<p>Given the equation:</p> <p>\(\begin{pmatrix} 0 & 1 \\ y & 5 \end{pmatrix} + \begin{pmatrix} 4 & -1 \\ 6 & x \end{pmatrix} = \begin{pmatrix} 4 & 0 \\ x & 7 \end{pmatrix}\)</p> <p>We can separate this into two equations by comparing corresponding elements:</p> <p>1. \(0 + 4 = 4\)</p> <p>2. \(1 - 1 = 0\)</p> <p>3. \(y + 6 = x\)</p> <p>4. \(5 + x = 7\)</p> <p>From equation 1 and 2, we directly confirm that both hold true. Now, solving for \(x\) from equation 4:</p> <p>From \(5 + x = 7\)</p> <p>\(x = 7 - 5\)</p> <p>Thus, \(x = 2\).</p> <p>Now substitute \(x\) into equation 3:</p> <p>\(y + 6 = 2\)</p> <p>Thus, \(y = 2 - 6\)</p> <p>So, \(y = -4\).</p> <p>Final values: \(x = 2, y = -4\).</p>
The image shows a matrix equation where matrix A is given, and we need to calculate the product of matrix A with another matrix. Matrix A is: \[ A = \begin{bmatrix} 3 & -5 & 6 \\ -2 & 4 & 2 \\ -1 & 0 & 3 \end{bmatrix} \] And it needs to be multiplied by the matrix: \[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \] This second matrix is the identity matrix \( I_3 \) of size 3x3. The matrix product of any matrix with the identity matrix of the appropriate size is the original matrix itself. That's because the identity matrix acts like the number 1 for matrix multiplication. Therefore, the product of matrix A with this identity matrix is matrix A unchanged: \[ A \cdot I_3 = \begin{bmatrix} 3 & -5 & 6 \\ -2 & 4 & 2 \\ -1 & 0 & 3 \end{bmatrix} \]
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