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Precalculus · A.REI.C.9 · N.VM.C.10

Inverse of a 2 by 2 matrix, and solving with it

Dividing by a number undoes multiplying by it. Matrices have no division, so is there a matrix that undoes multiplying by a given one, and how do you find it?

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When you reach for it

For numbers, the equation 3x = 12 is solved by multiplying both sides by \displaystyle \frac{1}{3}, the number that undoes multiplying by 3, because \displaystyle \frac{1}{3} \cdot 3 = 1. A linear system written as a matrix equation AX = B looks just as simple, but there is no dividing by a matrix. What can exist is a matrix that undoes A: its inverse, A^{-1}.

You reach for the inverse when a square system AX = B has to be solved, especially when the same coefficient matrix A meets many right sides B, as in a code where every block of a message is multiplied by one key matrix. For a 2 \times 2 matrix the inverse comes from the determinant in one line; for a larger one it comes from row reduction.

The lesson runs over two days. Day 1 is the inverse itself: what it is, the 2 \times 2 formula and why it works, and when there is none. Day 2 puts it to work: solving AX = B, a 3 \times 3 inverse by row reduction, and the inverse of a product.

The procedure

Day 1: what an inverse is

The identity matrix I of size n has 1s on the main diagonal and 0s elsewhere, as in I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. By the row-by-column rule, AI = IA = A for every n \times n matrix A, so I plays the part of the number 1.

A square matrix A is invertible when there is a matrix A^{-1} of the same size with

AA^{-1} = A^{-1}A = I

Both products are required, because matrix multiplication is not commutative. A matrix with no inverse is called singular. The -1 is a name, not an exponent you can compute with: A^{-1} is never the matrix of reciprocals of the entries.

An inverse is unique. Suppose B and C both satisfy the definition. Then, using AC = I, the associative property of multiplication, and BA = I,

B = BI = B(AC) = (BA)C = IC = C

So it is fair to speak of the inverse.

The 2 by 2 inverse

For A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} with \det A = ad - bc \ne 0,

A^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}

In words: swap the two entries on the main diagonal, change the signs of the other two, and divide every entry by the determinant.

Why it works. Multiply A by the swapped-and-negated matrix and watch the entries:

\begin{aligned} \begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} &= \begin{bmatrix} ad - bc & -ab + ba \\ cd - dc & -cb + da \end{bmatrix} \\ &= (ad - bc)\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \end{aligned}

The off-diagonal entries cancel because ab = ba and cd = dc, and both diagonal entries are the determinant. Multiplying in the other order gives da - bc, db - bd, -ca + ac and -cb + ad, the same matrix. A number factor can be moved through a matrix product, so dividing the second matrix by ad - bc divides the product by it, and both products become I.

No inverse when \det A = 0. Write D for the swapped-and-negated matrix, so AD = 0, the zero matrix, by the computation above. Suppose A had an inverse anyway. Then

D = ID = (A^{-1}A)D = A^{-1}(AD) = A^{-1}0 = 0

so d = -b = -c = a = 0 and A itself is the zero matrix. But 0 \cdot A^{-1} is the zero matrix, not I, which is impossible. So a 2 \times 2 matrix is invertible exactly when its determinant is not zero.

(2, 1) (1, 1)
M = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix},\; \det M = 1
 (2, 1) (1, 0)
\left(0, \tfrac{1}{2}\right)
N = \begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix},\; \det N = 0
Figure 1Left: M moves the square to a parallelogram, and M^{-1} moves it back. Right: N flattens the square onto a line, sending different points to the same place, so nothing can undo it.

The figure shows the reason in a picture. A determinant of 0 means the columns are parallel, so the matrix squashes the plane onto a line. Two different points such as (1, 0) and \displaystyle \left(0, \frac{1}{2}\right) then land on the same point, and no matrix can send that one point back to two places.

Day 2: solving AX = B with the inverse

If A is invertible, multiply both sides of AX = B on the left by A^{-1}:

A^{-1}(AX) = (A^{-1}A)X = IX = X, \quad\text{so}\quad X = A^{-1}B

The step is allowed because equal matrices times the same matrix on the same side stay equal, and the regrouping is the associative property. The column A^{-1}B really is a solution, since A(A^{-1}B) = (AA^{-1})B = B, and it is the only one, since every solution equals A^{-1}B. So an invertible coefficient matrix means exactly one solution. The side matters: for a 2 \times 2 system, BA^{-1} is a 2 \times 1 matrix times a 2 \times 2 one, which is not even defined.

