CP 3.6 - Singular and inverse matrices

CP 3.6 - Singular and inverse matrices

An inverse matrix reverses a transformation, but only when the original transformation has kept enough information. You will recognise singular matrices, calculate two by two and three by three inverses, and use inverse properties without losing the order of multiplication.

When a transformation can be reversed

You are told that (1224)\begin{pmatrix}1&2\\2&4\end{pmatrix} sent an unknown point to (2,4)(2,4). Can you recover the point? Both (2,0)(2,0) and (0,1)(0,1) give that image:

(1224)(20)=(24)=(1224)(01)\begin{pmatrix}1&2\\2&4\end{pmatrix}\begin{pmatrix}2\\0\end{pmatrix} =\begin{pmatrix}2\\4\end{pmatrix} =\begin{pmatrix}1&2\\2&4\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}

The transformation has merged different inputs. No single reversing rule can decide which one you started with.

You already know

The identity matrix leaves every vector unchanged. A determinant of zero means a linear transformation collapses area or volume to zero. Matrix multiplication usually depends on the order.

For this matrix, every output has its second coordinate twice its first, so the entire plane is flattened onto one line. Its determinant, 1(4)−2(2)=01(4)-2(2)=0, detects that loss of information.

Singular and non-singular matrices

A square matrix is singular if its determinant is zero, and non-singular if its determinant is non-zero. A singular matrix has no inverse. For example, (1224)\begin{pmatrix}1&2\\2&4\end{pmatrix} is singular, while (1225)\begin{pmatrix}1&2\\2&5\end{pmatrix}, with determinant 11, is non-singular. The zero matrix is singular; the identity matrix is non-singular.

An inverse undoes the transformation in either order. It restores every input, rather than just one conveniently chosen point.

Inverse matrix

For a non-singular square matrix M\mathbf{M}, its inverse M−1\mathbf{M}^{-1} satisfies

M−1M=MM−1=I\mathbf{M}^{-1}\mathbf{M}=\mathbf{M}\mathbf{M}^{-1}=\mathbf{I}

For example, (2003)−1=(120013)\begin{pmatrix}2&0\\0&3\end{pmatrix}^{-1}=\begin{pmatrix}\frac12&0\\0&\frac13\end{pmatrix}. Taking the reciprocal of every entry of a general matrix does not give its inverse.

For a parameter matrix, the test is still the determinant. For N=(122k)\mathbf{N}=\begin{pmatrix}1&2\\2&k\end{pmatrix}, det⁡N=k−4\det\mathbf{N}=k-4, so k=4k=4 is the singular case and every other real value is non-singular. The practical boundary is exact equality to zero, not whether the determinant is negative.

Key idea

Non-zero determinant means a unique reverse transformation exists. Zero determinant means some information has been lost and no inverse can restore it.

A negative determinant gives a reversal of orientation, which is different from a failure of reversibility.

Finding a two by two inverse

The two by two inverse uses a small rearrangement of the entries followed by one scale factor. Checking the determinant first prevents division by zero.

For M=(abcd)\mathbf{M}=\begin{pmatrix}a&b\\c&d\end{pmatrix}, multiplication gives

(abcd)(d−b−ca)=(ad−bc00ad−bc)=(ad−bc)I\begin{pmatrix}a&b\\c&d\end{pmatrix} \begin{pmatrix}d&-b\\-c&a\end{pmatrix} =\begin{pmatrix}ad-bc&0\\0&ad-bc\end{pmatrix} =(ad-bc)\mathbf{I}

The off-diagonal entries cancel. Dividing by ad−bcad-bc, when it is non-zero, therefore gives the identity.

Two by two inverse

If ad−bc≠0ad-bc\ne0,

(abcd)−1=1ad−bc(d−b−ca)\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1} =\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}

Swap the main-diagonal entries and negate the other two entries, then divide every entry by the determinant.

Not in the formula booklet: learn it.

The reciprocal determinant multiplies the entire matrix. Keep it outside the brackets until the last step if that makes your arithmetic clearer.

Worked example 1: a positive determinant

Question. Find A−1\mathbf{A}^{-1} for A=(2113)\mathbf{A}=\begin{pmatrix}2&1\\1&3\end{pmatrix}.

Plan. Find the determinant, rearrange the entries, then divide by the determinant.

det⁡A=2(3)−1(1)=5non-zero, so invertibleA−1=15(3−1−12)swap and negate\begin{aligned} \det\mathbf{A}&=2(3)-1(1)=5 && \text{non-zero, so invertible}\\ \mathbf{A}^{-1}&=\frac15\begin{pmatrix}3&-1\\-1&2\end{pmatrix} && \text{swap and negate} \end{aligned}

Why this step? The rearranged matrix multiplies A\mathbf{A} to 5I5\mathbf{I}, so the factor 15\frac15 is essential.

