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  • Using matrix method, solve the system of equations 3x + 2y – 2z = 3, x + 2y + 3z = 6, 2x – y + z = 2.

Using matrix method, solve the system of equations 3x + 2y – 2z = 3, x + 2y + 3z = 6, 2x – y + z = 2.

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\\Given,\\ \vspace{\baselineskip} 3x + 2y -2z = 3,\\ \\ \vspace{\baselineskip}x + 2y + 3z = 6,\\ \\ \vspace{\baselineskip}2x -y + z = 2\\ \vspace{\baselineskip} We know,\\ \vspace{\baselineskip} AX = B\\

\\\left[\begin{array}{ccc}3 & 2 & -2 \\ 1 & 2 & 3 \\ 2 & -1 & 1\end{array}\right]\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right]=\left[\begin{array}{l}3 \\ 6 \\ 2\end{array}\right]

\therefore \mathrm{X}=\mathrm{A}^{-1} \mathrm{~B} \\ \mathrm{A}^{-1}=\frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|} \text{Find } |A| \text{. Expand } |A| \text{ along Column 1.}

\begin{array}{c} |\mathrm{A}|=\mathrm{a}_{11}(-1)^{1+1}\left|\begin{array}{cc} \mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right|+\mathrm{a}_{21}(-1)^{2+1}\left|\begin{array}{cc} \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right| +\mathrm{a}_{31}(-1)^{3+1}\left|\begin{array}{cc} \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{22} & \mathrm{a}_{23} \end{array}\right| \\ |\mathrm{A}|=(3)\left|\begin{array}{cc} 2 & 3 \\ -1 & 1 \end{array}\right|-(1)\left|\begin{array}{cc} 2 & -2 \\ -1 & 1 \end{array}\right|+2\left|\begin{array}{cc} 2 & -2 \\ 2 & 3 \end{array}\right| \end{array}

 

\\\begin{aligned} &=3(2+3)-1(2-2)+2(6+4)\\ &=3(5)+2(10)\\ &=15+20\\ &=35\\ &\text { To find adj } A\\ &a_{11}=\left|\begin{array}{cc} 2 & 3 \\ -1 & 1 \end{array}\right|=2+3=5\\ &a_{12}=-\left|\begin{array}{ll} 1 & 3 \\ 2 & 1 \end{array}\right|=-(1-6)=5\\ &a_{13}=\left|\begin{array}{cc} 1 & 2 \\ 2 & -1 \end{array}\right|=-1-4=-5 \end{aligned}

 

\\\begin{array}{l} a_{21}=-\left|\begin{array}{cc} 2 & -2 \\ -1 & 1 \end{array}\right|=-(2-2)=0 \\ a_{22}=\left|\begin{array}{cc} 3 & -2 \\ 2 & 1 \end{array}\right|=3+4=7 \\ a_{23}=-\left|\begin{array}{cc} 3 & 2 \\ 2 & -1 \end{array}\right|=-(-3-4)=-7 \\ a_{32}=-\left|\begin{array}{cc} 3 & -2 \\ 1 & 3 \end{array}\right|=-(9+2)=-11 \\ a_{33}=\left|\begin{array}{cc} 3 & 2 \\ 1 & 2 \end{array}\right|=6-2=4 \end{array}

\\\begin{array}{l} \therefore \text { adj } A=\left[\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right]^{T}=\left[\begin{array}{ccc} 5 & 5 & -5 \\ 0 & 7 & -7 \\ 10 & -11 & 4 \end{array}\right]^{T}=\left[\begin{array}{ccc} 5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4 \end{array}\right] \\ \therefore A^{-1}=\frac{\operatorname{adj} A}{|A|}=\frac{\left[\begin{array}{ccc} 5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4 \end{array}\right]}{35}=\frac{1}{35}\left[\begin{array}{ccc} 5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4 \end{array}\right] \\ \text { Now, } x=A^{-1} B \end{array}

\\\begin{array}{l} {\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\frac{1}{35}\left[\begin{array}{ccc} 5 & 0 & 10 \\ 5 & 7 & -11 \\ -5 & 7 & 4 \end{array}\right]\left[\begin{array}{l} 3 \\ 6 \\ 2 \end{array}\right]} \\\\ \Rightarrow\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\frac{1}{35}\left[\begin{array}{c} 15+20 \\ 15+42-22 \\ -15+42+8 \end{array}\right] \\ \\\Rightarrow\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\frac{1}{35}\left[\begin{array}{l} 35 \\ 35 \\ 35 \end{array}\right] \\ \\\Rightarrow\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right]=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right] \end{array}

 

∴ x = 1, y = 1 and z = 1

 

 

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