A 3 by 3 inverse by row reduction

For a larger matrix there is no short formula to use by hand. Instead set the identity beside A and row reduce:

[A \mid I] \;\longrightarrow\; [I \mid A^{-1}]

Here is why. Column j of A^{-1} is the column X with AX equal to column j of I, since AA^{-1} = I is checked one column at a time. So finding A^{-1} means solving n systems with the same coefficient matrix A. Row operations depend only on the left part, so all n systems can be reduced at once with all n right sides beside A. When the left part has become I, each right column has become the solution of its system. If the left part instead produces a row of zeros, A has no inverse. The full method is the subject of Finding an inverse by row reducing A next to I.

The inverse of a product

If A and B are invertible and the same size, then AB is invertible and

(AB)^{-1} = B^{-1}A^{-1}

with the order reversed. Check it with the definition, regrouping by the associative property:

\begin{aligned} (AB)(B^{-1}A^{-1}) &= A(BB^{-1})A^{-1} \\ &= AIA^{-1} = AA^{-1} = I \end{aligned}

and in the same way (B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}B = I. Putting on socks and then shoes is undone by taking off the shoes first. The order A^{-1}B^{-1} would leave A and A^{-1} separated by B, with nothing to cancel.

Worked examples

Examples 1 to 3 go with Day 1, Examples 4 to 7 with Day 2.

Example 1

An inverse with integer entries

Find the inverse of A = \begin{bmatrix} 7 & 2 \\ 3 & 1 \end{bmatrix} and check it both ways.

Solution

The determinant is 7 \cdot 1 - 2 \cdot 3 = 1, which is not zero, so A is invertible. Swap the 7 and the 1, change the signs of the 2 and the 3, and divide by 1:

A^{-1} = \begin{bmatrix} 1 & -2 \\ -3 & 7 \end{bmatrix}

Check. The definition asks for both products.

AA^{-1} = \begin{bmatrix} 7 - 6 & -14 + 14 \\ 3 - 3 & -6 + 7 \end{bmatrix} = I
A^{-1}A = \begin{bmatrix} 7 - 6 & 2 - 2 \\ -21 + 21 & -6 + 7 \end{bmatrix} = I

A determinant of 1 is why no fractions appear.

Example 2

An inverse with fractions

Find the inverse of A = \begin{bmatrix} 2 & -5 \\ 4 & -6 \end{bmatrix}.

Solution
\det A = 2(-6) - (-5)(4) = -12 + 20 = 8

Swap the diagonal entries 2 and -6, and change the signs of -5 and 4, keeping track of the signs already there: -(-5) = 5. Then divide every entry by 8.

A^{-1} = \frac{1}{8}\begin{bmatrix} -6 & 5 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} -\frac{3}{4} & \frac{5}{8} \\ -\frac{1}{2} & \frac{1}{4} \end{bmatrix}

Check. It is easier to multiply before dividing:

\begin{bmatrix} 2 & -5 \\ 4 & -6 \end{bmatrix}\begin{bmatrix} -6 & 5 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix}

which is 8I, so dividing by 8 gives I.

Example 3

A matrix with no inverse

Show that A = \begin{bmatrix} 3 & -6 \\ -2 & 4 \end{bmatrix} has no inverse, first from the determinant and then directly from the definition.

Solution

\det A = 3 \cdot 4 - (-6)(-2) = 12 - 12 = 0, so the formula would divide by zero and, by the argument in the procedure, there is no inverse. The columns show it too: (-6, 4) = -2(3, -2).

Directly: suppose \begin{bmatrix} p & q \\ r & s \end{bmatrix} were an inverse. Then A times its first column \begin{bmatrix} p \\ r \end{bmatrix} must be the first column of I, which gives two equations:

\begin{aligned} 3p - 6r &= 1 \\ -2p + 4r &= 0 \end{aligned}

The second says p = 2r. Putting that into the first gives 6r - 6r = 1, that is 0 = 1. No numbers p and r work, so no inverse exists.

Example 4

Solving two systems with one inverse

Ebuka has two systems with the same left sides:

\begin{aligned} 3x + 4y &= 6 \\ 2x + 5y &= 11 \end{aligned} \qquad \begin{aligned} 3x + 4y &= 0 \\ 2x + 5y &= 7 \end{aligned}

Solve both with one inverse.

Solution

Both are AX = B with A = \begin{bmatrix} 3 & 4 \\ 2 & 5 \end{bmatrix}, and \det A = 15 - 8 = 7, so

A^{-1} = \frac{1}{7}\begin{bmatrix} 5 & -4 \\ -2 & 3 \end{bmatrix}

Then X = A^{-1}B for each right side:

\frac{1}{7}\begin{bmatrix} 5(6) - 4(11) \\ -2(6) + 3(11) \end{bmatrix} = \frac{1}{7}\begin{bmatrix} -14 \\ 21 \end{bmatrix} = \begin{bmatrix} -2 \\ 3 \end{bmatrix}
\frac{1}{7}\begin{bmatrix} 5(0) - 4(7) \\ -2(0) + 3(7) \end{bmatrix} = \frac{1}{7}\begin{bmatrix} -28 \\ 21 \end{bmatrix} = \begin{bmatrix} -4 \\ 3 \end{bmatrix}

So (x, y) = (-2, 3) for the first system and (-4, 3) for the second. Check. 3(-2) + 4(3) = 6 and 2(-2) + 5(3) = 11; 3(-4) + 4(3) = 0 and 2(-4) + 5(3) = 7. The inverse was found once and used twice.

