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Class 12 Determinants · Very Long Answer5 Marksmatrix_method_linear_systems

Class 12 Maths · Determinants · Very Long Answer #70

A shop sells pen varieties A,B,C. Meenu buys one of each for ₹21; Jeevan buys 4,3,2 respectively for ₹60; Shikha buys 6,2,3 respectively for ₹70. Let x,y,z be the respective prices in rupees. Solve by the matrix method: (1)x + (1)y + (1)z = 21; (4)x + (3)y + (2)z = 60; (6)x + (2)y + (3)z = 70.

✅ Step-by-Step Solution

Final Answer: x=5, y=8, z=8
  1. Write MX=B, where M=\(\begin{bmatrix}1 & 1 & 1\\4 & 3 & 2\\6 & 2 & 3\end{bmatrix}\), X=\(\begin{bmatrix}x\\y\\z\end{bmatrix}\), and B=\(\begin{bmatrix}21\\60\\70\end{bmatrix}\).
  2. The determinant is -5≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}5 & 0 & -10\\-1 & -3 & 4\\-1 & 2 & -1\end{bmatrix}\).
  3. Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}5 & -1 & -1\\0 & -3 & 2\\-10 & 4 & -1\end{bmatrix}\), hence M⁻¹=adj(M)/(-5)=\(\begin{bmatrix}-1 & 1/5 & 1/5\\0 & 3/5 & -2/5\\2 & -4/5 & 1/5\end{bmatrix}\).
  4. Multiply the inverse by the right side: X=M⁻¹B=\(\begin{bmatrix}5\\8\\8\end{bmatrix}\). The resulting entries satisfy MX=B by direct multiplication.
  5. Therefore x=5, y=8, z=8.

📋 CBSE Stepwise Marking Scheme

  • [1 mark] Write MX=B, where M=\(\begin{bmatrix}1 & 1 & 1\\4 & 3 & 2\\6 & 2 & 3\end{bmatrix}\), X=\(\begin{bmatrix}x\\y\\z\end{bmatrix}\), and B=\(\begin{bmatrix}21\\60\\70\end{bmatrix}\).
  • [1 mark] The determinant is -5≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}5 & 0 & -10\\-1 & -3 & 4\\-1 & 2 & -1\end{bmatrix}\).
  • [1 mark] Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}5 & -1 & -1\\0 & -3 & 2\\-10 & 4 & -1\end{bmatrix}\), hence M⁻¹=adj(M)/(-5)=\(\begin{bmatrix}-1 & 1/5 & 1/5\\0 & 3/5 & -2/5\\2 & -4/5 & 1/5\end{bmatrix}\).
  • [1 mark] Multiply the inverse by the right side: X=M⁻¹B=\(\begin{bmatrix}5\\8\\8\end{bmatrix}\). The resulting entries satisfy MX=B by direct multiplication.
  • [1 mark] Therefore x=5, y=8, z=8.

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