A determinant is a function whose input is a square matrix and whose output is a number. If \(A\) is a square matrix, then the determinant of \(A\) is represented by either \(| A |\) or \(\det ( A )\).
Do not confound the determinant symbol with absolute value. A determinant can be a negative number.
For a \(2 \times 2\) matrix, the determinant is \(\begin{vmatrix} a_1 & b_1\\ a_2 & b_2 \end{vmatrix}=a_1b_2-a_2b_1.\)
For a \(2\times 2\) matrix, the determinant equals the product of the two numbers on the main diagonal minus the product of the two numbers on the minor diagonal.Compute the determinant:
\(\begin{vmatrix} 3 & ~~~5\\ 1 & -2 \end{vmatrix}\)
Let \(A=\begin{vmatrix} a_1 & b_1 & c_1\\a_2 & b_2 & c_2\\a_3 & b_3 & c_3 \end{vmatrix}\) then
\[\det(A)= a_1b_2c_3 +a_2b_3c_1+a_3b_1c_2-a_3b_2c_1-a_2b_1c_3-a_1b_3c_2\]Find the determinant of \( \begin{vmatrix} -1 & 2 & 0\\ \enspace\,3 & 1 & 5\\ -2 & 0 & 1 \end{vmatrix}\)
Cramer's Rule utilizes the coefficient matrix of a system of linear equations and determinants to find the solution.
Consider the general system of two linear equations in two variables:
\(\begin{eqnarray*} a_1x + b_1y &=& c_1\\ a_2x + b_2y &=& c_2 \end{eqnarray*}\)The coefficient matrix is
\( \left[\begin{array}{rr|r} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ \end{array}\right]\)
Three determinants are defined in terms of the coefficient matrix.
The matrix \(D\) is formed from the first two columns: \(D=\begin{vmatrix}a_1 & b_1\\a_2 & b_2\end{vmatrix}\).
The matrix \(D_x\) is formed from the \(D\) matrix by replacing the \(x\)-column with the \(c\)-column: \(D_x=\begin{vmatrix}c_1 & b_1\\c_2 & b_2\end{vmatrix}\).
The matrix \(D_y\) is formed from the \(D\) matrix by replacing the \(y\)-column with the \(c\)-column: \(D_y=\begin{vmatrix}a_1 & c_1\\a_2 & c_2\end{vmatrix}\).
Then Cramer's Rule finds the solution of the system of equations as
\[x=\frac{D_x}{D}\qquad y=\frac{D_y}{D}\]
Use Cramer's Rule to solve the system:
\(\begin{eqnarray*} 5x + 3y &=& -2\\-3x + ~~y &=& ~~4 \end{eqnarray*}\)
Consider the general system of three linear equations is three variables:
\(\begin{eqnarray*} a_1x + b_1y +c_1z&=& d_1\\ a_2x + b_2y +c_2z&=& d_2\\ a_3x + b_3y +c_3z&=& d_3\\ \end{eqnarray*}\)
The coefficient matrix is:
\( \left[\begin{array}{rrr|r} a_1 & b_1 & c_1 & d_1\\ a_2 & b_2 & c_2 & d_2\\ a_3 & b_3 & c_3 & d_3 \end{array}\right]\)
The determinants are:
\[ D= \begin{vmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{vmatrix}\quad D_x= \begin{vmatrix} d_1 & b_1 & c_1\\ d_2 & b_2 & c_2\\ d_3 & b_3 & c_3 \end{vmatrix}\quad D_y= \begin{vmatrix} a_1 & d_1 & c_1\\ a_2 & d_2 & c_2\\ a_3 & d_3 & c_3 \end{vmatrix}\quad D_z= \begin{vmatrix} a_1 & b_1 & d_1\\ a_2 & b_2 & d_2\\ a_3 & b_3 & d_3 \end{vmatrix} \]
The solution by Cramer's Rule is:
\[x=\frac{D_x}{D}\qquad y=\frac{D_y}{D}\qquad z=\frac{D_z}{D}\]
Use Cramer's Rule to solve the system:
\(\begin{eqnarray*} 2x -5y + 2z&=& ~~0\\ ~~x + 3y +~~z&=& -1\\ y - 3z &=& ~~0\\ \end{eqnarray*}\)