Answer. A−1=15(3−1−12)\mathbf{A}^{-1}=\frac15\begin{pmatrix}3&-1\\-1&2\end{pmatrix}.

Check.

AA−1=15(6−1−2+23−3−1+6)=I\mathbf{A}\mathbf{A}^{-1} =\frac15\begin{pmatrix}6-1&-2+2\\3-3&-1+6\end{pmatrix} =\mathbf{I}

Worked example 2 changes one entry so that the determinant is negative. The inverse formula itself does not change.

Worked example 2: a negative determinant

Question. Find the inverse of B=(2110)\mathbf{B}=\begin{pmatrix}2&1\\1&0\end{pmatrix}.

Plan. Use the signed determinant in the reciprocal factor.

det⁡B=2(0)−1(1)=−1calculate firstB−1=−(0−1−12)divide by −1=(011−2)multiply every entry\begin{aligned} \det\mathbf{B}&=2(0)-1(1)=-1 && \text{calculate first}\\ \mathbf{B}^{-1}&=-\begin{pmatrix}0&-1\\-1&2\end{pmatrix} && \text{divide by }-1\\ &=\begin{pmatrix}0&1\\1&-2\end{pmatrix} && \text{multiply every entry} \end{aligned}

Why this step? Using ∣det⁡B∣|\det\mathbf{B}| would lose the sign and give a matrix whose product with B\mathbf{B} is −I-\mathbf{I}.

Answer. B−1=(011−2)\mathbf{B}^{-1}=\begin{pmatrix}0&1\\1&-2\end{pmatrix}.

Check. (2110)(011−2)=(1001)\begin{pmatrix}2&1\\1&0\end{pmatrix}\begin{pmatrix}0&1\\1&-2\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}.

Use the signed determinant in an inverse, even though geometrical areas use its modulus.

Watch out

Wrong: replace 1det⁡M\frac{1}{\det\mathbf{M}} by 1∣det⁡M∣\frac{1}{|\det\mathbf{M}|}. Right: retain the sign of the determinant. Only an area or volume scale factor discards that sign.

The next matrix changes the zero entry to a negative entry, while keeping the same method.

Finding a three by three inverse

For three by three matrices, the swap-and-negate shortcut becomes a systematic calculation with minors, signs and a transpose. Keeping those three stages separate makes mistakes much easier to spot.

The cofactor of an entry is its minor multiplied by the sign in the checkerboard (+−+−+−+−+)\begin{pmatrix}+&-&+\\-&+&-\\+&-&+\end{pmatrix}. The cofactor matrix C\mathbf{C} keeps those cofactors in their original positions. Its transpose, CT\mathbf{C}^{\mathrm{T}}, swaps rows and columns and is called the adjugate.

Three by three inverse

For a non-singular three by three matrix M\mathbf{M} with cofactor matrix C\mathbf{C},

M−1=1det⁡MCT\mathbf{M}^{-1}=\frac{1}{\det\mathbf{M}}\mathbf{C}^{\mathrm{T}}

Not in the formula booklet: learn it.

Why transpose? In MCT\mathbf{M}\mathbf{C}^{\mathrm{T}}, a diagonal entry is a row of M\mathbf{M} dotted with its own cofactors, so it is det⁡M\det\mathbf{M}. An off-diagonal entry uses another row's cofactors: it is a determinant with two equal rows, so it is zero. Thus MCT=(det⁡M)I\mathbf{M}\mathbf{C}^{\mathrm{T}}=(\det\mathbf{M})\mathbf{I}. Dividing by the determinant completes the reversal.

Worked example 3: minors, cofactors, transpose

Question. Find D−1\mathbf{D}^{-1}, where D=(110011101)\mathbf{D}=\begin{pmatrix}1&1&0\\0&1&1\\1&0&1\end{pmatrix}. Show the stages of your calculation.

Plan. Find the determinant and all nine minors, apply the cofactor signs, then transpose and divide.

det⁡D=1(1)−1(−1)+0=2\det\mathbf{D}=1(1)-1(-1)+0=2

The minors and then the cofactors are

minors=(1−1−111−1111)delete each row and columnC=(11−1−1111−11)apply the sign patternD−1=12(1−1111−1−111)transpose, then divide\begin{aligned} \text{minors}&=\begin{pmatrix}1&-1&-1\\1&1&-1\\1&1&1\end{pmatrix} && \text{delete each row and column}\\ \mathbf{C}&=\begin{pmatrix}1&1&-1\\-1&1&1\\1&-1&1\end{pmatrix} && \text{apply the sign pattern}\\ \mathbf{D}^{-1}&=\frac12\begin{pmatrix}1&-1&1\\1&1&-1\\-1&1&1\end{pmatrix} && \text{transpose, then divide} \end{aligned}

For example, the minor in row 11, column 22 is ∣0111∣=−1\begin{vmatrix}0&1\\1&1\end{vmatrix}=-1, so its cofactor is +1+1.

