Systems of Linear Equations and Matrices

 

A linear equation in the n variables x1, x2, …, xn is an equation of the form

a1x1 + a2x2 + … + anxn = b

where the symbols  a1, a2, … , an  and  b  represent constants (real numbers).

In a linear equation, all of the n variables appear to the first power, there are no products and no roots of variables, and the variables do not appear as arguments of functions such as rational functions, trigonometric functions, exponential functions, logarithmic functions, etc.

 

The following are examples of linear equations:

a)   3x + 4 = 0                                       b)   5x – 3y = 2

c)   2x – 8y + 7z = 6                              d)   4x + 2y – 3z + 5w = 9

e)   6x + 4y – 4z + 8w – 2t = 5               f)    7x1 + 5x2 – x3 + 2x4 – 9x5 + 3x6 + 4x7 + x8 = 1

 

An equation that is not linear is said to be a nonlinear equation.

The following are examples of non-linear equations:

a)   x2 - 9 = 0                                        b)   xy + 3x – 2y = 5

c)                                    d)  

e)                                          f)    y = ln(2x) + 3

g)    y – 1 = 3ex + t                                h)  y + sin x = cos z

 

A solution to a linear equation  a1x1 + a2x2 + … + anxn = b  is an n-tuple, (c1, c2, …, cn),  (or a sequence of n real numbers  c1, c2, …, cn),  that makes the equation true when we substitute  x1 = c1,  x2 = c2,  …,  xn = cn.  

 

Now  consider the following groups of equations:

1)   3x + 4y = 5                                     2)   2x  -  y + 3z = 9 

      5x – 2y = -9                                           x + 2y – 4z = -9

                                                                  5x + 3y – 2z = -5

 

3)   8x + 2y – 3z + 4w = 4                      4)   4x – 5y + 2z – 3w + 6t = 0

      -x + 4y + 5z  -  w = 8                           2x + 3y – 4z – 6w – 5t = 8

      2x – 3y – 6z + 5w = 6                            3x – 9y + 3z –  w + 4t = 10

        x  + y + 2z – 3w = 3

      3x – 2y  -  z +  w = 1

      4x + 5y – 8z + 2w = 5

 

The four groups of equations represent examples of systems of linear equations  or  linear systems.

More generally, a system of linear of equations consists of a finite set of linear equations of the form

 

   a11x1 + a12x2 + a13x3 + … + a1nxn  =  b1

   a21x1 + a22x2 + a23x3 + … + a2nxn  =  b2

   a31x1 + a32x2 + a33x3 + … + a3nxn  =  b3

            .

            .

            .

   am1x1 + am2x2 + am3x3 + … + amnxn  =  bm

 

This system is said to be an mΧn linear system since it consists of m linear equations in n variables.  The variables are also called unknowns, so we may sometimes say that this is a linear system of m equations in n unknowns.  In the four groups of equations that we presented above, we have that

1)   3x + 4y = 5                                                 2)   2x  -  y + 3z = 9 

      5x – 2y = -9                                                       x + 2y – 4z = -9

                                                                              5x + 3y – 2z = -5

      This is a 2Χ2 linear system                                

      (Two equations in two variables)                        This is a 3Χ3 linear system

      (Two equations in two unknowns)                     (Three equations in three variables)

                                                                             (Three equations in three unknowns)

 

 

3)   8x + 2y – 3z + 4w = 4                                  4)   4x – 5y + 2z – 3w + 6t = 0

      -x + 4y + 5z  -  w = 8                                       2x + 3y – 4z – 6w – 5t = 8

      2x – 3y – 6z + 5w = 6                                        3x – 9y + 3z –  w + 4t = 10

        x  + y + 2z – 3w = 3

      3x – 2y  -  z +  w = 1                                        This is a 3Χ5 linear system

      4x + 5y – 8z + 2w = 5                                        (Three equations in five variables)

                                                                              (Three equations in five unknowns)

      This is a 6Χ4 linear system

      (Six equations and four variables)

      (Six equations and four unknowns)

 

 

A solution to a linear system is an n-tuple, (c1, c2, …, cn),  (or a sequence of n real numbers  c1, c2, …, cn),  that makes every equation in the system true when we substitute  x1 = c1,  x2 = c2,  …,  xn = cn.  That is, the n-tuple is a solution to every equation in the system.

