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yulyashka [42]
2 years ago
13

Use Cramer’s rule to solve for x: x + 4y − z = −14 5x + 6y + 3z = 4 −2x + 7y + 2z = −17

Mathematics
1 answer:
V125BC [204]2 years ago
5 0

Looks like the system is

x + 4y - z = -14

5x + 6y + 3z = 4

-2x + 7y + 2z = -17

or in matrix form,

\mathbf{Ax} = \mathbf b \iff \begin{bmatrix} 1 & 4 & -1 \\ 5 & 6 & 3 \\ -2 & 7 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -14 \\ 4 \\ -17 \end{bmatrix}

Cramer's rule says that

x_i = \dfrac{\det \mathbf A_i}{\det \mathbf A}

where x_i is the solution for i-th variable, and \mathbf A_i is a modified version of \mathbf A with its i-th column replaced by \mathbf b.

We have 4 determinants to compute. I'll show the work for det(A) using a cofactor expansion along the first row.

\det \mathbf A = \begin{vmatrix} 1 & 4 & -1 \\ 5 & 6 & 3 \\ -2 & 7 & 2 \end{vmatrix}

\det \mathbf A = \begin{vmatrix} 6 & 3 \\ 7 & 2 \end{vmatrix} - 4 \begin{vmatrix} 5 & 3 \\ -2 & 2 \end{vmatrix} - \begin{vmatrix} 5 & 6 \\ -2 & 7 \end{vmatrix}

\det \mathbf A = ((6\times2)-(3\times7)) - 4((5\times2)-(3\times(-2)) - ((5\times7)-(6\times(-2)))

\det\mathbf A = 12 - 21 - 40 - 24 - 35 - 12 = -120

The modified matrices and their determinants are

\mathbf A_1 = \begin{bmatrix} -14 & 4 & -1 \\ 4 & 6 & 3 \\ -17 & 7 & 2\end{bmatrix} \implies \det\mathbf A_1 = -240

\mathbf A_2 = \begin{bmatrix} 1 & -14 & -1 \\ 5 & 4 & 3 \\ -2 & -17 & 2 \end{bmatrix} \implies \det\mathbf A_2 = 360

\mathbf A_3 = \begin{bmatrix} 1 & 4 & -14 \\ 5 & 6 & 4 \\ -2 & 7 & -17 \end{bmatrix} \implies \det\mathbf A_3 = -480

Then by Cramer's rule, the solution to the system is

x = \dfrac{-240}{-120} \implies \boxed{x = 2}

y = \dfrac{360}{-120} \implies \boxed{y = -3}

z = \dfrac{-480}{-120} \implies \boxed{z = 4}

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