Why this step? Negating that minor belongs to the cofactor stage; moving it to row 22, column 11 belongs to the transpose stage. Both are needed.

Answer. D−1=12(1−1111−1−111)\mathbf{D}^{-1}=\frac12\begin{pmatrix}1&-1&1\\1&1&-1\\-1&1&1\end{pmatrix}.

Check. Multiplication before the factor gives D(2D−1)=(200020002)\mathbf{D}(2\mathbf{D}^{-1})=\begin{pmatrix}2&0&0\\0&2&0\\0&0&2\end{pmatrix}, so the final product is I\mathbf{I}.

Worked example 4 changes only the lower-left entry, from 11 to 22. Several minors change, although the sign pattern stays fixed.

Worked example 4: an altered entry

Question. Find E−1\mathbf{E}^{-1}, where E=(110011201)\mathbf{E}=\begin{pmatrix}1&1&0\\0&1&1\\2&0&1\end{pmatrix}.

Plan. Recalculate the determinant and cofactors rather than editing the previous inverse by eye.

det⁡E=1(1)−1(−2)=3first-row expansionminors=(1−2−211−2111)nine small determinantsC=(12−2−1121−11)apply the signsE−1=13(1−1121−1−221)transpose and divide\begin{aligned} \det\mathbf{E}&=1(1)-1(-2)=3 && \text{first-row expansion}\\ \text{minors}&=\begin{pmatrix}1&-2&-2\\1&1&-2\\1&1&1\end{pmatrix} && \text{nine small determinants}\\ \mathbf{C}&=\begin{pmatrix}1&2&-2\\-1&1&2\\1&-1&1\end{pmatrix} && \text{apply the signs}\\ \mathbf{E}^{-1}&=\frac13\begin{pmatrix}1&-1&1\\2&1&-1\\-2&2&1\end{pmatrix} && \text{transpose and divide} \end{aligned}

Why recalculate? A single changed entry contributes to several minors, so an inverse rarely changes in only one position.

Answer. E−1=13(1−1121−1−221)\mathbf{E}^{-1}=\frac13\begin{pmatrix}1&-1&1\\2&1&-1\\-2&2&1\end{pmatrix}.

Check. The first row of E\mathbf{E} times the three columns of the integer matrix gives (3,0,0)(3,0,0); the other rows give (0,3,0)(0,3,0) and (0,0,3)(0,0,3).

A permitted calculator can calculate three by three inverses. Enter each row carefully, use the matrix inverse operation, and retain exact fractions. If a question requires all stages, show the determinant, cofactors and transpose; calculator output alone cannot demonstrate the process.

Watch out

Wrong: M−1=C/det⁡M\mathbf{M}^{-1}=\mathbf{C}/\det\mathbf{M}. Right: transpose the cofactor matrix first. In the last example, (E−1)12=−13(\mathbf{E}^{-1})_{12}=-\frac13, whereas the untransposed cofactor in that position would give 23\frac23.

The same single-entry variation makes a useful final practice of the complete process.

Undoing products in reverse order

If you put on socks and then shoes, undoing the process starts with the shoes. Matrix inverses follow the same order reversal, because the inverse factors must cancel beside their partners.

For non-singular matrices of the same size,

(B−1A−1)(AB)=B−1(A−1A)B=I(\mathbf{B}^{-1}\mathbf{A}^{-1})(\mathbf{A}\mathbf{B}) =\mathbf{B}^{-1}(\mathbf{A}^{-1}\mathbf{A})\mathbf{B} =\mathbf{I}

Properties of inverses

For non-singular square matrices A\mathbf{A}, B\mathbf{B} of the same size, and a non-zero scalar kk,

(AB)−1=B−1A−1,(A−1)−1=A(\mathbf{A}\mathbf{B})^{-1}=\mathbf{B}^{-1}\mathbf{A}^{-1},\qquad (\mathbf{A}^{-1})^{-1}=\mathbf{A} (kA)−1=1kA−1,det⁡(A−1)=1det⁡A(k\mathbf{A})^{-1}=\frac1k\mathbf{A}^{-1},\qquad \det(\mathbf{A}^{-1})=\frac1{\det\mathbf{A}}

Not in the formula booklet: learn it.