 

EXAMPLES

1)  The ordered pair (5, -1) is the solution of the system    x – 2y = 7

                                                                              3x + 7y = 8    because when we substitute x = 5 and y = –1 into each equation, we get a true equality.  In other words, (5, –1) is a solution of each of the two equations.

 

2)   The triplet, or 3-tuple, (2, –3, 1) is a solution of the system      4x + 3y – 2z = –3

                                                                                                2x  –  y + 5z = 12     because when we substitute   x = 2,  y = –3, and z = 1 into each equation, we get a true equality.  That is, the triplet (2, –3, 1) is a solution of each of the two equations.

 

3)   The quintuplet, or 5-tuple, (3, –4, –1, 0, 5) is a solution of the system     –x1 + 4x2 – 3x3 + 6x4 – 2x5 = –26

                                                                                                              3x1 - 5x2 – 2x3 + 7x4 + 4x5 =  11

                                                                                                                x1 + 2x2 + 5x3 - 5x4 – 3x5 = –25

                                                                                                            -5x1 +  x2  – 4x3 + 8x4 + 5x5 =  10

                                                                                                              2x1 –  x2  +  x3  + 9x4 – 2x5 =  –1  

because when we substitute  x1 = 3,   x2 = –4, x3 = -1, x4 = 0, and x5 = 5 into each equation, we get a true equality.  That is, the quintuplet (3, –4, –1, 0, 5) is a solution of each of the five equations.

 

 

It turns out that not all systems of linear equations have solutions.  Some systems have exactly one solution, others have infinitely many solutions, and others have no solutions at all.  For example:

1)  The system     x – 2y = 7

                         3x + 7y = 8    has exactly one solution. 

    The ordered pair (5, -1) is the only ordered pair that is a solution of each of the two equations, simultaneously.

 

2)   The system    –x1 + 4x2 – 3x3 + 6x4 – 2x5 = –26

            3x1 - 5x2 – 2x3 + 7x4 + 4x5 =  11

                             x1 + 2x2 + 5x3 - 5x4 – 3x5 = –25

                         -5x1 +  x2  – 4x3 + 8x4 + 5x5 =  10

                           2x1 –  x2  +  x3  + 9x4 – 2x5 = –1  

      has exactly one solution, which is the quintuplet (3, –4, –1, 0, 5).  This is the only set of five numbers that is a solution of each of the five equations simultaneously.

 

3)  Notice that the linear equation  2x – y = 4 has infinitely many solutions.  We can view this as a system of one equation in two unknowns.  To find all solutions, all we need to do is solve for y to obtain  y = 2x – 4.  Then the solutions are given by  x = t and y = 2t – 4, where t could be any real number.  So,

      when t = 0, the ordered pair (0, – 4) is a solution;      when t = 1, the ordered pair (1, –2) is a solution;

      when t = -2, the ordered pair (-2, -8) is a solution;   when t = 3.75, the ordered pair (3.75, 3.50) is a solution…

      and so on... Every time we choose a value of t, we get an ordered pair that is a solution to the equation. 

 

4)   The system  4x + 3y – 2z = –3

                         2x  –  y + 5z = 12    has infinitely many solutions.  (Exercise 6 by the Gauss-Jordan method) 

       We will show that the solutions are given by  ,

       Where t represents any real number.

       For example, if t = 0, the triplet    is a solution to the system.

       If t = 10, the triplet    is a solution to the system.

       If t =, the triplet    is a solution to the system…

       and so on. 

 

In this chapter we will develop methods for solving systems of linear equations.

 

5)   The system   has no solution, since we will not be able to find two real numbers with sum equal to 1 and at the same time with sum equal to 3.  A set of equations like the one shown in this example is a said to be a set of contradictory equations.

