Determinants and Inverses

Stage 13 of 23 Strand 2 of 5 7 lessons

7 illustrated lessons, each teaching the why before the how.

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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

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A4253

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

A4253

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

det A = ad − bcthe two diagonal productsdet A = 12 − 10det A = 2

For A = (a b; c d), det A = ad − bc, one diagonal product minus the other: det A = 2.

B2412

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

Now you

6353

What is the determinant of this matrix?

1326

Does this matrix have an 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, changes the signs of the other two entries and divides by the determinant

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Aabcddbca→

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

dbcad−b−ca→

Change the signs of the other two entries, b and c.

1 ÷ det Ad−b−ca×

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

A3512A⁻¹2−5−13→

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

A3512A⁻¹2−5−13I1001×=

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

Now you

A3512

This matrix has determinant 1. Which matrix is its inverse?

A5364

The determinant here is 2. What goes in row 1, column 1 of the inverse?

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

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A3512xyb114×=

For 3x + 5y = 11 and x + 2y = 4, the coefficients make A, x and y make one column, and 11 and 4 make b.

A3512xyb114×=

Row 1 multiplies out to 3x + 5y = 11, the first equation.

A X = bmultiply both sides on the left by the inverse
A⁻¹A is I, and I X is just X

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

A⁻¹2−5−13b11421×=

Work out A⁻¹b: 2 × 11 − 5 × 4 = 2 and −11 + 3 × 4 = 1, so x = 2 and y = 1.

3(2) + 5(1) = 11both equations hold2 + 2(1) = 4

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

Now you

A5221xyb136×=

What is x?

A1428xyb58×=

This pair cannot be solved this way. What is det A?

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

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M213045106

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

M213045106

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

M213045106

Delete the row and column through 1: that 2 × 2 has determinant −5.

M213045106

Delete the row and column through 3: that 2 × 2 has determinant −4.

2(24), 1(−5), 3(−4)signs alternate: plus, minus, plusdet M = 2(24) − 1(−5) + 3(−4)det M = 48 + 5 − 12det M = 41

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

Now you

662434112

Expanding along row 1, what sign goes with the entry in column 1?

022031342

What is the determinant of the 2 × 2 left in gold?

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

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A123014560

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.

cofactors−2420−518−1545−41adjugate−2418520−15−4−541→

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

1 ÷ det Aadjugate−2418520−15−4−541×

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

A123014560A⁻¹−2418520−15−4−541×

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

Now you

416444363

What sign does the cofactor in row 2, column 3 take?

651565435

What is the cofactor of the entry in row 3, column 2?

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

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x + 2y + 3z = 14y + 4z = 14a missing term is a coefficient of 05x + 6y = 17

There are three equations in three unknowns, and two of them have a term missing.

A123014560xyzb141417×=

Put the coefficients into A, and the three equations become one: A X = b.

A123014560xyzb141417×=

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

A X = bmultiply on the left, as at two unknowns

Multiply both sides on the left by A⁻¹, exactly as with two unknowns.

A⁻¹−2418520−15−4−541b141417123×=

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

A123014560

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

Now you

A123014560xyzb141417123×=→

X = A⁻¹ b came out as this column. What is x?

A123246411xyzb6125×=

Why can this system not be solved by an inverse?

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

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x + y + z = 3each equation is a flat plane in space2x + y − z = 2

A linear equation in x, y and z describes a plane, so three equations describe three planes.

A123014560

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

B11121−1320

Here row 3 is row 1 plus row 2, so det B is 0.

x + y + z = 32x + y − z = 23 + 2 = 5, so it holds automatically3x + 2y = 5

The right-hand sides add the same way, so the equations are consistent: only two are independent, and a whole line of points solves them.

3x + 2y = 5same left-hand side, a different right-hand side3x + 2y = 9

Change 5 to 9 and the equations are inconsistent: no point lies on all three planes.

one point
det = 0, consistent line
det = 0, inconsistent no point

The determinant decides whether there is one point; when it is 0, consistency decides between a line and no point.

Now you

B11121−1320

det B is 0 because row 3 is row 1 plus row 2, and the right-hand sides add the same way. How do the planes meet?

3x + 2y = 5both cannot hold3x + 2y = 9

det B is 0 and the equations are inconsistent. How do the three planes meet?

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