Find the inverse of the matrix A = 1 & 1 3 & 1 .

Question

Find the inverse of the matrix A = 1 & 1 3 & 1 .

Asked on April 24, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-04-24T10:43:14.053Z

5. Find the inverse of the matrix $\begin{pmatrix} 1 & 1 \\ 3 & 1 \end{pmatrix}$. Hence, determine the point of intersection of the lines $y+x=7$ and $3x+y=15$. Step 1: Find the inverse of the matrix $A = \begin{pmatrix} 1 & 1 \\ 3 & 1 \end{pmatrix}$. The determinant of $A$ is given by $\det(A) = (1)(1) - (1)(3) = 1 - 3 = -2$. The inverse matrix $A^{-1}$ is calculated as: $$ A^{-1} = \frac{1}{\det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} $$ $$ A^{-1} = \frac{1}{-2} \begin{pmatrix} 1 & -1 \\ -3 & 1 \end{pmatrix} $$ $$ A^{-1} = \begin{pmatrix} -\frac{1}{2} & \frac{1}{2} \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix} $$ The inverse of the matrix is $\boxed{\begin{pmatrix} -\frac{1}{2} & \frac{1}{2} \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix}}$. Step 2: Write the system of linear equations in matrix form. The given equations are: $x+y=7$ $3x+y=15$ This can be written as $AX=B$: $$ \begin{pmatrix} 1 & 1 \\ 3 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7 \\ 15 \end{pmatrix} $$ Step 3: Use the inverse matrix to solve for $X$. $$ X = A^{-1}B $$ $$ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -\frac{1}{2} & \frac{1}{2} \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix} \begin{pmatrix} 7 \\ 15 \end{pmatrix} $$ Step 4: Perform the matrix multiplication. $$ x = \left(-\frac{1}{2}\right)(7) + \left(\frac{1}{2}\right)(15) = -\frac{7}{2} + \frac{15}{2} = \frac{8}{2} = 4 $$ $$ y = \left(\frac{3}{2}\right)(7) + \left(-\frac{1}{2}\right)(15) = \frac{21}{2} - \frac{15}{2} = \frac{6}{2} = 3 $$ The point of intersection is $(x, y) = \boxed{(4, 3)}$. That's 2 down. 3 left today — send the next one.

Accepted Answer · 2026-04-24T10:43:14.053Z

5. Find the inverse of the matrix $\begin{pmatrix} 1 & 1 \\ 3 & 1 \end{pmatrix}$. Hence, determine the point of intersection of the lines $y+x=7$ and $3x+y=15$. Step 1: Find the inverse of the matrix $A = \begin{pmatrix} 1 & 1 \\ 3 & 1 \end{pmatrix}$. The determinant of $A$ is given by $\det(A) = (1)(1) - (1)(3) = 1 - 3 = -2$. The inverse matrix $A^{-1}$ is calculated as: $$ A^{-1} = \frac{1}{\det(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} $$ $$ A^{-1} = \frac{1}{-2} \begin{pmatrix} 1 & -1 \\ -3 & 1 \end{pmatrix} $$ $$ A^{-1} = \begin{pmatrix} -\frac{1}{2} & \frac{1}{2} \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix} $$ The inverse of the matrix is $\boxed{\begin{pmatrix} -\frac{1}{2} & \frac{1}{2} \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix}}$. Step 2: Write the system of linear equations in matrix form. The given equations are: $x+y=7$ $3x+y=15$ This can be written as $AX=B$: $$ \begin{pmatrix} 1 & 1 \\ 3 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7 \\ 15 \end{pmatrix} $$ Step 3: Use the inverse matrix to solve for $X$. $$ X = A^{-1}B $$ $$ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -\frac{1}{2} & \frac{1}{2} \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix} \begin{pmatrix} 7 \\ 15 \end{pmatrix} $$ Step 4: Perform the matrix multiplication. $$ x = \left(-\frac{1}{2}\right)(7) + \left(\frac{1}{2}\right)(15) = -\frac{7}{2} + \frac{15}{2} = \frac{8}{2} = 4 $$ $$ y = \left(\frac{3}{2}\right)(7) + \left(-\frac{1}{2}\right)(15) = \frac{21}{2} - \frac{15}{2} = \frac{6}{2} = 3 $$ The point of intersection is $(x, y) = \boxed{(4, 3)}$. That's 2 down. 3 left today — send the next one.

Find the inverse of the matrix A = 1 & 1 3 & 1 .