 

6)   The system     has no solution, since the equations    are contradictory equations (why?).

 

In general, any system containing contradictory equations has no solutions and every system not containing contradictory equations, but with more variables than equations, has infinitely many solutions. 

 

MATRICES

 

Consider the following linear system: 

 2x  -  y + 3z = 5      The rectangular array of numbers consisting of the coefficients of the variables

   x + 4y – 2z = 3       together with the constants on the right hand side of the equations is called the

 3x + 6y + 2z = 8       AUGMENTED MATRIX  for the system.

 

That is, the augmented matrix for the given system is:

 

.  If we leave out the right hand side constants and write only the coefficients of the variables,

the resulting rectangular array is called the MATRIX OF COEFFICIENTS.  The matrix of coefficients for the

given system is

 

In general, a MATRIX  is a rectangular array of numbers.  A matrix consists of a finite number of rows, m, and a finite number of columns, n.

 

For the general mΧn linear system

 

 

. 

 

Notice that the matrix of coefficients consists of m rows (also called row vectors) and n columns (also called column vectors).  The number of columns corresponds to the number of variables in the linear system and the number of rows corresponds to the number of equations in the linear system.

The augmented matrix consists of m rows and (n+1) columns because it contains the additional column of constants b1, b2, …, bm.

 

 

The coefficients of the variables have double subscripts of the form ij.  The first subscript, i, represents the number of the row where the element  is located.  The second subscript, j, represents the number of the column where the element  is located.  For example, the element  is located in the third row of column 2, (or equivalently, in the second column of row 3).  The element  is located in the second row of column 3, (or equivalently, in the third column of row 2).  In every matrix, the number of rows by the number of columns is referred to as the size of the matrix.  In general, a matrix with m rows and n columns is said to be an mΧn matrix, where mΧn is the size of the matrix.

 

EXERCISES

1)    For the linear system:  2x + 5y – 4z = 8                                               A)  Write the augmented matrix.

                                          4x – 7y + 3z = 9                                              B)  Give the size of the augmented matrix.

                                                                                                                  C)  Write the matrix of coefficients.

                                          D) Give the size of the matrix of coefficients.      E)  Write the constant vector.

                                          F)  Give the size of the constant vector.

 

2)  For the linear system:    5x – 2y + 8z – 2w = 10                 A)  Write the augmented matrix.

                                                   y – 6z + 3w = 13                   B)  Give the size of the augmented matrix.

                                          3x +        4z  –  w = 20                  C)  Write the matrix of coefficients.

                                          -x – 5y +  z + 7w = -8                  D)  Give the size of the matrix of coefficients.

                              2x + 3y                 = 14                  E)   Write the constant vector.

                                              5z – 6w = -3                 F)   Give the size of the constant vector.

 

 3)   For the linear system      x +  y  –  z = 0                             A)  Write the augmented matrix.

                                          3x – 2y +  z = 4                             B)  Give the size of the augmented matrix.

                                          –x + 4y – 2z = 6                            C)  Write the matrix of coefficients.

D)    Give the size of the matrix of coefficients.

                                          E)  Write the constant vector.          F)   Give the size of the constant vector.

 

4)  Write the linear system that corresponds to the augmented matrix

    

 

5)  Write the linear system that corresponds to the augmented matrix

           

 

6)  Write the linear system that corresponds to the augmented matrix

     

 

SOLVING LINEAR SYSTEMS USING MATRICES

We will go through the process step by step using examples.

Consider the following 2Χ2 linear system:    2x +  y  =  1

                                                                    x – 3y = -10

 

       SYSTEM                                                                          AUGMENTED MATRIX

                                                         

We can interchange the two equations.                                      Interchange the two rows

                                                    

Multiply an equation by a nonzero constant.                               Multiply a row by a nonzero constant

So, multiply the first equation by -2.                                          So, multiply the first row by -2.

                                                    

Add a multiple of one equation to another equation.                   Add a multiple of one row to another row.

