Example Question - matrix
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Matrix Equation Solution
<p>Let A = \begin{pmatrix} 0 & 1 \\ y & 5 \end{pmatrix}, B = \begin{pmatrix} 4 & -1 \\ 6 & x \end{pmatrix}, C = \begin{pmatrix} 4 & 0 \\ x & 7 \end{pmatrix}.</p> <p>Then, we have:</p> <p>A + B = C.</p> <p>Thus, we can write:</p> <p>\begin{pmatrix} 0 + 4 & 1 - 1 \\ y + 6 & 5 + x \end{pmatrix} = \begin{pmatrix} 4 & 0 \\ x & 7 \end{pmatrix}.</p> <p>From this, we can derive the equations:</p> <p>0 + 4 = 4,</p> <p>1 - 1 = 0,</p> <p>y + 6 = x,</p> <p>5 + x = 7.</p> <p>Solve the last equation for x:</p> <p>x = 7 - 5 = 2.</p> <p>Then substitute x back into the third equation:</p> <p>y + 6 = 2.</p> <p>y = 2 - 6 = -4.</p> <p>The values of y and x are:</p> <p>x = 2,</p> <p>y = -4.</p>
Determinants of Matrices with Variable Elements
Since this image contains determinant calculations and operations on matrices with variables, I will provide the solution to find the determinant based on the visible matrices in the image. <p>\text{Let's calculate the determinant of the first visible matrix:}</p> <p>\begin{vmatrix} a-1 & 3 & 5 \\ -1 & 3-5 & 5 \\ 1 & 3 & 5 \end{vmatrix}</p> <p>\text{Expanding along the first column:}</p> <p>(a-1)\begin{vmatrix} 3-5 & 5 \\ 3 & 5 \end{vmatrix} -(-1)\begin{vmatrix} -1 & 5 \\ 1 & 5 \end{vmatrix} + 1\begin{vmatrix} -1 & 3-5 \\ 1 & 3 \end{vmatrix}</p> <p>=(a-1)((3-5)(5)-5(3)) -(-1)((-1)(5)-5(1)) + (1)((-1)(3)-3(3-5))</p> <p>=(a-1)(-10+15) -(-1)(-5) + (1)(-3+6)</p> <p>=(a-1)(5) -5 + 3</p> <p>=5a-5 -5 +3</p> <p>=5a-7</p> <p>\text{Without more context or additional operations indicated for the other matrices, the calculation of their determinants isn't completed in the image.}</p>
Matrix Method for Counting Battery Production
<p>The question asks to find the number P of batteries in a certain box out of a total Q using a matrix method. This is a mathematical problem involving systems of equations that can be expressed in matrix form.</p> <p>Let the number of batteries on Monday be M and on Tuesday be T. We have:</p> <p>M = 960</p> <p>T = 1016</p> <p>Assuming there is a linear relationship between the days and the number of batteries, we can express this as a system of equations, with x representing the constant difference in daily production, and P being the production on an unknown day (using a 0-based index for days where Monday is day 0):</p> <p>M = P + 0*x</p> <p>T = P + 1*x</p> <p>We can then solve for P and x using matrix operations:</p> <p>\[ \begin{bmatrix}1 & 0\\1 & 1\end{bmatrix} * \begin{bmatrix}P\\x\end{bmatrix} = \begin{bmatrix}960\\1016\end{bmatrix} \]</p> <p>To find the values of P and x, we can use matrix inversion or other methods:</p> <p>\[ \begin{bmatrix}P\\x\end{bmatrix} = \begin{bmatrix}1 & 0\\1 & 1\end{bmatrix}^{-1} * \begin{bmatrix}960\\1016\end{bmatrix} \]</p> <p>By finding the inverse of the coefficient matrix and multiplying it by the production matrix, we can find P. However, the question lacks sufficient information to calculate an exact answer without additional context or assumptions about the pattern of battery production.</p>
Solving a System of Linear Equations Using the Gauss-Jordan Method
<p>To solve the system of equations using the Gauss-Jordan method, we need to represent the system as an augmented matrix and then use row operations to get the matrix in reduced row echelon form.</p> <p>The augmented matrix for the system is:</p> <p>\[\begin{bmatrix} 3 & -0.1 & -0.2 & | & 7.85 \\ 0.1 & 7 & -0.3 & | & -19.3 \\ 0.3 & -0.2 & 10 & | & 71.4 \end{bmatrix}\]</p> <p>Step 1: Use the first row to eliminate the entries in the first column of the second and third rows.</p> <p>\[\begin{bmatrix} 3 & -0.1 & -0.2 & | & 7.85 \\ 0 & 7.0033 & -0.2933 & | & -19.475 \\ 0 & -0.13 & 9.94 & | & 70.345 \end{bmatrix}\]</p> <p>Step 2: Make the leading coefficient of the second row as 1.</p> <p>\[\begin{bmatrix} 3 & -0.1 & -0.2 & | & 7.85 \\ 0 & 1 & -0.0419 & | & -2.7817 \\ 0 & -0.13 & 9.94 & | & 70.345 \end{bmatrix}\]</p> <p>Step 3: Use the second row to eliminate the second column entries in the first and third rows.</p> <p>\[\begin{bmatrix} 3 & 0 & -0.1983 & | & 7.7239 \\ 0 & 1 & -0.0419 & | & -2.7817 \\ 0 & 0 & 9.9355 & | & 70.0891 \end{bmatrix}\]</p> <p>Step 4: Make the leading coefficient of the third row as 1.</p> <p>\[\begin{bmatrix} 3 & 0 & -0.1983 & | & 7.7239 \\ 0 & 1 & -0.0419 & | & -2.7817 \\ 0 & 0 & 1 & | & 7.0522 \end{bmatrix}\]</p> <p>Step 5: Use the third row to eliminate the third column entries in the first and second rows.</p> <p>\[\begin{bmatrix} 3 & 0 & 0 & | & 8.4617 \\ 0 & 1 & 0 & | & -2.5108 \\ 0 & 0 & 1 & | & 7.0522 \end{bmatrix}\]</p> <p>Step 6: Make the leading coefficient of the first row as 1.</p> <p>\[\begin{bmatrix} 1 & 0 & 0 & | & 2.8206 \\ 0 & 1 & 0 & | & -2.5108 \\ 0 & 0 & 1 & | & 7.0522 \end{bmatrix}\]</p> <p>Now, the matrix is in reduced row echelon form, and we can read the solutions for \(x_1\), \(x_2\), and \(x_3\) directly:</p> <p>\[x_1 = 2.8206\]</p> <p>\[x_2 = -2.5108\]</p> <p>\[x_3 = 7.0522\]</p> <p>The solution to the system of linear equations using the Gauss-Jordan method is \(x_1 = 2.8206\), \(x_2 = -2.5108\), and \(x_3 = 7.0522\).</p>
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