The last identity follows by taking determinants of AA−1=I\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}: their product is 11. The scalar rule follows because kk and 1k\frac1k cancel. Neither rule is entry-by-entry division.

Worked example 5: reversing a product

Question. Given A=(1101)\mathbf{A}=\begin{pmatrix}1&1\\0&1\end{pmatrix} and B=(2001)\mathbf{B}=\begin{pmatrix}2&0\\0&1\end{pmatrix}, find (AB)−1(\mathbf{A}\mathbf{B})^{-1} using the separate inverses.

Plan. Find A−1\mathbf{A}^{-1} and B−1\mathbf{B}^{-1}, then multiply them in reverse order.

A−1=(1−101),B−1=(12001)\mathbf{A}^{-1}=\begin{pmatrix}1&-1\\0&1\end{pmatrix},\qquad \mathbf{B}^{-1}=\begin{pmatrix}\frac12&0\\0&1\end{pmatrix} (AB)−1=B−1A−1reverse the order=(12−1201)row by column\begin{aligned} (\mathbf{A}\mathbf{B})^{-1} &=\mathbf{B}^{-1}\mathbf{A}^{-1} && \text{reverse the order}\\ &=\begin{pmatrix}\frac12&-\frac12\\0&1\end{pmatrix} && \text{row by column} \end{aligned}

Why this order? The original product applies B\mathbf{B} first and A\mathbf{A} second. Its inverse must apply A−1\mathbf{A}^{-1} first.

Answer. (AB)−1=(12−1201)(\mathbf{A}\mathbf{B})^{-1}=\begin{pmatrix}\frac12&-\frac12\\0&1\end{pmatrix}.

Check. AB=(2101)\mathbf{A}\mathbf{B}=\begin{pmatrix}2&1\\0&1\end{pmatrix}, whose two by two inverse agrees.

Worked example 6 changes only the order of the original factors. Keeping the same inverse order would now be wrong.

Worked example 6: the other product

Question. For the same matrices, find (BA)−1(\mathbf{B}\mathbf{A})^{-1}.

Plan. Reverse the order of this product, giving A−1B−1\mathbf{A}^{-1}\mathbf{B}^{-1}.

(BA)−1=(1−101)(12001)correct reversed order=(12−101)multiply\begin{aligned} (\mathbf{B}\mathbf{A})^{-1} &=\begin{pmatrix}1&-1\\0&1\end{pmatrix} \begin{pmatrix}\frac12&0\\0&1\end{pmatrix} && \text{correct reversed order}\\ &=\begin{pmatrix}\frac12&-1\\0&1\end{pmatrix} && \text{multiply} \end{aligned}

Why this step? The −1-1 is multiplied by the lower-right 11, not by 12\frac12. Matrix factors cannot be exchanged casually.

Answer. (BA)−1=(12−101)(\mathbf{B}\mathbf{A})^{-1}=\begin{pmatrix}\frac12&-1\\0&1\end{pmatrix}.

Check. BA=(2201)\mathbf{B}\mathbf{A}=\begin{pmatrix}2&2\\0&1\end{pmatrix}; multiplication by the answer gives I\mathbf{I}.

Here is a pleasant special case: an inverse can equal the original matrix. Reflection in y=xy=x has matrix R=(0110)\mathbf{R}=\begin{pmatrix}0&1\\1&0\end{pmatrix}, and R2=I\mathbf{R}^2=\mathbf{I}. Reflecting twice brings every point back. The inverse of R\mathbf{R} is therefore R\mathbf{R} itself, although R≠I\mathbf{R}\ne\mathbf{I}.

Watch out

Wrong: (AB)−1=A−1B−1(\mathbf{A}\mathbf{B})^{-1}=\mathbf{A}^{-1}\mathbf{B}^{-1}. Right: B−1A−1\mathbf{B}^{-1}\mathbf{A}^{-1}. Write the reversal before inserting any numbers.

The point at the start cannot be recovered uniquely: the matrix is singular. Every non-singular case has one inverse, and its identity-product check gives you a reliable way to verify it.

Quick Recap

  • A square matrix is singular exactly when its determinant is zero.
  • An inverse satisfies MM−1=M−1M=I\mathbf{M}\mathbf{M}^{-1}=\mathbf{M}^{-1}\mathbf{M}=\mathbf{I}.
  • For two by two inverses, swap the main diagonal, negate the other entries, and divide by the signed determinant.
  • For three by three inverses, calculate cofactors, transpose, and divide by the signed determinant.
  • Undo a product in reverse order: (AB)−1=B−1A−1(\mathbf{A}\mathbf{B})^{-1}=\mathbf{B}^{-1}\mathbf{A}^{-1}.

The opening transformation cannot be undone: it sends distinct points to the same image and has determinant zero.