So, add equation 1 to equation 2.                                              So, add row 1 to row 2.

                                                    

At this stage, we can use the second equation, 7y = 21, to solve for y and obtain y = 3.  With this value, we can go back to equation 1, substitute y = 3, and solve for x to obtain that

 x = -1.  A process like this is called backward substitution method.

 

It is customary to continue as follows:

Multiply the second equation by .                                           Multiply the second row by .

                                                    

Add  -6 times equation 2 to equation 1.                                    Add  -6 times row 2 to row 1.

                                                    

Multiply - times equation 1.                                                   Multiply - times row 1.

                                                    

Notice that resulting augmented matrix is the augmented matrix for the system at left and clearly displays the correct solution x = -1, y = 3.

The operations that we performed using the rows of the augmented matrix are known as elementary row operations.  Also, the resulting augmented matrix is said to in reduced row-echelon form and the process of solving linear systems of equations by converting the original augmented matrix to row-echelon form is called the Gauss-Jordan Elimination Method.

 

GAUSS-JORDAN ELIMINATION METHOD

The following is a second example using the Gauss-Jordan Elimination to solve a 3Χ3 system:  2x – y + 3z = 5

                                                                                                                                             3x + y – 2z = 5

                                                                                                                                               x + 2y – z = 0

 

       SYSTEM                                                                          AUGMENTED MATRIX

                                                  

Interchange equations 1 and 3.                                                  Interchange rows 1 and 3.

                                          

Multiply equation 1 by -3 and add it to equation 2.                   Multiply row 1 by -3 and add it to row 2.

Multiply equation 1 by -2 and add it to equation 3.                   Multiply row 1 by -2 and add it to row 3.

                                          

Add 2 times equation 2 to 5 times equation 1.                           Add 2 times row 2 to 5 times row 1.

Add -1 times equation 2 to equation 3.                                     Add -1 times row 2 to row 3.

                                          

Multiply the third equation by .                                               Multiply the third row by .

                                          

Add  3 times equation 3 to equation 1.                                      Add  3 times row 3 to row 1.

Add -1 times equation 3 to equation 2.                                     Add -1 times row 3 to row 1.

                                          

Multiply  times equation 1.                                                     Multiply  times row 1.

Multiply  - times equation 2.                                                  Multiply  - times row 1.

                                          

The resulting augmented matrix is the augmented matrix for the system at left clearly displays the correct solution x = 2,  y = -1,  z = 0.

 

BACKWARD SUBSTITUTION METHOD

The following example illustrates the backward substitution method to solve a linear system.

 

       SYSTEM                                                                          AUGMENTED MATRIX

                                       

Interchange equations 1 and 2.                                                  Interchange rows 1 and 2.

                                       

Multiply equation 1 by -3 and add it to equation 2.                   Multiply row 1 by -3 and add it to row 2.

Add equation 1 to equation 3.                                                   Add row 1 to row 3.

Multiply equation 1 by -4 and add it to equation 4.                   Multiply row 1 by -4 and add it to row 4.

                                       

Add 3 times equation 2 to 4 times equation 3.                           Add 3 times row 2 to 4 times row 3.

Add -7 times equation 2 to 4 times equation 4.                         Add -7 times row 2 to 4 times row 4.

                                       

Add 25 times equation 3 to 13 times equation 4.                       Add 25 times row 3 to 13 times row 4.

                                   

The resulting augmented matrix is the augmented matrix for the system at left.

From equation 4, (row 4), it is clear that w = 8.  If we go backward, one equation at a time, we can use w = 8 in equation 3 and then solve for z and get  z = 5.  Then replace w and z in equation 2 and solve for  y to get y = -1.  Finally, replace y, w, and z in equation 1 and solve for x to get  x = -2.

 

EXERCISES

 

1)   Solve by the backward substitution method:   10x + 3y =  2

                                                                           -7x – 2y = -2

      If the system has no solutions, explain why.  If the system has infinite solutions, find them.

