Matrices become something more ...

You've learned about using matrices to represent and solve linear systems, either through Gaussian elimination or by finding the inverse of the coefficient matrix, if it exists.

Now we'll see how matrices can have aspects that can move you beyond just solving linear systems. The determinant concept is kind of odd at first.

What is a determinant? — A 2-D linear system

We'll begin with a 2-equations, 2-unknown (2-D) problem and look for some general (not dependent on the actual numbers) solutions. The system is:

$$ \begin{align} a_1 x + b_1 y = c_1 \\[5pt] a_2 x + b_2 y = c_2 \end{align}$$

where $a_1$, $a_2$, $b_1$, and $b_2$ are the numerical coefficients of the variables $x$ and $y$. Here is that system in matrix form:

$$ \left( \begin{matrix} a_1 & b_1 \\[5pt] a_2 & b_2 \end{matrix} \right) \left( \begin{matrix} x \\[5pt] y \end{matrix} \right) = \left( \begin{matrix} c_1 \\[5pt] c_2 \end{matrix} \right)$$

Now if we solve each equation for $y$, we get these two equations, each from one of the original linear equations:

$$y = \frac{c_1 - a_1 x}{b_1} \; \; \color{#E90F89}{\text{and}} \; \; y = \frac{c_2 - a_2 x}{b_2}$$

If each right-hand side is equal to $y$, then each must be equal to the other (transitive property):

$$\frac{c_1 - a_1 x}{b_1} = \frac{c_2 - a_2 x}{b_2}$$

Now cross multiply (or multiply both sides by $b_1 b_2$):

$$b_2 c_1 - a_1 b_2 x = b_1 c_2 - b_1 a_2 x$$

Then group terms containing $x$ on the left side:

A shortcut to the 2 × 2 determinant

The determinant of a 2×2 matrix is the difference of the products along its two diagonals. It looks like this:

Examples — 2×2 determinants

In the top example, the linear system represented by the matrix is not linearly independent because one row can be formed by applying a linear operation on the other. For example, if the top row is multiplied by -1, the result would be the bottom row. There is really only one piece of information there to find two variables, so the system can't be solved, and this is indicated by the zero determinant.

The other two matrices can be solved because their determinants are not zero.

Practice problems

Calculate the determinants of these 2×2 systems of equations

1. $$ \begin{align} 2x - 4y &= 6 \\[5pt] -x + 3 y &= -2 \end{align}$$

   Solution

   
   

   The determinant is

   $$\begin{align} \text{det} \left( \begin{matrix} 2 & -4 \\ -1 & 4 \end{matrix} \right) &= 2(4) -(1-)(-4) \\[5pt] &= 8 - 4 \\[5pt] &= 4 \end{align}$$

   The determinant is not zero, so this system of equations has a solution.

2. $$ \begin{align} x + 3y &= 7 \\[5pt] x + 2y &= 10 \end{align}$$

   Solution

   
   

   The determinant is

   $$\begin{align} \text{det} \left( \begin{matrix} 1 & 3 \\ 1 & 2 \end{matrix} \right) &= 1(2) - 1(3) \\[5pt] &= 2 - 3 \\[5pt] &= -1 \end{align}$$

   The determinant is not zero, so this system of equations has a solution.

How can determinants be used to solve linear systems:
Cramer's rule

With the idea of determinants of 2×2 matrices in hand, let's return to the expressions for $x$ and $y$ in our 2-equations, 2 unknowns system above. They were:

$$ \begin{align} x &= \frac{b_2 c_1 - b_1 c_2}{a_1 b_2 - a_2 b_1} \\[8pt] y &= \frac{a_1 c_2 - a_2 c_1}{a_1 b_2 - a_2 b_1} \end{align}$$

We've already shown that the identical denominators are the determinant of the matrix of coefficients for this system. The numerators can be written as determinants of 2 × 2 matrices, too.

Those matrices consist of the coefficient matrix with the first column substituted with the result vector $(c_1, \; c_2)$ for the $x$-coordinate, and with $(c_1, \; c_2)$ substituted for the second column for the $y$-coordinate. Here's a picture:

Geometric interpretation of the 2×2 determinant

The 2×2 determinant can be interpreted geometrically as the area of the parallelogram of the matrix formed by two row vectors. Here's the setup: Take vectors $(a, b)$ and $(b, c)$ and locat them at the origin. Then the parallelogram that they define is shown below:

Now the area of that parallelogram is just the length of a base multiplied by the height. If we choose vector $(a, b)$ to be the base, then its length is easy to find, $|(a, b)| = \sqrt{a^2 + b^2}$, but the height is a little trickier. It's all made a lot simpler if we just determine the angle between vector $(a, b)$ and the x-axis, then rotate the whole parallelogram by that much in the clockwise direction.

