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Class 12 Determinants · Very Long Answer5 Marksmatrix_method_linear_systems

Class 12 Maths · Determinants · Very Long Answer #58

Given A=\(\begin{bmatrix}1 & 2 & -3\\3 & 2 & -2\\2 & -1 & 1\end{bmatrix}\), find A⁻¹ and use it to solve by the matrix method: (1)x + (2)y + (-1)z = 6; (3)x + (2)y + (-2)z = 3; (2)x + (-1)y + (1)z = 2.

✅ Step-by-Step Solution

Final Answer: x=1, y=5, z=5
  1. The supplied A has det(A)=7 and A⁻¹=\(\begin{bmatrix}0 & 1/7 & 2/7\\-1 & 1 & -1\\-1 & 5/7 & -4/7\end{bmatrix}\). However, the printed system has coefficient matrix M=\(\begin{bmatrix}1 & 2 & -1\\3 & 2 & -2\\2 & -1 & 1\end{bmatrix}\), not A. Retain both and solve MX=B with B=[6,3,2]ᵀ.
  2. The determinant is -7≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}0 & -7 & -7\\-1 & 3 & 5\\-2 & -1 & -4\end{bmatrix}\).
  3. Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}0 & -1 & -2\\-7 & 3 & -1\\-7 & 5 & -4\end{bmatrix}\), hence M⁻¹=adj(M)/(-7)=\(\begin{bmatrix}0 & 1/7 & 2/7\\1 & -3/7 & 1/7\\1 & -5/7 & 4/7\end{bmatrix}\).
  4. Multiply the inverse by the right side: X=M⁻¹B=\(\begin{bmatrix}1\\5\\5\end{bmatrix}\). The resulting entries satisfy MX=B by direct multiplication.
  5. Therefore x=1, y=5, z=5.

📋 CBSE Stepwise Marking Scheme

  • [1 mark] The supplied A has det(A)=7 and A⁻¹=\(\begin{bmatrix}0 & 1/7 & 2/7\\-1 & 1 & -1\\-1 & 5/7 & -4/7\end{bmatrix}\). However, the printed system has coefficient matrix M=\(\begin{bmatrix}1 & 2 & -1\\3 & 2 & -2\\2 & -1 & 1\end{bmatrix}\), not A. Retain both and solve MX=B with B=[6,3,2]ᵀ.
  • [1 mark] The determinant is -7≠0. The signed-minor cofactor matrix is \(\begin{bmatrix}0 & -7 & -7\\-1 & 3 & 5\\-2 & -1 & -4\end{bmatrix}\).
  • [1 mark] Transpose that cofactor matrix: adj(M)=\(\begin{bmatrix}0 & -1 & -2\\-7 & 3 & -1\\-7 & 5 & -4\end{bmatrix}\), hence M⁻¹=adj(M)/(-7)=\(\begin{bmatrix}0 & 1/7 & 2/7\\1 & -3/7 & 1/7\\1 & -5/7 & 4/7\end{bmatrix}\).
  • [1 mark] Multiply the inverse by the right side: X=M⁻¹B=\(\begin{bmatrix}1\\5\\5\end{bmatrix}\). The resulting entries satisfy MX=B by direct multiplication.
  • [1 mark] Therefore x=1, y=5, z=5.

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