Example 5

Decoding a message

Zarema and Chinedu send coded messages. Each letter becomes a number, A = 1, B = 2, and so on to Z = 26; the numbers are written two at a time as columns, and each column is multiplied by the key K = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}. Chinedu receives 80, 47, 99, 61. Decode it.

Solution

Zarema sent C = KM, where the columns of M are the letter pairs. Multiplying on the left by K^{-1} gives M = K^{-1}C. Here \det K = 4 - 3 = 1, so

K^{-1} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}

Put both received columns into one matrix and decode them together:

\begin{aligned} K^{-1}\begin{bmatrix} 80 & 99 \\ 47 & 61 \end{bmatrix} &= \begin{bmatrix} 160 - 141 & 198 - 183 \\ -80 + 94 & -99 + 122 \end{bmatrix} \\ &= \begin{bmatrix} 19 & 15 \\ 14 & 23 \end{bmatrix} \end{aligned}

Reading down the columns gives 19, 14, 15, 23, the letters S, N, O, W. Check. K times the column (19, 14) is (38 + 42, 19 + 28) = (80, 47), the first two numbers received. A key with determinant 1 has an inverse with integer entries, so decoding never produces fractions.

Example 6

A 3 by 3 inverse

Find the inverse of A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 2 \\ 1 & 2 & 3 \end{bmatrix} by row reducing [A \mid I].

Solution
\left[\begin{array}{ccc|ccc} 1 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 2 & 0 & 1 & 0 \\ 1 & 2 & 3 & 0 & 0 & 1 \end{array}\right]

Clear the 1 below the first pivot with R_3 - R_1 \to R_3, giving row 3 as 0, 1, 3 \mid -1, 0, 1. Clear the 1 below the second pivot with R_3 - R_2 \to R_3:

\left[\begin{array}{ccc|ccc} 1 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 2 & 0 & 1 & 0 \\ 0 & 0 & 1 & -1 & -1 & 1 \end{array}\right]

The third pivot is already 1. Clear above it with R_2 - 2R_3 \to R_2, which makes row 2 0, 1, 0 \mid 2, 3, -2, and then clear above the second pivot with R_1 - R_2 \to R_1:

\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & -1 & -3 & 2 \\ 0 & 1 & 0 & 2 & 3 & -2 \\ 0 & 0 & 1 & -1 & -1 & 1 \end{array}\right]

The left part is I, so

A^{-1} = \begin{bmatrix} -1 & -3 & 2 \\ 2 & 3 & -2 \\ -1 & -1 & 1 \end{bmatrix}

Check. Row 3 of A times the columns of A^{-1} gives -1 + 4 - 3 = 0, -3 + 6 - 3 = 0 and 2 - 4 + 3 = 1, which is row 3 of I; rows 1 and 2 check the same way.

Example 7

Reversing the order

Let A = \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix} and B = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}. Find (AB)^{-1} directly, then compare it with B^{-1}A^{-1} and with A^{-1}B^{-1}.

Solution
AB = \begin{bmatrix} 2 + 2 & 1 + 2 \\ 2 + 3 & 1 + 3 \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ 5 & 4 \end{bmatrix}

Its determinant is 16 - 15 = 1, so (AB)^{-1} = \begin{bmatrix} 4 & -3 \\ -5 & 4 \end{bmatrix}.

Each factor also has determinant 1: A^{-1} = \begin{bmatrix} 3 & -2 \\ -1 & 1 \end{bmatrix} and B^{-1} = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}. Then

B^{-1}A^{-1} = \begin{bmatrix} 3 + 1 & -2 - 1 \\ -3 - 2 & 2 + 2 \end{bmatrix} = \begin{bmatrix} 4 & -3 \\ -5 & 4 \end{bmatrix}
A^{-1}B^{-1} = \begin{bmatrix} 3 + 2 & -3 - 4 \\ -1 - 1 & 1 + 2 \end{bmatrix} = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}

The reversed product matches (AB)^{-1}, as the procedure proved; the same order does not.

