Using elementary transformations, find the inverse of the matrix A=843211122 use it to solve the following system of linear equations: 8x + 4y + 3z = 19 2x + y + z = 5 and x + 2y + 2z = 7.

Given matrix is A=843211122 Here, we use elementary row transformations, so we consider A=IA⇒843211122=100010001A On applying R3→R3-2R2, we get 843211-300=1000100-21AOn applying R3→-13R3, we get 843211100=100010023-13A On applying R3↔R1, we get 100211843=023-13010100A On applying R3→R3- R2, we get 10021100-1=023-130101-40A On applying R2→R2-2R1, we get 10021100-1=023-130-13231-40A On applying R2→R2-2R1, we get 10021100-1=023-130-13231-40A On applying R2→R2+R3, we get 10001000-1=023-131-133231-40A On applying R3→(-1)R3, we get 100010001=023-130-13323-140A Thus, A-1=023-131-13323-140 Given system of equations can be written in matrix form as AX = B,where, A=843211122,A=xyzand B=1957 ⇒X=A-1B=023-131-13323-1401957 ⇒xyz=0+103-7319-653+143-19+20+0=121On comparing the corresponding elements, we getx = 1, y = 2 and z = 1.

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Using elementary transformations, find the inverse of the matrix $A = [\begin{matrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{matrix}]$ use it to solve the following system of linear equations:
8x + 4y + 3z = 19
2x + y + z = 5
and x + 2y + 2z = 7.

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Detailed Solution

$Given matrix is A = [\begin{matrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{matrix}] Here, we use elementary row transformations, so we consider A = IA \Rightarrow [\begin{matrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A On applying R_{3} \to R_{3} - 2 R_{2}, we get [\begin{matrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ - 3 & 0 & 0 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & - 2 & 1 \end{matrix}] A$ $On applying R_{3} \to - \frac{1}{3} R_{3}, we get [\begin{matrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 0 & 0 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & \frac{2}{3} & - \frac{1}{3} \end{matrix}] A On applying R_{3} \leftrightarrow R_{1}, we get [\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 1 \\ 8 & 4 & 3 \end{matrix}] = [\begin{matrix} 0 & \frac{2}{3} & - \frac{1}{3} \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}] A On applying R_{3} \to R_{3} - R_{2}, we get [\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 1 \\ 0 & 0 & - 1 \end{matrix}] = [\begin{matrix} 0 & \frac{2}{3} & - \frac{1}{3} \\ 0 & 1 & 0 \\ 1 & - 4 & 0 \end{matrix}] A On applying R_{2} \to R_{2} - 2 R_{1}, we get [\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 1 \\ 0 & 0 & - 1 \end{matrix}] = [\begin{matrix} 0 & \frac{2}{3} & - \frac{1}{3} \\ 0 & - \frac{1}{3} & \frac{2}{3} \\ 1 & - 4 & 0 \end{matrix}] A On applying R_{2} \to R_{2} - 2 R_{1}, we get [\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 1 \\ 0 & 0 & - 1 \end{matrix}] = [\begin{matrix} 0 & \frac{2}{3} & - \frac{1}{3} \\ 0 & - \frac{1}{3} & \frac{2}{3} \\ 1 & - 4 & 0 \end{matrix}] A On applying R_{2} \to R_{2} + R_{3}, we get [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{matrix}] = [\begin{matrix} 0 & \frac{2}{3} & - \frac{1}{3} \\ 1 & - \frac{13}{3} & \frac{2}{3} \\ 1 & - 4 & 0 \end{matrix}] A On applying R_{3} \to (- 1) R_{3}, we get [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 0 & \frac{2}{3} & - \frac{1}{3} \\ 0 & - \frac{13}{3} & \frac{2}{3} \\ - 1 & 4 & 0 \end{matrix}] A Thus, A^{- 1} = [\begin{matrix} 0 & \frac{2}{3} & - \frac{1}{3} \\ 1 & - \frac{13}{3} & \frac{2}{3} \\ - 1 & 4 & 0 \end{matrix}]$

Given system of equations can be written in matrix form as AX = B,

$where, A = [\begin{matrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{matrix}], A = [\begin{matrix} x \\ y \\ z \end{matrix}] and B = [\begin{matrix} 19 \\ 5 \\ 7 \end{matrix}] \Rightarrow X = A^{- 1} B = [\begin{matrix} 0 & \frac{2}{3} & - \frac{1}{3} \\ 1 & - \frac{13}{3} & \frac{2}{3} \\ - 1 & 4 & 0 \end{matrix}] [\begin{matrix} 19 \\ 5 \\ 7 \end{matrix}] \Rightarrow [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 0 + \frac{10}{3} - \frac{7}{3} \\ 19 - \frac{65}{3} + \frac{14}{3} \\ - 19 + 20 + 0 \end{matrix}] = [\begin{matrix} 1 \\ 2 \\ 1 \end{matrix}]$