left[\begin{matrix}4 & -3\\3 & -5\end{matrix}\right]\), B = \(\left[\begin{matrix}3\\7\end{matrix}\right]\), X=\(\left[\begin{matrix}x\\y\end{matrix}\right]\).

💡 Concept Hints

  1. Identify whether this needs determinant expansion, an inverse, or the absolute determinant formula for area.
  2. For a determinant use signed cofactors. For a system first check invertibility; for an area take half the absolute determinant.

✅ Step-by-Step Solution

Final Answer: \(\left[\begin{matrix}- \frac{6}{11}\\- \frac{19}{11}\end{matrix}\right]\)
  1. For a 2×2 matrix use det A=a₁₁a₂₂−a₁₂a₂₁.
  2. \(-20\) − \(-9\) = \(-11\).
  3. First check whether this determinant is zero.
  4. The cofactor matrix is \(\left[\begin{matrix}-5 & -3\\3 & 4\end{matrix}\right]\).
  5. Transpose the cofactor matrix: adj A=\(\left[\begin{matrix}-5 & 3\\-3 & 4\end{matrix}\right]\).
  6. Divide by det A: A⁻¹=\(\left[\begin{matrix}\frac{5}{11} & - \frac{3}{11}\\\frac{3}{11} & - \frac{4}{11}\end{matrix}\right]\).
  7. Multiplication verifies AA⁻¹=A⁻¹A=I.
  8. Multiply X=A⁻¹B=\(\left[\begin{matrix}- \frac{6}{11}\\- \frac{19}{11}\end{matrix}\right]\).
  9. Substituting this vector into the original equations gives AX=\(\left[\begin{matrix}3\\7\end{matrix}\right]\)=B. Hence the system has a unique solution.

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