Solve the system using inverse matrices. {[9x+6y=0],[x+y=-1]:}

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Solve the system using inverse matrices.  {[9x+6y=0],[x+y=-1]:}

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Write Equations in Matrix Form: First, we need to write the system of equations in matrix form. The system is: $$ \begin{align*} 9x + 6y &= 0 \ x + y &= -1 \end{align*} $$ This can be written in matrix form as: $$ \begin{bmatrix} 9 & 6 \ 1 & 1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ -1 \end{bmatrix} $$ Let's denote the coefficient matrix as A, the variable matrix as X, and the constant matrix as B: $$ A = \begin{bmatrix} 9 & 6 \ 1 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \ y \end{bmatrix}, \quad B = \begin{bmatrix} 0 \ -1 \end{bmatrix} $$Find Determinant of A: Next, we need to find the inverse of matrix A, denoted as $ A^{-1} $. To find the inverse, we need to calculate the determinant of A and then use it to find the adjugate matrix. The determinant of A, denoted as det(A), is: $$ \text{det}(A) = (9)(1) - (6)(1) = 9 - 6 = 3 $$Find Adjugate of A: Since the determinant is not zero, A is invertible. Now we need to find the adjugate of A, which involves finding the matrix of cofactors and then transposing it. The matrix of cofactors for A is: $$ \text{Cof}(A) = \begin{bmatrix} 1 & -6 \ -1 & 9 \end{bmatrix} $$ Now we transpose it to get the adjugate matrix, denoted as adj(A): $$ \text{adj}(A) = \begin{bmatrix} 1 & -1 \ -6 & 9 \end{bmatrix} $$Find Inverse of A: Now we can find the inverse of A by dividing the adjugate of A by the determinant of A: $$ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) = \frac{1}{3} \begin{bmatrix} 1 & -1 \ -6 & 9 \end{bmatrix} = \begin{bmatrix} 1/3 & -1/3 \ -2 & 3 \end{bmatrix} $$Solve for X: We can now solve for X by multiplying the inverse of A with B: $$ X = A^{-1}B = \begin{bmatrix} 1/3 & -1/3 \ -2 & 3 \end{bmatrix} \begin{bmatrix} 0 \ -1 \end{bmatrix} = \begin{bmatrix} (1/3)(0) + (-1/3)(-1) \ (-2)(0) + (3)(-1) \end{bmatrix} = \begin{bmatrix} 1/3 \ -3 \end{bmatrix} $$Final Solution: The solution to the system of equations is: $$ x = 1/3, \quad y = -3 $$

Solve the system using inverse matrices. $\newline$ $\left\{\begin{array}{c} 9 x+6 y=0 \\ x+y=-1 \end{array}\right.$

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Solve the system using inverse matrices.\newline{9x+6y=0x+y=−1 \left\{\begin{array}{c} 9 x+6 y=0 \\ x+y=-1 \end{array}\right. {9x+6y=0x+y=−1​

Solve the system using inverse matrices. $\newline$ $\left\{\begin{array}{c} 9 x+6 y=0 \\ x+y=-1 \end{array}\right.$