Teaching Notes · Math for Social Sciences

An Introduction to Matrix Algebra

Every set of notes so far has worked with one equation, or at most a small system of two, solved by hand. A matrix is simply a rectangular grid of numbers, and once a problem is written in matrix form, an entire system, however many equations and unknowns it has, can be solved by following the same short list of mechanical steps every time. This closing set of notes introduces matrices as a way to organize numbers, shows how a system of equations becomes a single matrix equation, solves a system by reducing a matrix directly, and ends with the matrix inverse, a tool that turns solving a system into little more than a multiplication.

Matrices as a way to organize numbers

A row of numbers is a row vector, and a column of numbers is a column vector; a general matrix is simply a grid with both rows and columns. Suppose the prices, in dollars per unit, of three products A, B, and C are collected into a price vector P = [2 3 4], and the quantities purchased of each are collected into a column vector Q = [7, 5, 11]. Multiplying a row by a column means pairing up entries in order and adding the products: PQ = [(2·7) + (3·5) + (4·11)] = [73], the total cost of the purchase, 73 dollars, produced in a single matrix multiplication rather than three separate multiplications added up by hand afterward.

Writing a system of equations as AX = B

The same multiplication rule turns a whole system of equations into a single matrix equation. For the system 2x₁ + 5x₂ = 4 and 8x₁ + 3x₂ = 7, collect the coefficients into a matrix A, the unknowns into a column X, and the right hand side into a column B:

A = [[2, 5], [8, 3]], X = [x₁, x₂], B = [4, 7]

and the system is exactly AX = B. Nothing about the equations has changed, only the notation; but that notation is what makes the row reduction and inverse methods below possible, since both work directly on the coefficient matrix A rather than on the equations as written.

Solving a system by reducing a matrix

A matrix can be transformed into an equivalent one, meaning one that represents the same solution, using three elementary row operations: interchanging two rows, multiplying a row by a nonzero number, and adding a multiple of one row to another. Applied patiently to an augmented matrix, a coefficient matrix with the right hand side attached as an extra column, these operations reduce the system to one where each unknown can simply be read off.

At an electronics sale, a customer spending a combined total of 3500 pays 80 percent of whatever is left after the refrigerator for the television, and 85 percent of whatever is left after the television for the refrigerator. Writing T and R for the two payable prices, the two conditions become T + 0.8R = 2800 and 0.85T + R = 2975. As an augmented matrix, [[1, 0.8 | 2800], [0.85, 1 | 2975]]; subtracting 0.85 times the first row from the second clears the first entry of the second row, leaving [0, 0.32 | 595], so R = 595/0.32 ≈ 1,859.38. Substituting back into the first equation gives T = 2800 − 0.8(1859.38) ≈ 1,312.50, both payable prices falling out of the same two row operations.

The inverse of a matrix

If A is a square matrix and some matrix C satisfies CA = I, the identity matrix, with 1s down its diagonal and 0s everywhere else, then C is called an inverse of A, written A⁻¹, and A is said to be invertible. The inverse is found by exactly the row reduction technique above, applied to an augmented matrix [A | I] instead of [A | B]: whatever sequence of row operations reduces A to I on the left automatically turns I into A⁻¹ on the right.

For A = [[1, 1], [0, 2]], start from [A | I] and scale the second row by one half, then subtract the new second row from the first:

Solving AX = B with the inverse

Once A⁻¹ is known, the system AX = B solves itself: multiplying both sides on the left by A⁻¹ gives A⁻¹AX = A⁻¹B, and since A⁻¹A = I, this reduces to X = A⁻¹B. For the system 3x + y = 10 and 5x + 2y = 17, A = [[3, 1], [5, 2]] has inverse A⁻¹ = [[2, −1], [−5, 3]] (found by the same row reduction method, or by the shortcut formula for a 2 by 2 matrix). Multiplying,

X = A⁻¹B = [[2, −1], [−5, 3]] [10, 17] = [(2·10) + (−1·17), (−5·10) + (3·17)] = [3, 1]

so x = 3 and y = 1, a solution that checks immediately against both original equations: 3(3) + 1 = 10 and 5(3) + 2(1) = 17. The real advantage of this method shows up once a system grows past two equations: the inverse of a 3 by 3 or larger coefficient matrix is found by exactly the same row reduction procedure, and from there every system that shares that same coefficient matrix, however many different right hand sides B it is paired with, is solved by the same one multiplication.

A matrix is, in the end, nothing more than a bookkeeping device: a way of keeping an entire system of numbers organized so that the same handful of row operations can be applied uniformly, rather than improvised fresh for every new system. That economy of effort, one method covering any number of equations, is exactly why matrix algebra closes out this course rather than opening it.

Source: Ernest F. Haeussler Jr., Richard S. Paul, and Richard J. Wood, Introductory Mathematical Analysis for Business, Economics, and the Life and Social Sciences, 14th edition (Pearson Canada, 2019), chapter 6, covering matrices, systems of equations in matrix form, elementary row operations, and the inverse of a matrix.