 

2)   Solve by the Gauss-Jordan method:     4x –  5y = 6

                                                               -8x + 10y = 5

      If the system has no solutions, explain why.  If the system has infinite solutions, find them.

 

3)   Solve by the backward substitution method:   -3x +  7y =  5

                                                                             6x – 14y = -10

      If the system has no solutions, explain why.  If the system has infinite solutions, find them.

 

4)   Solve by the Gauss–Jordan method:  x  +  y  +  z = 3

                                                                       y  + 2z = 2

                                                               x + 2y  + 4z = 6

       If the system has no solutions, explain why.  If the system has infinite solutions, find them.

 

5)   Solve by the Gauss–Jordan method:   2x – 5y + 3z = 26

                                                               -x + 2y  -  z  = -7

                                                                 x  +  y  -  z  =  0

       If the system has no solutions, explain why.  If the system has infinite solutions, find them.

 

6)   Solve by the Gauss-Jordan method:  4x + 3y – 2z = -3

                                                               2x -  y + 5z = 12

       If the system has no solutions, explain why.  If the system has infinite solutions, find them.

 

7)   Solve by the backward substitution method:  2x + 3y      -   w = -4

                                                                          -x +  y - z + 2w =  2

                                                                               -2y + z  +  w = 11

                                                                            x        + z + 4w = 19

       If the system has no solutions, explain why.  If the system has infinite solutions, find them.

 

8)   Solve by the backward substitution method:  2x – 3y  + 5z = 4

                                                                            x + 2y  – 3z = 0

                                                                        -6x + 9y – 15z = 7      

       If the system has no solutions, explain why.  If the system has infinite solutions, find them.

 

9)   Solve by the Gauss-Jordan method:     3x + 4y – 2z – 5w + 6t =  12

                                                               -2x – 3y  +  z – 3w + 4t = -58

                                                                   x + 2y – 4z +  w  - 5t = -10

                                                                 -x –  y  – 3z + 4w + 2t  =  2

                                                                 4x + 5y  -  z + 2w – 8t  = 24  

      If the system has no solutions, explain why.  If the system has infinite solutions, find them.

 

OPERATIONS WITH MATRICES

MATRIX ADDITION:  Matrix addition is done by adding corresponding entries.  Therefore, in order for addition to be possible, the matrices must be of exactly the same size. Example, a 3Χ4 matrix can be added only to another 3Χ4 matrix. However, a 3Χ4 matrix cannot be added to a 3Χ3 matrix or to any matrix that is not 3Χ4 because not all the entries are corresponding. The general expression for addition of matrices is:

 

EXAMPLES

 

 

 

 

 MULTIPLICATION OF A SCALAR TIMES A MATRIX:

Let k be a scalar (any real number).

 

 

 

EXAMPLE

 

 

MULTIPLICATION OF A MATRIX TIMES ANOTHER MATRIX:

Multiplication of a matrix times another matrix requires that the number of columns of the matrix on the left is equal to the number of rows of the matrix on the right.  A matrix M with size mΧn can be multiplied times another matrix N if the size of N is nΧk and M is on the left.  That is,.  The size of the product matrix is mΧk.  For example, a 3Χ4 matrix can be multiplied by a 4Χ5 and the size of the product is 3Χ5.  A 3Χ4 can be multiplied times other matrices if the sizes are 4Χ1 or 4Χ2 or 4Χ3 or 4Χ4 or 4Χ5 or 4Χq, where q is a positive integer. 

In the product matrix, , a typical element  is obtained from row i of matrix M and column j of matrix N by adding the products of the corresponding entries.

