In the figure below we multiply the matrix I, consisting of 1's on the main diagonal and 0's elsewhere, by the column vector x to get a column vector y.

• Since by our calculation y = x, we see that I x = x.
• Since x could be any vector at all., we see that multiplication by the matrix I leaves any vector x unchanged.
• This vector is thus the identity vector for multiplication of 3-row matrices.

We can represent the system of two equations at the top of the figure below as a matrix calculation, as indicated.

• In the matrix form the equation can be represented by the symbolic equation A x = y, where A is the 2 x 2 matrix.
• If we can solve this equation for x, which is the unknown column vector [x, y]`, we will have the solutions x and y of the equations.

To solve an equation of the form A x = y, we are inclined to divide by A.

• However, matrix multiplication involves the somewhat complex task of multiplying rows by columns, and is not clear what we would have to do to divide matrices.

It is better to think of multiplying by the inverse matrix A^-1.

• The product of two inverses is the identity, so when we multiply both sides by A^-1 we obtain simply I x = x = A^-1 y.

The only problem is, we don't know yet how to get A^-1.

Video File #01

We will return to the problem of finding A^-1 later. For now we will review the solution of the set of simultaneous equations, then develop a matrix method for finding the solution.

Our strategy will be to eliminate x from the second equation, then to eliminate y from the first.

• We eliminate x by multiplying the first equation by 2 and the second equation by 3, then adding.
• The second equation becomes 22 y = 36, and is easily solved for y.
• At the right of the figure we have rewritten the two equations, with the second equation replaced by the resulting equation 22 y = 36.
• We have then rewritten the two equations once more after dividing the second equation by 22.

This last version of the equations is repeated at the top of the figure below.

• If from the first equation we subtract 10 times the second, we will have eliminated y from the first, as indicated in the second version of the equations.

• This subtraction is indicated at the right of the figure below.

• The third version of the equations simply divides the first by 6, obtaining a solution for x.

In the figure below we indicate a matrix version of the two equations.

• It is understood that the numbers in the first column of the matrix represent the coefficients of x while the second column represents the coefficients of y; the third column represents the numbers to the right of the = sign.
• Thus the two rows of the matrix represent the two equations.

We proceed to 'reduce' the matrix, using the same steps we used in solving the system of equations.

• The second matrix at left below represents the first equation multiplied by 2 and the second by -3.
• In the third matrix the first row represents the first equation while the second row represents what we get when we add the first equation to the second.
• This second row represents the equation 22 y = 36.
• In the fourth matrix we have divided the second row by 22, equivalent to dividing the second equation by 22.
• The fifth matrix (top right in the figure) is the same as the fourth.
• The sixth matrix is obtained by adding -10 times the second row to the first, equivalent to adding -10 times the second equation to the first.
• In the final matrix we have divided the first row by 6.

Our final matrix represents the equations x = 3/11, y = 18/11.  These equations are the solution of the system.

Video File #02

In the figure below we see how the matrix equation A x = b is represented by product of two matrices, then by a single matrix which by the process of row reduction is reduced to an identity matrix plus one column.

• This new matrix is equivalent to the matrix equation at the bottom of the figure, which is equivalent to the final equation at bottom right.
• This final equation gives us the values of x and y.

In the figure below we reduce the original matrix to an equivalent upper-triangular form, from which we will be easily able to determine the solutions of the equations represented by the original matrix.

We consider the first column to be the coefficients of x, the second to be the coefficients of y, the third to be the coefficients of z, and the fourth to be the numbers to the right of the = sign.

We will begin by eliminating the x coefficients of the second and third equations, then we will eliminate the y coefficient of the third line.

• To eliminate the x coefficients in the second and third equations, we subtract half the first row from the second and 1/4 of the first row from the third.
• To then eliminate the y coefficients in the third row, we subtract 1/3 of the second row from the third.

• It was important that the second row have 0 as the x coefficient before subtracting it from third, since otherwise we would have 'messed up' the 0 at the beginning of the third row.

We can now easily find the solutions x, y and z.

• The third equation now represents 10 z = 675, which we can easily solve for z.
• The second equation represents 12 y + 3 z = 600; substituting our newly found value for z we will obtain an equation from which we can obtain y.
• Substituting our values of y and z into the equation corresponding to the first row we can easily obtain x.

Video File #03