Under a rotation like that, the lengths of the vectors would be preserved, so our parallelogram would be the samd — and have the same area. Here's how that would look. We'll re-label our vectors as $(a', b')$ and $(c', d')$ to show that they are different, but that their lengths are preserved.

The

Now notice that vector $(a, b)$ becomes $(a', 0)$. The area of our parallelogram is just $A = a' \times d'$.

Now let's try that with some real vectors just to see how it works. Let's let $\vec a = (2, 1)$ and $\vec b = (1, 3)$. Here's the diagram:

Practice problems — Cramer's rule

Solve these 2×2 systems of equations using Cramer's rule. (These are the same systems as in the practice problems above).

1. $$ \begin{align} 2x - 4y &= 6 \\[5pt] -x + 3 y &= -2 \end{align}$$

   Solution

   
   

   The relevant determinants are

   $$\text{det} A = \left( \begin{matrix} 2 & -4 \\ -1 & 3 \end{matrix} \right) = 2 $$

   $$\text{det} A_x = \left( \begin{matrix} 6 & -4 \\ -2 & 3 \end{matrix} \right) = 10 $$

   $$\text{det} A_y = \left( \begin{matrix} 2 & 6 \\ -1 & -2 \end{matrix} \right) = 2 $$

   Then the x and y components of our solution are

   $$\begin{align} x &= \frac{\text{det} (A_x)}{\text{det} (A)} = \frac{10}{-2} = 5 \\[5pt] y &= \frac{\text{det} (A_y)}{\text{det} (A)} = \frac{2}{-2} = 1 \end{align}$$

   Our solution is $(x, y) = (5, 1)$.

   ---

2. $$ \begin{align} x + 3y &= 7 \\[5pt] x + 2y &= 10 \end{align}$$

   Solution

   
   

   The relevant determinants are

   $$\text{det} A = \left( \begin{matrix} 1 & 3 \\ 1 & 2 \end{matrix} \right) = -1 $$

   $$\text{det} A_x = \left( \begin{matrix} 7 & 3 \\ 10 & 2 \end{matrix} \right) = -16 $$

   $$\text{det} A_y = \left( \begin{matrix} 1 & 7 \\ 1 & 10 \end{matrix} \right) = 3 $$

   Then the x and y components of our solution are

   $$\begin{align} x &= \frac{\text{det} (A_x)}{\text{det} (A)} = \frac{-16}{-1} = 16 \\[5pt] y &= \frac{\text{det} (A_y)}{\text{det} (A)} = \frac{3}{-1} = -3 \end{align}$$

   Our solution is $(x, y) = (16, -3)$.

   ---

Determinants of larger matrices

You might be saying to yourself, "Self, this is nice, but I can solve a two-equation — two unknown problem like these faster the old fashioned way." And you'd be right. In order to make Cramer's rule really useful, we'd like to apply it to very large systems of linear equations, ones with dimensions in the hundreds or thousands.

A trick for finding 3×3 determinants

Here is the formula for the determinant of any 3 × 3 matrix. It looks daunting, but don't worry, there's an easy way to do it that doesn't require memorization of a formula.

$$\text{det} \left( \begin{matrix} a & b & c \\ e & f & g \\ i & j & k \end{matrix} \right) = afk + bgi + cej - [ifc + jga + keb]$$

The way to remember this method is illustrated below. First, rewrite the first two columns of the matrix, (a, e, i) and (b, f, j), to the right of the matrix. Next, draw in the red diagonals as shown below:

The first three terms of the determinant are the products of the elements along all of these red diagonals: afk + bgi + cej. Then draw in the blue diagonals in the opposite direction (lower left to upper right), as shown. The next three terms are the products along those diagonals subtracted from the first part: -ifc - jga - keb. It's simple.

Some examples of 3×3 determinants

In this example, the determinant is zero, meaning that the system of 3 equations and 3 unknowns represented by this matrix can't be solved. Notice that the third row of the matrix is just the sum of the first two. It's a linear combination of the first two, so it's not linearly independent.

Here are two more examples, just so you can practice finding 3×3 determinants:

The latter two matrices represent solvable systems of equations – three equations and three unknowns – because they have non-zero determinants.