 

EXAMPLE:  Examine the following product

 

 

The multiplication shows a 2Χ4 matrix on the left times a 4Χ3 matrix. The number of columns of the matrix on the left (4) is equal to the number of rows of the second matrix (4). The product matrix is of size 2Χ3. The multiplication procedure is clearly shown below

2(1) + 5(4) + 0(5) – 1(0) = 22  ;    2(-3) + 5(2) + 0(1) – 1(-2) = 6  ;       2(0) + 5(-1) + 0(3) – 1(6) = -11

3(1) - 4(4) + 1(5) – 2(0) = -8  ;   3(-3) - 4(2) + 1(1) – 2(-2) = -12  ;   3(0) - 4(-1) + 1(3) – 2(6) = -5

 

 

EXAMPLE:  Practice the following product

 

 

This time, the matrix on the left is a 4Χ2 matrix and it is being multiplied times a 2Χ5 matrix. The number of columns of the matrix on the left (2) is equal to the number of rows of the second matrix (2). The product matrix is of size 4Χ5.  Please verify that the product is correct.

 

If A is a matrix and B is matrix, then the product AB is possible if the number of columns of matrix A is equal to the number of rows of matrix B.  Otherwise, the product is said to be undefined.  Notice that matrix multiplication is not commutative.

 

EXAMPLES:

 

1)

      

 

      .

       Clearly,  AB   BA

 

2)

       

 

          

 

        Clearly,  AB   BA       

 

EXERCISES

 

1)   Consider the following matrices

 

  ;

 

Perform the following operations.  If an operation cannot be performed, explain why.

(a)   E + H                                                                               (b)    H – J

(c)   B + E                                                                                (d)   5E + 3J

(e)   AC                                                                                    (f)    BD

(g)   CE                                                                                    (h)   DH

(i)    EI                                                                                     (j)    FG

(k)   DC + EB                                                                          (l)    HG

 

      (a)  Find

      (b)  Find , where n is a positive integer.

 

 

 

EQUALITY OF MATRICES

 

Two matrices are said to be equal if the have exactly the same size and corresponding entries are equal.

That is,

 

EXAMPLE

 

 

 

THE  MATRIX  EQUATION:   AX = B

 

 

 

 

Using matrix symbols, we write the mΧn linear system as  AX = B.

 

SQUARE  MATRICES,  TRIANGULAR MATRICES, DIAGONAL  MATRICES,  IDENTITY  MATRICES

 

A matrix for which the number of rows equals the number of columns, m = n, is said to be a square matrix.

A square matrix with  for every i > j  (0’s under the main diagonal) is said to be upper triangular matrix.

A square matrix with  for every i <j  (0’s above the main diagonal) is said to be lower triangular matrix.

A square matrix which is either upper triangular or lower triangular is said to be a triangular matrix.

A square matrix with non-zero real numbers along its diagonal and zero’s everywhere else is said to be a diagonal matrix.

A diagonal matrix with 1’s along its diagonal is said to be an identity matrix.

EXAMPLES

Square Matrices

Upper Triangular Matrix

Lower Triangular Matrix

 

Diagonal Matrices

Identity Matrices

 

IDENTITY  MATRICES  AND  INVERSE  MATRICES

 

Identity matrices have the symbol I.  If A is a square matrix with exactly the same size as I, then

AI = A  and  IA = A.

 

EXERCISES

 

 

 

Many square matrices have the following property:

If  A  is a square matrix, then there exists another matrix  B  of exactly the same size as  A,  such that  AB = I  and  BA = I.  The matrix B is said to be the inverse of A.  The symbol for the inverse of matrix A is .  Therefore,  and  .  Matrices for which the inverse exists are called invertible matrices  (also called  Nonsingular Matrices).

 

Remember the matrix equation: AX = B.  Matrix multiplication is associative, so if  A  is a nonsingular matrix with inverse  , the following is true:

 AX = B.  Since A = I, then by the associative property,  I X = B.  Since  I X = X, then  X = B  is the solution to the system.

 

We now describe a method for finding  for a nonsingular matrix A.

 

FINDING  THE  INVERSE  OF  A  NONSINGULAR  MATRIX

 

Suppose that A is an nΧn nonsingular matrix.  We start with the matrix [A | I ], where I is the identity matrix of exactly the same size as A.  Use elementary row operations on [A | I ]  until the matrix A is transformed into I  ( that is,  A is in reduced row echelon form).  The resulting matrix has the form   and the inverse of A appears on the right hand side, next to I.

