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Augmented matrix

From Wikipedia, the free encyclopedia
Matrix formed by appending columns of two other matrices
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Inlinear algebra, anaugmented matrix(A|B){\displaystyle (A\vert B)} is ak×(n+1){\displaystyle k\times (n+1)}matrix obtained by appending ak{\displaystyle k}-dimensional column vectorB{\displaystyle B}, on the right, as a further column to ak×n{\displaystyle k\times n}-dimensional matrixA{\displaystyle A}. This is usually done for the purpose of performing the sameelementary row operations on the augmented matrix(A|B){\displaystyle (A\vert B)} as is done on the original oneA{\displaystyle A} when solving a system of linear equations byGaussian elimination.

For example, given the matricesA{\displaystyle A} and column vectorB{\displaystyle B}, whereA=[132201522],B=[431],{\displaystyle A={\begin{bmatrix}1&3&2\\2&0&1\\5&2&2\end{bmatrix}},\quad B={\begin{bmatrix}4\\3\\1\end{bmatrix}},}the augmented matrix(A|B){\displaystyle (A\vert B)} is(A|B)=[132420135221].{\displaystyle (A|B)=\left[{\begin{array}{ccc|c}1&3&2&4\\2&0&1&3\\5&2&2&1\end{array}}\right].}

For a given numbern{\displaystyle n} of unknowns, the number of solutions to a system ofk{\displaystyle k} linear equations depends only on therank of the matrix of coefficientsA{\displaystyle A} representing the system and the rank of the corresponding augmented matrix(A|B){\displaystyle (A\vert B)} where the components ofB{\displaystyle B} consist of the right hand sides of thek{\displaystyle k} successive linear equations. According to theRouché–Capelli theorem, any system of linear equations

AX=B{\displaystyle AX=B}

whereX=(x1,,xn)T{\displaystyle X=(x_{1},\dots ,x_{n})^{T}} is then{\displaystyle n}-component column vector whose entries are the unknowns of the system isinconsistent (has no solutions) if therank of the augmented matrix(A|B){\displaystyle (A\vert B)} is greater than the rank of thecoefficient matrixA{\displaystyle A} . If, on the other hand, the ranks of these two matrices are equal, the system must have at least one solution. The solution is unique if and only if the rank equals the number of variablesn{\displaystyle n}. Otherwise the general solution hasj{\displaystyle j} free parameters wherej{\displaystyle j} is the difference between the number of variablesn{\displaystyle n} and the rank. In such a case there as an affine space of solutions of dimension equal to this difference.

The inverse of a nonsingular square matrixA{\displaystyle A} of dimensionn×n{\displaystyle n\times n} may be found by appending then×n{\displaystyle n\times n}identity matrixI{\displaystyle \mathbf {I} } to the right ofA{\displaystyle A} to form then×2n{\displaystyle n\times 2n} dimensional augmented matrix(A|I){\displaystyle (A\vert \mathbf {I} )}. Applying elementary row operations to transform the left-handn×n{\displaystyle n\times n} block to theidentity matrixI{\displaystyle \mathbf {I} }, the right-handn×n{\displaystyle n\times n} block is then the inverse matrixA1{\displaystyle A^{-1}}

Example of finding the inverse of a matrix

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LetA{\displaystyle A} be the square 2×2 matrixA=[1350].{\displaystyle A={\begin{bmatrix}1&3\\-5&0\end{bmatrix}}.}

To find the inverse ofA{\displaystyle A} we form the augmented matrix(A|I2){\displaystyle (A\vert \mathbf {I} _{2})} whereI2{\displaystyle \mathbf {I} _{2}} is the2×2{\displaystyle 2\times 2}identity matrix. We then reduce the part of(A|I2){\displaystyle (A\vert \mathbf {I} _{2})} corresponding toA{\displaystyle A} to the identity matrix usingelementary row operations on(A|I2){\displaystyle (A\vert \mathbf {I} _{2})} .(A|I2)=[13105001]{\displaystyle (A\vert \mathbf {I} _{2})=\left[{\begin{array}{cc|cc}1&3&1&0\\-5&0&0&1\end{array}}\right]}(I|A1)=[100150113115],{\displaystyle (I|A^{-1})=\left[{\begin{array}{cc|cc}1&0&0&-{\frac {1}{5}}\\0&1&{\frac {1}{3}}&{\frac {1}{15}}\end{array}}\right],}the right part of which is the inverseA1{\displaystyle A^{-1}}.

Existence and number of solutions

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Consider the system of equationsx+y+2z=2x+y+z=32x+2y+2z=6.{\displaystyle {\begin{aligned}x+y+2z&=2\\x+y+z&=3\\2x+2y+2z&=6.\end{aligned}}}

The coefficient matrix isA=[112111222],{\displaystyle A={\begin{bmatrix}1&1&2\\1&1&1\\2&2&2\\\end{bmatrix}},}and the augmented matrix is(A|B)=[112211132226].{\displaystyle (A|B)=\left[{\begin{array}{ccc|c}1&1&2&2\\1&1&1&3\\2&2&2&6\end{array}}\right].}

Since both of these have the same rank, namely 2, there exists at least one solution; and since their rank is less than the number of unknowns, the latter being 3, there are an infinite number of solutions.

In contrast, consider the systemx+y+2z=3x+y+z=12x+2y+2z=5.{\displaystyle {\begin{aligned}x+y+2z&=3\\x+y+z&=1\\2x+2y+2z&=5.\end{aligned}}}

The coefficient matrix isA=[112111222],{\displaystyle A={\begin{bmatrix}1&1&2\\1&1&1\\2&2&2\\\end{bmatrix}},}and the augmented matrix is(A|B)=[112311112225].{\displaystyle (A|B)=\left[{\begin{array}{ccc|c}1&1&2&3\\1&1&1&1\\2&2&2&5\end{array}}\right].}

In this example the coefficient matrix has rank 2 while the augmented matrix has rank 3; so this system of equations has no solution. Indeed, an increase in the number of linearly independent rows has made the system of equationsinconsistent.

Solution of a linear system

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As used in linear algebra, an augmented matrix is used to represent thecoefficients and the solution vector of each equation set.For the set of equationsx+2y+3z=03x+4y+7z=26x+5y+9z=11{\displaystyle {\begin{aligned}x+2y+3z&=0\\3x+4y+7z&=2\\6x+5y+9z&=11\end{aligned}}}the coefficients and constant terms give the matricesA=[123347659],B=[0211],{\displaystyle A={\begin{bmatrix}1&2&3\\3&4&7\\6&5&9\end{bmatrix}},\quad B={\begin{bmatrix}0\\2\\11\end{bmatrix}},}and hence give the augmented matrix(A|B)=[1230347265911].{\displaystyle (A|B)=\left[{\begin{array}{ccc|c}1&2&3&0\\3&4&7&2\\6&5&9&11\end{array}}\right].}

Note that the rank of the coefficient matrix, which is 3, equals the rank of the augmented matrix, so at least one solution exists; and since this rank equals the number of unknowns, there is exactly one solution.

To obtain the solution, row operations can be performed on the augmented matrix to obtain the identity matrix on the left side, yielding[100401010012],{\displaystyle \left[{\begin{array}{ccc|r}1&0&0&4\\0&1&0&1\\0&0&1&-2\\\end{array}}\right],}so the solution of the system is(x,y,z) = (4, 1, −2).

References

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  • Marcus, Marvin; Minc, Henryk (1992).A survey of matrix theory and matrix inequalities. Dover books on advanced mathematics. New York: Dover Publ. p. 31.ISBN 978-0-486-67102-4.
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