A matrix can serve as a device for representing and solving a system of equations. To express a system in matrix form, we extract the coefficients of the variables and the constants, and these become the entries of the matrix. We use a vertical line to separate the coefficient entries from the constants, essentially replacing the equal signs. When a system is written in this form, we call it an augmented matrix. Show
2×22\times 2 system of equations. 3x+ 4y=74x−2y=5\begin{array}{l}3x+4y=7\\ 4x - 2y=5\end{array} We can write this system as an augmented matrix: [344 −2 ∣ 7 5]\left[\begin{array}{rr}\qquad 3& \qquad 4\\ \qquad 4& \qquad -2\end{array}\text{ }|\text{ }\begin{array}{r}\qquad 7\\ \qquad 5\end{array}\right] We can also write a matrix containing just the coefficients. This is called the coefficient matrix. [344−2 ]\left[\begin{array}{cc}3& 4\\ 4& -2\end{array}\right] A three-by-three system of equations such as 3 x−y−z=0 x+y=5 2x−3z=2 \begin{array}{l}3x-y-z=0\qquad \\ \text{ }x+y=5\qquad \\ \text{ }2x - 3z=2\qquad \end{array} has a coefficient matrix [3−1−111020 −3]\left[\begin{array}{rrr}\qquad 3& \qquad -1& \qquad -1\\ \qquad 1& \qquad 1& \qquad 0\\ \qquad 2& \qquad 0& \qquad -3\end{array}\right] and is represented by the augmented matrix [3−1−1 1102 0−3 ∣ 0 52]\left[\begin{array}{rrr}\qquad 3& \qquad -1& \qquad -1\\ \qquad 1& \qquad 1& \qquad 0\\ \qquad 2& \qquad 0& \qquad -3\end{array}\text{ }|\text{ }\begin{array}{r}\qquad 0\\ \qquad 5\\ \qquad 2\end{array}\right] Notice that the matrix is written so that the variables line up in their own columns: x-terms go in the first column, y-terms in the second column, and z-terms in the third column. It is very important that each equation is written in standard form ax+by+cz=dax+by+cz=d so that the variables line up. When there is a missing variable term in an equation, the coefficient is 0. How To: Given a system of equations, write an augmented matrix.
Example 1: Writing the Augmented Matrix for a System of EquationsWrite the augmented matrix for the given system of equations. x+2y−z=3 2x−y+2z =6 x−3y+3z=4\begin{array}{l}\text{ }x+2y-z=3\qquad \\ \text{ }2x-y+2z=6\qquad \\ \text{ }x - 3y+3z=4\qquad \end{array} SolutionThe augmented matrix displays the coefficients of the variables, and an additional column for the constants. [12−12−121−33 ∣ 364 ]\left[\begin{array}{rrr}\qquad 1& \qquad 2& \qquad -1\\ \qquad 2& \qquad -1& \qquad 2\\ \qquad 1& \qquad -3& \qquad 3\end{array}\text{ }|\text{ }\begin{array}{r}\qquad 3\\ \qquad 6\\ \qquad 4\end{array}\right] Try It 1Write the augmented matrix of the given system of equations. 4x−3y=113x+2y=4\begin{array}{l}4x - 3y=11\\ 3x+2y=4\end{array} Solution Writing a System of Equations from an Augmented MatrixWe can use augmented matrices to help us solve systems of equations because they simplify operations when the systems are not encumbered by the variables. However, it is important to understand how to move back and forth between formats in order to make finding solutions smoother and more intuitive. Here, we will use the information in an augmented matrix to write the system of equations in standard form. Example 2: Writing a System of Equations from an Augmented Matrix FormFind the system of equations from the augmented matrix. [ 1−3−52 −5−4−354 ∣ −25 6]\left[\begin{array}{rrr}\qquad 1& \qquad -3& \qquad -5\\ \qquad 2& \qquad -5& \qquad -4\\ \qquad -3& \qquad 5& \qquad 4\end{array}\text{ }|\text{ }\begin{array}{r}\qquad -2\\ \qquad 5\\ \qquad 6\end{array}\right] SolutionWhen the columns represent the variables xx , yy , and zz , [1 −3−52−5 −4−354 ∣ −25 6]→x−3y−5z=−22x−5 y−4z=5−3x+5y+4z=6\left[\begin{array}{rrr}\qquad 1& \qquad -3& \qquad -5\\ \qquad 2& \qquad -5& \qquad -4\\ \qquad -3& \qquad 5& \qquad 4\end{array}\text{ }|\text{ }\begin{array}{r}\qquad -2\\ \qquad 5\\ \qquad 6\end{array}\right]\to \begin{array}{l}x - 3y - 5z=-2\qquad \\ 2x - 5y - 4z=5\qquad \\ -3x+5y+4z=6\qquad \end{array} Try It 2Write the system of equations from the augmented matrix. [ 1−112−1 3011∣51−9]\left[\begin{array}{ccc}1& -1& 1\\ 2& -1& 3\\ 0& 1& 1\end{array}|\begin{array}{c}5\\ 1\\ -9\end{array}\right] Solution Licenses and AttributionsHow do you find the solution of a matrix?The given equation can be written in a matrix form as AX = D and then by obtaining A-1 and multiplying it on both sides we can solve the given problem. = X − 1 Y − 1 = ( Y X ) − 1 .
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