Determinants and Inverses

The Determinant of a 2 × 2 Matrix #

One diagonal product, less the other.

The determinant of a 2 × 2 matrix is ad − bc, and it is zero exactly when no inverse exists

Multiply down the main diagonal: 4 × 3 = 12.

Then multiply the other diagonal: 2 × 5 = 10.

The determinant takes one diagonal product from the other: det A = 2.

When the two products match, det = 0 and the matrix has no inverse.

The Inverse of a 2 × 2 Matrix #

Swap the diagonal, flip two signs, divide.

The inverse of a 2 × 2 matrix swaps the main diagonal, flips the other two signs and divides by the determinant

Swap the two entries on the main diagonal: a and d change places.

Flip the signs of the other two entries.

Last, divide every entry by the determinant, which is ad − bc.

Here ad − bc = 1, so the inverse of (3 5; 1 2) is (2 −5; −1 3).

Check it by multiplying back: a matrix times its inverse gives I.

Solving Simultaneous Equations with a Matrix Inverse #

Two equations, one matrix, one step.

A pair of linear equations is one matrix equation, and the inverse solves it in a single step

The coefficients make A, the unknowns make one column, and the answers make another.

Row 1 multiplies out to 3x + 5y = 11, which is the equation you started with.

Multiply on the left by the inverse and the left side becomes just X.

Work out the product: the column that comes out is x = 2 and y = 1.

Both equations hold, and elimination would find the same pair.

The Determinant of a 3 × 3 Matrix #

Three smaller determinants, with alternating signs.

A 3 × 3 determinant expands along a row as three 2 × 2 determinants with alternating signs

Expand along row 1: each entry there multiplies a smaller determinant.

Delete the row and column through 2: the 2 × 2 that is left has determinant 24.

Delete them through 1: that 2 × 2 has determinant −5.

Delete them through 3: that 2 × 2 has determinant −4.

Multiply each entry by its own 2 × 2, alternate the signs, and add.

The Inverse of a 3 × 3 Matrix #

Minors, then signs, then a transpose.

A 3 × 3 inverse is the transposed matrix of cofactors divided by the determinant

Every entry has a minor: delete its row and column, and keep the 2 × 2 that is left.

Give each minor the sign from this pattern and it becomes a cofactor.

Reflect the cofactors in the main diagonal: that transpose is the adjugate.

Divide the adjugate by the determinant, which is 1 here, and the inverse is done.

Row 1 times column 1 gives 1, and every other row-column pair gives 0: that is I.

Solving Three Equations with a Matrix Inverse #

Three unknowns, one product, no elimination.

Three equations in three unknowns are one matrix equation, solved in a single step by the inverse

There are three equations and three unknowns, and a term is missing here and there.

Stack the coefficients into A and the three become one equation, A X = b.

Row 1 multiplies out to x + 2y + 3z = 14, the equation it was built from.

The step is the one you already know: multiply on the left by the inverse.

One product reads x = 1, y = 2 and z = 3 — all three unknowns at once.

It works only while det A is not 0, and here that determinant is 1.

The Geometry of Three Planes #

A point, a line, or nowhere at all.

Three linear equations in three unknowns are three planes, meeting at a point, along a line, or nowhere

One linear equation in x, y and z draws a plane, so three of them draw three.

When det A is not 0 the inverse exists, and the three cross at a single point.

Here row 3 is row 1 plus row 2, so det B is 0 and one equation repeats the others.

The totals add up too, so only two equations are independent and a whole line of solutions is left.

Change 5 to 9 and no point can satisfy all three, so the planes never meet at once.

The determinant decides whether there is one point, and the totals decide between a line and nothing.