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Q.(a) Find the inverse of the matrix : A=[−1123−11−134]A = \begin{bmatrix} -1 & 1 & 2 \\ 3 & -1 & 1 \\ -1 & 3 & 4 \end{bmatrix} and hence show that AA−1=IAA^{-1} = I.

(OR)
(b) Using matrix method, solve the following system of equations for xx, yy and zz : x−y+z=4x - y + z = 4 2x+y−3z=02x + y - 3z = 0 x+y+z=2x + y + z = 2
CBSECBSE Class XII Board 2023Subjective· 5mImportance★★★★★
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  1. A−1=110[−723−13−2782−2]A^{-1}=\tfrac{1}{10}\begin{bmatrix}-7&2&3\\-13&-2&7\\8&2&-2\end{bmatrix};
  2. x=2, y=−1, z=1x=2,\ y=-1,\ z=1.

Part (a) — Inverse of A=[−1123−11−134]A=\begin{bmatrix}-1&1&2\\3&-1&1\\-1&3&4\end{bmatrix}

A−1=1det⁡A adj(A)A^{-1}=\dfrac{1}{\det A}\,\text{adj}(A), where adj(A)\text{adj}(A) is the transpose of the cofactor matrix.

  1. det⁡A=−1[(−1)(4)−(1)(3)]−1[(3)(4)−(1)(−1)]+2[(3)(3)−(−1)(−1)]\det A=-1\big[(-1)(4)-(1)(3)\big]-1\big[(3)(4)-(1)(-1)\big]+2\big[(3)(3)-(-1)(-1)\big] =−1(−7)−1(13)+2(8)=7−13+16=10.=-1(-7)-1(13)+2(8)=7-13+16=10.
  2. Cofactors: C11=−7, C12=−13, C13=8, C21=2, C22=−2, C23=2, C31=3, C32=7, C33=−2.C_{11}=-7,\ C_{12}=-13,\ C_{13}=8,\ C_{21}=2,\ C_{22}=-2,\ C_{23}=2,\ C_{31}=3,\ C_{32}=7,\ C_{33}=-2.
  3. Adjoint (transpose of cofactor matrix): adj(A)=[−723−13−2782−2].\text{adj}(A)=\begin{bmatrix}-7&2&3\\-13&-2&7\\8&2&-2\end{bmatrix}.
  4. A−1=110[−723−13−2782−2].A^{-1}=\dfrac{1}{10}\begin{bmatrix}-7&2&3\\-13&-2&7\\8&2&-2\end{bmatrix}. …

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