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Wize High School Algebra II Textbook (Common Core) > Systems of Linear Equations (Extension Topic)

Solving a System of Linear Equations Using Matrices

Solving Linear SystemsPrevious Section
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Method of Inverse

This is a method used to solve a system of linear equations using the inverse of the coefficient matrix

Steps

1. Write the system of linear equations as 𝐴x⃗=𝑏𝐴\vec{x} = 𝑏
  • AA is the coefficient matrix
  • x⃗\vec x is the solution vector
  • bb is the column vector representing constants on the right-hand side of the equation
2. Find A−1A^{-1}

3. The solution is x⃗=A−1b\vec x=A^{-1}b
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Wize Concept Clarifier:

Solve the system of linear equations using the method of inverses.
x+2y+z=1x+3y+2z=23x+2y=3\begin{array}{c} x+2y+z&=&1\\ x+3y+2z&=&2\\ 3x+2y&=&3 \end{array}

The augmented matrix is [121113223203]\left[\begin{array}{rrr|r} 1&2&1&1\\ 1&3&2&2\\ 3&2&0&3 \end{array}\right].
So, the coefficient matrix is A=[121132320]A=\left[\begin{array}{rrr} 1&2&1\\ 1&3&2\\ 3&2&0 \end{array}\right] and the constant matrix is b=[123]b=\begin{bmatrix} 1\\2\\3 \end{bmatrix}.

1. Rewriting the system, we get [121132320][xyz]=[123]\begin{bmatrix}1&2&1\\1&3&2\\3&2&0 \end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}
2. We need to find A−1A^{-1}:
[121132320∣100010001]\left[\begin{array}{l} 1 & 2 & 1 \\ 1 &3 & 2 \\ 3 & 2 & 0 \\ \end{array}\right.\left|\begin{array}{l} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}\right]

→R2: R2−R1R3: R3−3R1[1210110−4−3∣100−110−301] \xrightarrow{\begin{array}{l} R_2:\ R_2-R_1\\ R_3:\ R_3-3R_1 \end{array}} \left[\begin{array}{rrr} 1 & 2 & 1 \\ 0 &1 & 1 \\ 0 & -4 & -3 \\ \end{array}\right.\left|\begin{array}{rrr} 1&0&0\\ -1&1&0\\ -3&0&1 \end{array}\right]


→R1: R1−2R2R3: R3+4R2[10−1011001∣3−20−110−741]\xrightarrow{ \begin{array}{l} R_1:\ R_1-2R_2\\ R_3:\ R_3+4R_2 \end{array}}\left[\begin{array}{rrr} 1 & 0 & -1 \\ 0 &1 & 1 \\ 0 & 0 & 1 \\ \end{array}\right.\left|\begin{array}{rrr} 3&-2&0\\ -1&1&0\\ -7&4&1 \end{array}\right]

→R1: R1+R3R2: R2−R3[100010001∣−4216−3−1−741]\xrightarrow{ \begin{array}{l} R_1:\ R_1+R_3\\ R_2:\ R_2-R_3 \end{array}}\left[\begin{array}{l} 1 & 0 & 0 \\ 0 &1 & 0 \\ 0 & 0 & 1 \\ \end{array}\right.\left|\begin{array}{rrr} -4&2&1\\ 6&-3&-1\\ -7&4&1 \end{array}\right]

So, the inverse is A−1=[−4216−3−1−741]A^{-1}=\left[\begin{array}{rrr} -4&2&1\\ 6&-3&-1\\ -7&4&1 \end{array}\right]

3. The solution is x⃗=[xyz]=[−4216−3−1−741][123]=[3−34]\vec x=\begin{bmatrix} x\\y\\z \end{bmatrix} = \left[\begin{array}{rrr} -4&2&1\\ 6&-3&-1\\ -7&4&1 \end{array}\right] \begin{bmatrix} 1\\2\\3 \end{bmatrix} = \begin{bmatrix} 3\\ -3\\ 4 \end{bmatrix}

Therefore, the system has a unique solution (3,−3,4)(3,-3,4).
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Practice: Method of Inverse

Given the matrix A=[2130101−12]A=\left[\begin{array}{rrr} 2&1&3\\ 0&1&0\\ 1&-1&2 \end{array}\right],
a.) find the inverse A−1A^{-1}.


b.) using the inverse found in part a.), solve the system of linear equations 2x+y+3z=3y=−5x−y+2z=4\begin{array}{c} 2x+y+3z&=&3\\ y&=&-5\\ x-y+2z&=&4 \end{array}.

Practice Question: Method of Inverse

Let [1−10213−20−1]\left[\begin{array}{c} 1&-1&0\\ 2&1&3\\ -2&0&-1 \end{array}\right]be the inverse of the matrix [abcdefghi]\left[\begin{array}{c} a&b&c\\ d&e&f\\ g&h&i \end{array}\right].

Find the values of zz in the unique solution to the system of linear equations:
ax+by+cz=1dx+ey+fz=−2gx+hy+iz=−3\begin{array}{c} ax+by+cz=1\\ dx+ey+fz=-2\\ gx+hy+iz=-3 \end{array}

Practice: Matrix Inverse Application

Let A=[120−1]A=\left[\begin{array}{c} 1&2\\0&-1 \end{array}\right] and suppose that 3AB−[1−120]=I23AB- \left[\begin{array}{c} 1&-1\\ 2&0 \end{array}\right] =I_2.
Find the matrix BB.