 

EXAMPLES

      

 

                     

 

                            

      

                        

      

                                                             is  the  inverse  matrix.

 

 

 

     

 

The expression  (ad – bc)  is called the  determinant of the 2Χ2 matrix M.  Notice that if the determinant has value zero, then the inverse matrix  is undefined.  In such a case, we say that the given matrix  M  is not invertible.  A matrix that is not invertible is said to be a  singular matrix.

More about determinants follows later  (see Cramer’s Rule).

 

EXERCISES

 

 

 

 

 

 

 

    This is called a 5Χ5 Hilbert matrix.

 

SOLVING  LINEAR  SYSTEMS  USING  THE  INVERSE  OF THE  MATRIX  OF  COEFFICIENTS 

 

We mentioned before that the matrix equation  AX = B  has solution   if and only if  A  is a nonsingular matrix.

EXAMPLES:

 

 

     

 

 

     

 

EXERCISES

 

In each case below, a system of the form  AX = B  is given.  Find the inverse  (if it exists) of the matrix of coefficients and use it to solve the given system by the method .

If the system has no solution, explain why.  If the system has infinitely many solutions, find all such solutions.

 

 

 

 

 

DETERMINANTS

 

We mentioned before that for a 2Χ2 matrix of the form  , the expression  (ad – bd)  is called the determinant of the matrix.

More generally, the determinant is a function with domain equal to the set of square matrices . 

If A represents a square matrix, then the determinant of A is denoted by det(A).  Sometimes, we may use | A | to represent the determinant of A.

 

 

MINORS AND COFACTORS

 

In every square matrix  , the minor   of the entry  is the determinant of the (n – 1)Χ(n – 1) matrix formed by deleting row i and column j from the matrix .  Notice that the deleted row i and column j contain the element .  The cofactor  of the entry  is given by 

 

EXAMPLE

 

The cofactors are:

 

COMPUTATION  OF  THE  DETERMINANT  OF  AN  nΧn  MATRIX

 

 

 

 

      This computation is based on the cofactor expansion of column 1. 

      We could use also the cofactor expansion of row 1 and obtain the same value of  Det (A).

      .

       Det (A) can be obtained by using the cofactor expansion of any row or any column of A.

 

 

PROPERTIES  OF  DETERMINANTS

 

1)   If A is a square (nΧn) triangular matrix, then Det (A) is the product of the entries in the main diagonal of A.

      

2)    If A is a square (nΧn) matrix containing a row of zeros, then Det (A) = 0.

       

3)    If A is a square (nΧn) matrix containing a column of zeros, then Det (A) = 0.

      

 

4)    If A is a square (nΧn) matrix that contains two identical rows or two identical columns, then Det (A) = 0.

      

 

5)    If A is a square (nΧn) matrix and two rows of A are interchanged, then the resulting matrix has determinant equal to the negative of  Det (A).

6)    If A is a square (nΧn) matrix and one row of A is multiplied times a nonzero constant k, then the resulting matrix has determinant equal to kDet (A).

7)    If A is a square (nΧn) matrix and a multiple of one row is added to another row of A, then the resulting matrix has determinant equal to the  Det (A).

8)    If A is a square (nΧn) singular matrix, then the  Det (A)  =  0.  The converse is also true:  If A is a square (nΧn) matrix and Det (A)  =  0,  then A is a singular matrix.

11)   If A and B are square (nΧn) matrices of exactly the same order, then Det (AB)  =  Det (A) Det (B).

 

 

 

       whenever A is a square nΧn matrix.

     

14)  If  A is an nΧn  square matrix, then for every nΧ1 vector of constants B, the linear system  AX = B  has a unique solution.

 

EXERCISES

 

In each case below, compute the determinant of the given matrix.

 

 

 

8)  Prove that if A is an nΧn triangular matrix, then

 

 

10)   Compute the determinant of the given matrices:

      

      

 

 

 

 

13)   Which of the following matrices are singular?

       

 

14)  Which of the following linear systems have a unique solution?

