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Q.Find the inverse of the matrix A=[20−1510013]A = \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}. OR Solve the system of equations by matrix method: 3x−2y+3z=83x - 2y + 3z = 8, 2x+y−z=12x + y - z = 1, 4x−3y+2z=44x - 3y + 2z = 4.

Uttar Pradesh UpmspUP Board (UPMSP) Intermediate 2025Subjective· 8mImportance★★★★★
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∣A∣=1|A|=1; forming the cofactors and transposing gives adj AA, so A−1=A^{-1}= adj AA.

Concept. For an invertible matrix, A−1=1∣A∣ adj(A)A^{-1}=\dfrac{1}{|A|}\,\text{adj}(A), where adj(A)(A) is the transpose of the cofactor matrix.

A=[20−1510013].A=\begin{bmatrix}2&0&-1\\5&1&0\\0&1&3\end{bmatrix}.

Determinant (expand along row 1):

∣A∣=2(1⋅3−0⋅1)−0+(−1)(5⋅1−1⋅0)=2(3)−5=6−5=1.|A|=2(1\cdot3-0\cdot1)-0+(-1)(5\cdot1-1\cdot0)=2(3)-5=6-5=1.

Since ∣A∣=1≠0|A|=1\ne0, A−1A^{-1} exists.

Cofactors:

C11=+3, C12=−15, C13=+5,C_{11}=+3,\ C_{12}=-15,\ C_{13}=+5,

C21=−1, C22=+6, C23=−2,C_{21}=-1,\ C_{22}=+6,\ C_{23}=-2,

C31=+1, C32=−5, C33=+2.C_{31}=+1,\ C_{32}=-5,\ C_{33}=+2.

Cofactor matrix =[3−155−16−21−52]=\begin{bmatrix}3&-15&5\\-1&6&-2\\1&-5&2\end{bmatrix}; its transpose gives …

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