Given that $$ \frac{1}{21}\left(\begin{array}{cccc} 4 & 51 & 49 & -28 \\ -9 & -57 & -63 & 42 \\ -6 & -66 & -63 & 42 \\ 4 & 9 & 7 & -7 \end{array}\right) $$ is the inverse of $$ \left(\begin{array}{cccc} 3 & -1 & 4 & 6 \\ 1 & 2 & -1 & 2 \\ 2 & -2 & 3 & -2 \\ 5 & 0 & 4 & 1 \end{array}\right) $$ solve the following systems of equations: $$ \text { (a) } \begin{aligned} &3 w-x+4 y+6 z=19 \\ &w+2 x-y+2 z=11 \\ &2 w-2 x+3 y-2 z=-11 \\ &5 w+4 y+z=2 \end{aligned} $$ $$ \text { (b) } \begin{aligned} &2 \alpha-2 \beta+3 \gamma-2 \delta=11 \\ &\quad \alpha+2 \beta-\gamma+2 \delta=-2 \\ &\quad 5 \alpha+4 \gamma+\delta=15 \\ &3 \alpha-\beta+4 \gamma+6 \delta=1 \end{aligned} $$

Step by step solution

01

Write down the given inverse matrix and systems of equations

We are given the following inverse matrix and systems of equations to solve: Inverse matrix: $$ A^{-1} = \frac{1}{21}\left(\begin{array}{cccc} 4 & 51 & 49 & -28 \\\ -9 & -57 & -63 & 42 \\\ -6 & -66 & -63 & 42 \\\ 4 & 9 & 7 & -7 \end{array}\right) $$ System (a): $$ \begin{aligned} &3 w-x+4 y+6 z=19 \\\ &w+2 x-y+2 z=11 \\\ &2 w-2 x+3 y-2 z=-11 \\\ &5 w+4 y+z=2 \end{aligned} $$ System (b): $$ \begin{aligned} &2 \alpha-2 \beta+3 \gamma-2 \delta=11 \\\ &\quad \alpha+2 \beta-\gamma+2 \delta=-2 \\\ &\quad 5 \alpha+4 \gamma+\delta=15 \\\ &3 \alpha-\beta+4 \gamma+6 \delta=1 \end{aligned} $$

02

Multiply the inverse matrix by the right-hand side column vectors

For system (a), create a column vector with the constants on the right-hand side of the equations and multiply it by the inverse matrix A^{-1}: $$ A^{-1} \begin{bmatrix} 19 \\ 11 \\ -11 \\ 2 \end{bmatrix} = \begin{bmatrix} w \\ x \\ y \\ z \end{bmatrix} $$ For system (b), create a column vector with the constants on the right-hand side of the equations and multiply it by the inverse matrix A^{-1}: $$ A^{-1} \begin{bmatrix} 11 \\ -2 \\ 15 \\ 1 \end{bmatrix} = \begin{bmatrix} \alpha \\ \beta \\ \gamma \\ \delta \end{bmatrix} $$

03

Compute the solutions for systems (a) and (b)

After multiplying the inverse matrix A^{-1} with the right-hand side column vectors, we obtain the solutions for systems (a) and (b): System (a) solution: $$ \begin{bmatrix} w \\ x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \\ -3 \\ 4 \end{bmatrix} $$ System (b) solution: $$ \begin{bmatrix} \alpha \\ \beta \\ \gamma \\ \delta \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -1 \\ 0 \end{bmatrix} $$ So, the solutions for systems (a) and (b) are: System (a): w = 2, x = 0, y = -3, z = 4 System (b): α = 1, β = 2, γ = -1, δ = 0

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