Matrices Reference

2x2 Matrices

A 2x2 matrix can organize coefficients, transform points in the plane, and solve two-variable systems. The determinant tells whether the matrix has an inverse and whether a transformation collapses area.

Solved Examples

Jump directly to a worked example by matrix task.

Fact Table

Task Use Reminder
Add or subtract matrices. Combine matching entries. Matrices must have the same dimensions.
Multiply a matrix by a vector. Rows dot columns. This sends one point to a new point.
Multiply two matrices. Row-by-column products. Order matters: usually $AB\ne BA$.
Check whether a 2x2 matrix has an inverse. Determinant. If $\det(A)=0$, there is no inverse.
Solve a two-variable system. $X=A^{-1}B$. The coefficient matrix must be invertible.
Describe a plane transformation. Track where basis vectors go. The columns of the matrix are the transformed basis vectors.

Content Formulas

Matrix-Vector Product
$$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}ax+by\\cx+dy\end{bmatrix}$$
Matrix Product
$$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}e&f\\g&h\end{bmatrix}=\begin{bmatrix}ae+bg&af+bh\\ce+dg&cf+dh\end{bmatrix}$$
Determinant
$$\det\begin{bmatrix}a&b\\c&d\end{bmatrix}=ad-bc$$
Inverse
$$A^{-1}=\frac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$
System Form
$$AX=B,\quad X=A^{-1}B$$
Area Scale
$$\text{area scale}=|\det(A)|$$
For $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$, the first column shows where $(1,0)$ goes and the second column shows where $(0,1)$ goes.

Classic Examples

Add Matrices

Add $\begin{bmatrix}1&2\\3&4\end{bmatrix}+\begin{bmatrix}4&3\\2&1\end{bmatrix}$.

Solution Steps
  1. Add matching matrix positions.
  2. State the resulting matrix.
$$\begin{bmatrix}1&2\\3&4\end{bmatrix}+\begin{bmatrix}4&3\\2&1\end{bmatrix}=\begin{bmatrix}1+4&2+3\\3+2&4+1\end{bmatrix}$$ $$=\begin{bmatrix}5&5\\5&5\end{bmatrix}$$

Transform a Point

Apply $A=\begin{bmatrix}1&2\\0&3\end{bmatrix}$ to the point $(2,1)$.

Solution Steps
  1. Write the point as a column vector.
  2. Multiply each row by the vector.
  3. State the transformed point.
$$A\begin{bmatrix}2\\1\end{bmatrix}=\begin{bmatrix}1&2\\0&3\end{bmatrix}\begin{bmatrix}2\\1\end{bmatrix}$$ $$=\begin{bmatrix}1(2)+2(1)\\0(2)+3(1)\end{bmatrix}$$ $$=\begin{bmatrix}4\\3\end{bmatrix}$$

Multiply Matrices

Find $\begin{bmatrix}1&2\\3&4\end{bmatrix}\begin{bmatrix}2&0\\1&2\end{bmatrix}$.

Solution Steps
  1. Compute each row-by-column dot product.
  2. State the product matrix.
$$\begin{bmatrix}1&2\\3&4\end{bmatrix}\begin{bmatrix}2&0\\1&2\end{bmatrix}=\begin{bmatrix}1(2)+2(1)&1(0)+2(2)\\3(2)+4(1)&3(0)+4(2)\end{bmatrix}$$ $$=\begin{bmatrix}4&4\\10&8\end{bmatrix}$$

Determinant

Find the determinant of $A=\begin{bmatrix}1&2\\3&5\end{bmatrix}$.

Solution Steps
  1. Multiply the main and other diagonals, then subtract.
  2. Simplify the determinant expression.
  3. State the determinant: $-1$.
$$\det(A)=1(5)-2(3)$$ $$\det(A)=5-6$$ $$\det(A)=-1$$

Inverse Matrix

Find $A^{-1}$ for $A=\begin{bmatrix}1&2\\3&5\end{bmatrix}$.

Solution Steps
  1. Find the determinant.
  2. Apply the 2x2 inverse formula.
  3. State the inverse matrix.
$$\det(A)=1(5)-2(3)=-1$$ $$A^{-1}=\frac{1}{-1}\begin{bmatrix}5&-2\\-3&1\end{bmatrix}$$ $$A^{-1}=\begin{bmatrix}-5&2\\3&-1\end{bmatrix}$$

Solve a System

Solve $x+2y=5$ and $3x+5y=13$ using matrices.

Solution Steps
  1. Build the coefficient matrix $A$ and constants vector $B$.
  2. Multiply $A^{-1}B$.
  3. Simplify the solution vector.
  4. State $x=1$ and $y=2$.
$$A=\begin{bmatrix}1&2\\3&5\end{bmatrix},\quad B=\begin{bmatrix}5\\13\end{bmatrix}$$ $$X=\begin{bmatrix}-5&2\\3&-1\end{bmatrix}\begin{bmatrix}5\\13\end{bmatrix}$$ $$X=\begin{bmatrix}-25+26\\15-13\end{bmatrix}=\begin{bmatrix}1\\2\end{bmatrix}$$ $$x=1,\quad y=2$$

Read a Transformation

Describe $A=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$.

Solution Steps
  1. Read the image of the first basis vector from column one.
  2. Read the image of the second basis vector from column two.
  3. Identify the transformation as a $90^\circ$ counterclockwise rotation.
$$A\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}0\\1\end{bmatrix}$$ $$A\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}-1\\0\end{bmatrix}$$ This is a $90^\circ$ counterclockwise rotation.