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Question 146 of 146

Q.If A=(2−3532−411−2)A = \begin{pmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{pmatrix}, find A−1A^{-1}. Use it to solve the system of equations 2x−3y+5z=112x - 3y + 5z = 11, 3x+2y−4z=−53x + 2y - 4z = -5, x+y−2z=−3x + y - 2z = -3. OR Using elementary row transformations, find the inverse of the matrix A=(123257−2−4−5)A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{pmatrix}.

Uttar Pradesh UpmspCBSE Class XII Board 2018Subjective· 6mImportance★★★★★
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A−1=(01−2−29−23−15−13)A^{-1}=\begin{pmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{pmatrix} and the system solves to x=1, y=2, z=3x=1,\ y=2,\ z=3.

Concept. A−1=1det⁡A adj AA^{-1}=\dfrac{1}{\det A}\,\text{adj}\,A; a system AX=BAX=B has solution X=A−1BX=A^{-1}B.

Why this method. With A−1A^{-1} known, the solution is one matrix multiplication.

Working. A=(2−3532−411−2)A=\begin{pmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{pmatrix}.

det⁡A=2(0)+3(−2)+5(1)=−1.\det A=2(0)+3(-2)+5(1)=-1.

Cofactors give adj A=(0−122−9231−513)\text{adj}\,A=\begin{pmatrix}0&-1&2\\2&-9&23\\1&-5&13\end{pmatrix}, so

A−1=1−1 adj A=(01−2−29−23−15−13).A^{-1}=\frac{1}{-1}\,\text{adj}\,A=\begin{pmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{pmatrix}.

With B=(11−5−3)B=\begin{pmatrix}11\\-5\\-3\end{pmatrix}, X=A−1BX=A^{-1}B:

x=0(11)+1(−5)−2(−3)=1,x=0(11)+1(-5)-2(-3)=1,

y=−2(11)+9(−5)−23(−3)=−22−45+69=2,y=-2(11)+9(-5)-23(-3)=-22-45+69=2, …

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