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IfA=\begin{bmatrix} 5 &-1 &4 \\ 2 &3 &5 \\ 5& -2 &6 \end{bmatrix} , find A^{-1}and use it to solve the following system of equations :

5x - y + 4z = 5

2x + 3y + 5z = 2

5x - 2y + 6z = - 1

 

 

 
 
 
 
 

Answers (1)

Given:

A=\begin{bmatrix} 5 &-1 &4 \\ 2 &3 &5 \\ 5& -2 &6 \end{bmatrix} 

|\mathrm{A}|=5(18+10)+1(12-25)+4(-4-15) = 140 -13 - 76 = 51

Cofactors:

\\ \mathrm{A}_{11}=28 \ \ \ \ \mathrm{A}_{12}=13 \ \ \ \ \mathrm{A}_{13}=-19 \\ \mathrm{A}_{21}=-2 \ \ \ \ \mathrm{A}_{22}=10 \ \ \ \ \mathrm{A}_{23}=5 \\ \mathrm{A}_{31}=-17 \ \ \ \mathrm{A}_{32}=-17 \ \ \ \mathrm{A}_{33}=17 \\

\mathrm{Adj A}=\left[\begin{array}{ccc} 28 & 13 & -19 \\ -2 & 10 & 5 \\ -17 & -17 & 17 \end{array}\right]^T

\mathrm{A}^{-1}=\frac{AdjA}{|A|}

\mathrm{A}^{-1}=\frac{1}{51}\left[\begin{array}{ccc} 28 & -2 & -17 \\ 13 & 10 & -17 \\ -19 & 5 & 17 \end{array}\right]

Now Given lineear equations-

5x - y + 4z = 5

2x + 3y + 5z = 2

5x - 2y + 6z = - 1

The system of lineear equations can be written as

\mathrm{AX}=\mathrm{B}

\\ \mathrm{X}=\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right], \mathrm{B}=\left[\begin{array}{r} 5 \\ 2 \\ -1 \end{array}\right]

\mathrm{X}=\mathrm{A}^{-1} \mathrm{B}

X=\frac{1}{51}\left[\begin{array}{ccc} 28 & -2 & -17 \\ 13 & 10 & -17 \\ -19 & 5 & 17 \end{array}\right]\left[\begin{array}{r} 5 \\ 2 \\ -1 \end{array}\right]

\\ \left[\begin{array}{r} x \\ y \\ z \end{array}\right]=\frac{1}{51}\left[\begin{array}{r} 153 \\ 102 \\ -102 \end{array}\right]

\\ \left[\begin{array}{r} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} 3 \\ 2 \\ -2 \end{array}\right]\\ \\ x=3, y=2, z=-2

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Safeer PP

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