Cramer's rule with a 3×3 system of equations

Cramer's rule extends nicely to 3×3 and larger square matrices. Let's use it to solve a 3×3 system:

(Narrow screens: scroll wide equations left ↔ right)

$$ \begin{align} 3x + 2y - z &= 4 \\[5pt] x - 3y + z &= -2 \\[5pt] 2x + 2y + z &= 9 \end{align} \; \color{magenta}{\xrightarrow{\text{matrix form}}} \; \left( \begin{matrix} 3 & 2 & -1 \\ 1 & -3 & 1 \\ 2 & 2 & 1 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) = \left( \begin{matrix} 4 \\ -2 \\ 9 \end{matrix} \right)$$

Now the key to doing these problems by hand is to take your time and stay organized. First, we'll find the determinant of the coefficient matrix, which we'll call $A$. It will be the denominator in our expressions for $x$, $y$ and $z$.

$$ \begin{align} \text{det}(A) &= \text{det}\begin{bmatrix} 3 & 2 & -1 \\ 1 & -3 & 1 \\ 2 & 2 & 1 \end{bmatrix} \\[5pt] &= 3(-3)\cdot 1 + 2 \cdot 1 \cdot 2 + (-1) 1 \cdot 2 - 2(-3)(-1) - 2 \cdot 1 \cdot 3 - 1 \cdot 1 \cdot 2 \\[5pt] &= -9 + 4 - 2 - 6 - 6 - 2 = -21 \end{align}$$

Now the determinants of the matrices substituted with the result vector (4, -2, 9) (pink highlights):

$$ \begin{align} \text{det}(A_1) &= \text{det}\begin{bmatrix} \color{magenta}{4} & 2 & -1 \\ \color{magenta}{-2} & -3 & 1 \\ \color{magenta}{9} & 2 & 1 \end{bmatrix} \\[5pt] &= 4(-3)\cdot 1 + 2 \cdot 1 \cdot 9 + (-1)(-2) \cdot 2 - 9(-3)(-1) - 2 \cdot 1 \cdot 4 - 1(-2)2 \\[5pt] &= -12 + 18 + 4 - 27 - 8 + 4 = -21 \end{align}$$

$$ \begin{align} \text{det}(A_2) &= \text{det}\begin{bmatrix} 3 & \color{magenta}{4} & -1 \\ 1 & \color{magenta}{-2} & 1 \\ 2 & \color{magenta}{9} & 1 \end{bmatrix} \\[5pt] &= 3(-2)\cdot 1 + 4 \cdot 1 \cdot 2 + (-1) 1 \cdot 9 - 2(-2)(-1) - 9 \cdot 1 \cdot 3 - 1 \cdot 1 \cdot 4 \\[5pt] &= -6 + 8 - 9 - 4 - 27 - 4 = -42 \end{align}$$

$$ \begin{align} \text{det}(A) &= \text{det}\begin{bmatrix} 3 & 2 & \color{magenta}{4} \\ 1 & -3 & \color{magenta}{-2} \\ 2 & 2 & \color{magenta}{9} \end{bmatrix} \\[5pt] &= 3(-3)\cdot 9 + 2(-2)2 + 4 \cdot 1 \cdot 2 - 2(-3)4 - 2(-2)3 - 9 \cdot 1 \cdot 2 \\[5pt] &= -81 - 8 + 8 + 24 + 12 - 18 = -63 \end{align}$$

Now we just do the divisions and arrive at the solution:

$$(x, y, z) = (1, 2, 3)$$

Cramer's rule can be extended to larger matrices & systems of equations so long as we have n linearly independent equations to find n unknowns. It's also fairly easy to program to solve large systems by computer algorithm.

Geometric interpretation of the 3×3 determinant

By extension from our parallelogram example above, the determinant of a 3×3 matrix is the volume of the 3-D parallelogram (a parallelepiped) traced out by the three row vectors that form our 3×3 matrix.

A parallelepiped is defined by making copies of the three planes defined by pairs of vectors $\vec v_1 \text{ and } \vec v_2$ $\vec v_1 \text{ and } \vec v_3$ and $\vec v_2 \text{ and } \vec v_3$. The solid is formed by repeating these three plane sections (parallelograms) in the same manner that we repeated the vectors above to make our parallelogram.

As a simple example, consider a parallelepiped formed by vectors along the three 3D axes: $\vec v_1 = (1, 0, 0)$, $\vec v_2 = (0, 2, 0)$ and $\vec v_3 = (0, 0, 2)$. The determinant of the matrix formed by these three row vectors is

$$ \text{det} \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix}\right) = 1(2)(2) = 4 \text{ units}^3$$

That's the correct volume for this 1 × 2 × 2 rectangular solid.

Properties of determinants

There are several properties of determinants that are useful to keep in mind. Here they are.