       (a)   5x – 4y = 6                (b)   2x - 6y = 4                      (c)   x  +  3y = -1           (d)   4x - 7y = 19

                x + 3y = 5                      -x + 3y = -2                       -4x – 12y =  4                    3x + 2y = 7

 

      (e)   6x – 2y + 3z =   21    (f)  3x + 4y – 2z =    0       (g)   x + 3y  –  6z =   4       (h)  -2x + y – z = 3

             3x + 5y – 2z = -11          2x – 3y +  z  = -14           2x – 6y  +  4z = -4               2x – y + z = 0

              -x – 3y + 4z =   11         4x + 2y – 3z = -15           3x + 9y – 18z = 12               3x + y + z = 3

 

      (i)   2x – 6y  +  z – 3w = 10       (j)    x – 3y + 2z  –  w = 4       (k)   5x1 – 4x2 + 2x3 – 3x4 + 2x5 =  1

              x + 2y  -  z + 3w =  2             -x + 2y  - z  +  w = 6              -x1           - 2x3 + 2x4 – 3x5 =  3

                    3y + 2z – 4w = -1             3x - 6y + 3z – 3w = 8                x1 + 2x2           + 2x4 – 4x5 =  4

            3x  –  y – 4z          = -5             2x + 3y – 5z          = 3              3x1  -  x2 + 3x3 – 3x4 + 2x5 = -4

                                                                                                               2x1 + 3x2          + 2x4  -  x5 = 9

 

CRAMER’S  RULE

 

One more method to solve linear systems of equations is by Cramer’s Rule.

      

 

 

Let  represent the nΧn matrix formed by replacing the ith column of A with the columns vector B.

Then           where  i = 1, 2, 3, …, n

 

PROOF:   The matrix form of the given linear system is  AX = B. 

                 .

                 

                                     

                

 

             

 

                                

 

                                

                                   … …

                                   … …

                               

 

                                 Therefore,

                                 

 

                               

EXAMPLES

 

1)  Solve by Cramer’s Rule:     3x – 4y =  1

                                               2x + 5y = 16

     Solution:

                       

 

                       

     The solution to the system is  x = 3,  y = 2,  or the ordered pair  (3, 2).

 

2)   Solve by Cramer’s Rule:     4x – 2y + 3z =   3

                                                2x  - y  + 4z = -1

                                                -x + 3y - 2z =  3

      Solution:

                       

 

                       

 

                       

 

 

EXERCISES

 

1)   Solve by Cramer’s rule

       (a)   5x – 4y = 6                (b)   2x - 6y = 4                      (c)   x  +  3y = -1           (d)   4x - 7y = 19

                x + 3y = 5                      -x + 3y = -2                       -4x – 12y =  4                    3x + 2y = 7

 

2)   Solve by Cramer’s rule

       (a)   6x – 2y + 3z =   21    (b)  3x + 4y – 2z =    0       (c)   x + 3y  –  6z =   4       (d)  -2x + y – z = 3

              3x + 5y – 2z = -11          2x – 3y +  z  = -14           2x – 6y  +  4z = -4               2x – y + z = 0

              -x – 3y + 4z =   11          4x + 2y – 3z = -15           3x + 9y – 18z = 12               3x + y + z = 3

 

3)   Solve by Cramer’s rule

      (a)   2x – 6y  +  z – 3w = 10       (b)    x – 3y + 2z  –  w = 4       (c)   5x1 – 4x2 + 2x3 – 3x4 + 2x5 =  1

               x + 2y  -  z + 3w =  2              -x + 2y  - z  +  w = 6              -x1           - 2x3 + 2x4 – 3x5 =  3

                     3y + 2z – 4w = -1             3x - 6y + 3z – 3w = 8                x1 + 2x2           + 2x4 – 4x5 =  4

             3x  –  y – 4z          = -5             2x + 3y – 5z          = 3              3x1  -  x2 + 3x3 – 3x4 + 2x5 = -4

                                                                                                                2x1 + 3x2          + 2x4  -  x5 = 9