If \[\begin{bmatrix} 2x - y & x + 2y \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 2 & 3 \end{bmatrix}\]
then the value of $x$ and $y$ will be:
We are given the matrix equation:
\[\begin{bmatrix} 2x - y & x + 2y \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 2 & 3 \end{bmatrix}\]
Step 1: Compare corresponding elements of the matrices.
From the equation, we can equate the corresponding elements of both matrices. 1. From the top-left element: \[ 2x - y = 1 \text{(Equation 1)}. \] 2. From the top-right element: \[ x + 2y = 3 \text{(Equation 2)}. \] 3. From the bottom-left element: \[ 2 = 2 \text{(This is true, so no further information here)}. \] 4. From the bottom-right element: \[ 3 = 3 \text{(This is true, so no further information here)}. \]
Step 2: Solve the system of equations.
We now have the system of equations: 1. \(2x - y = 1\) 2. \(x + 2y = 3\) Let's solve this system. From Equation 1, solve for \(y\): \[ y = 2x - 1. \] Substitute this into Equation 2: \[ x + 2(2x - 1) = 3, \] \[ x + 4x - 2 = 3, \] \[ 5x = 5, \] \[ x = 1. \] Now substitute \(x = 1\) into the expression for \(y\): \[ y = 2(1) - 1 = 1. \]
Step 3: Conclusion.
The solution to the system is \(x = 1\) and \(y = 1\), so the correct answer is (A) \(x = 1, y = 1\).
If \[ A = \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix} \] then find \( A^{-1} \). Hence, solve the system of linear equations: \[ x - 2y = 10, \] \[ 2x - y - z = 8, \] \[ -2y + z = 7. \]