1. The determinant of a diagonal matrix is the product of all of the diagonal elements.

For a matrix with elements

$$\left( \begin{matrix} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ a_{n1} & a_{n2} & \dots & a_{nn} \end{matrix} \right)$$

where only the elements $a_{mm}$ are not zero, the determinant is just the product of those diagonal elements. For a 2×2 or 3×3 matrix, this is easy to show. Any diagonal product other than that of the main diagonal will contain at least one zero, and thus will be zero. Here are two examples, first a 2×2:

$$\text{det}\left( \begin{matrix} 2 & 0 \\ 0 & -3 \end{matrix} \right) = 2(-3) - 0(0) = -3$$

then a 3×3:

$$ \begin{align} &\text{det}\left( \begin{matrix} -1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 2 \end{matrix} \right) \\[5pt] &=-1(3)(2) + 0 \cdot 0 \cdot 0 + 0 \cdot 0 \cdot 0 \\[5pt] &-[0 \cdot 3 \cdot 0 + 0 \cdot 0 \cdot -1 + 2 \cdot 0 \cdot 0] \\[5pt] &= -6 \end{align}$$

It's easy to see that in any such determinant, the only non-zero product will be along the diagonal of the matrix.

2. The determinant of a block-diagonal matrix is the product of the determinants of the blocks.

A block-diagonal matrix has a form like this (I'll stick to a 4×4 here):

$$M = \begin{bmatrix} x_1 & x_2 & 0 & 0 \\ x_3 & x_4 & 0 & 0 \\ 0 & 0 & x_5 & x_6 \\ 0 & 0 & x_7 & x_8 \end{bmatrix}$$

Many large matrices may have – or may be coaxed into having – such a form. The nonzero elements form square "blocks" that share the diagonal of the larger matrix. In this case, instead of computing the determinant of the whole matrix, we can just calculate the determinants of each block and multiply them. Here's a real example using a 3×3 matrix with blocks of size 1×1 and 2×2.

$$\text{det} \left( \begin{matrix} \color{red}{2} & \color{#CCC}{0} & \color{#CCC}{0} \\ \color{#CCC}{0} & 2 & -1 \\ \color{#CCC}{0} & 3 & 1 \end{matrix} \right) = \color{red}{2} \cdot [2(1) - 3(-1)] = 10 $$

Here we took the determinant of the 1×1 block ( 2 ) and multiplied it by the determinant of the 2×2 block to get the determinant of the 3×3 [2·1 - 3·(-1)] = 5.

3. When any two rows or any two columns of a matrix are swapped, the determinant changes sign.

Let's take the simplest case, a 2×2 matrix,

$$\text{det} \left( \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} \right) = a_{11}a_{22} - a_{21}a_{12}$$

7. The determinant of a product of two matrices is the product of the individual determinants.

(Narrow screens: scroll wide equations left ↔ right)

That is, $\text{det}AB = \text{det}A \cdot \text{det}B$ Again we'll just do this for the 2×2 case. We know that

$$ \begin{align} \text{det} \left( \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} \right) &= a_{11} a_{22} - a_{21} a_{12} \\[5pt] \text{det} \left( \begin{matrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{matrix} \right) &= b_{11} b_{22} - b_{21} b_{12} \end{align}$$

Now let's multiply matrices $A$ and $B$:

$$A \cdot B = \left( \begin{matrix} a_{11}b_{11} + a_{12}b_{21} & a_{11}b_{12} + a_{12}b_{22} \\ a_{21}b_{11} + a_{22}b_{21} & a_{21}b_{12} + a_{22}b_{22} \end{matrix} \right)$$

Then the determinant of the product is

$$ \begin{align} \text{det}(AB) &= (a_{11}b_{11} + a_{12}b_{21})(a_{21}b_{12} + a_{22}b_{22}) - [(a_{21}b_{11} + a_{22}b_{21})(a_{11}b_{12} + a_{12}b_{22})] \\[5pt] &= \cancel{a_{11}a_{21}b_{11}b_{12}} + a_{11}a_{22}b_{11}b_{22} + a_{12}a_{21}b_{11}b_{12} + \cancel{a_{12}a_{22}b_{21}b_{22}} \\[5pt] &\phantom{00} - \cancel{a_{11}a_{21}b_{11}b_{12}} - a_{12}a_{21}b_{11}b_{12} - a_{11}a_{22}b_{12}b_{21} - \cancel{a_{12}a_{22}b_{21}b_{22}} \\[5pt] &= a_{11}a_{22}(b_{11}b_{22}-b_{12}b_{21}) - a_{12}a_{21}(b_{11}b_{22}-b_{12}b_{21}) \\[5pt] &= (a_{11}a_{22}-a_{12}a_{21})(b_{11}b_{22}-b_{12}b_{21}) \\[5pt] &= \text{det}A \cdot \text{det}B \end{align}$$