Boundaries

  • Square matrices only. A 2 \times 3 matrix has no inverse: no single matrix can make both products the same identity (Exercise 21).
  • The determinant must not be zero. The 2 \times 2 formula divides by ad - bc, and when it is 0 there is no inverse at all, not just no formula.
  • The 2 \times 2 formula is for 2 \times 2 only. Larger matrices use [A \mid I], or the cofactor methods of later lessons.
  • Multiply on the correct side. AX = B gives X = A^{-1}B; XA = B gives X = BA^{-1}. Matrix multiplication is not commutative, so the two are different.
  • A^{-1} is not the matrix of reciprocals, and (AB)^{-1} is B^{-1}A^{-1}, not A^{-1}B^{-1}.
  • The inverse method needs exactly one solution. A system with no solution or infinitely many has a singular coefficient matrix, and row reduction is the tool that sorts those cases out.

Where this goes next

Writing out X = A^{-1}B entry by entry for a 2 \times 2 system turns each unknown into a ratio of two determinants, which is Cramer's rule. Finding an inverse by row reducing A next to I develops the [A \mid I] method for any size, and When a matrix or transformation can be undone reads the figure above as geometry: a matrix can be undone exactly when it flattens nothing, which is when its determinant is not zero.

Exercises

Exercises 1 to 10 go with Day 1 (Examples 1 to 3), Exercises 11 to 20 with Day 2 (Examples 4 to 7), and the reasoning group draws on both.

Day 1: finding inverses

Find the inverse, or show there is none.

  1. \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}
  2. \begin{bmatrix} 4 & 1 \\ 6 & 2 \end{bmatrix}
  3. \begin{bmatrix} -3 & 5 \\ 2 & -4 \end{bmatrix}
  4. \begin{bmatrix} 0 & 3 \\ -2 & 1 \end{bmatrix}
  5. \displaystyle \begin{bmatrix} \frac{1}{2} & \frac{1}{3} \\ 1 & 1 \end{bmatrix}
  6. \begin{bmatrix} 6 & -9 \\ -4 & 6 \end{bmatrix} and \begin{bmatrix} 5 & -2 \\ 3 & -1 \end{bmatrix}
  7. \begin{bmatrix} 3 & 0 \\ 0 & -5 \end{bmatrix}

Day 1: using the formula

  1. For which values of k does \begin{bmatrix} k & 6 \\ 2 & k + 1 \end{bmatrix} have no inverse?
  2. Find the inverse of \begin{bmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{bmatrix} for any angle t.
  3. The inverse of a matrix A is \begin{bmatrix} 4 & -1 \\ -7 & 2 \end{bmatrix}. Find A.

Day 2: solving systems

Solve with an inverse matrix.

  1. 5x + 2y = 4, 7x + 3y = 5 (the inverse is Exercise 1)
  2. 4x + y = 3, 6x + 2y = 2 (the inverse is Exercise 2)
  3. 2x - 3y = 1, 4x + y = 9
  4. 3x + 2y = 2, 5x + 4y = 1
  5. Find the inverse of A = \begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 1 & 2 & 2 \end{bmatrix} by row reducing [A \mid I].
  6. Use Exercise 15 to solve x + 2y + z = 3, y + z = 1, x + 2y + 2z = 5.

Day 2: codes and products

  1. Zarema changes the key to \begin{bmatrix} 3 & 5 \\ 1 & 2 \end{bmatrix}, with letters numbered as in Example 5. Chinedu receives 59, 20, 97, 37. Decode it.
  2. Ebuka encodes the word GO with the key of Example 5. What numbers does he send?
  3. Let A = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} and B = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}. Find (AB)^{-1} directly and as B^{-1}A^{-1}.
  4. Solve XA = C for the 2 \times 2 matrix X, where A = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix} and C = \begin{bmatrix} 1 & 0 \\ 4 & 2 \end{bmatrix}.

Reasoning

  1. Explain why a 2 \times 3 matrix A cannot have an inverse. (What sizes are AB and BA?)
  2. Show that if A is an invertible 2 \times 2 matrix, then \displaystyle \det(A^{-1}) = \frac{1}{\det A}, and explain why (A^{-1})^{-1} = A.
  3. Ebuka solves AX = B by writing X = BA^{-1} for a 2 \times 2 matrix A and a 2 \times 1 column B. Explain what goes wrong.
  4. Let A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} with ad - bc = 0, not all entries zero. Show that AX = \begin{bmatrix} 0 \\ 0 \end{bmatrix} has a solution other than X = \begin{bmatrix} 0 \\ 0 \end{bmatrix}, and explain why this rules out an inverse.
  5. Chinedu suggests the key \begin{bmatrix} 4 & 6 \\ 2 & 3 \end{bmatrix}. Show that the letter pairs CD and FB are sent as the same numbers, and explain why this key cannot work.
Answers
  1. \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}. The determinant is 15 - 14 = 1.
  2. \displaystyle \begin{bmatrix} 1 & -\frac{1}{2} \\ -3 & 2 \end{bmatrix}. The determinant is 8 - 6 = 2, and \displaystyle \frac{1}{2}\begin{bmatrix} 2 & -1 \\ -6 & 4 \end{bmatrix} simplifies to this.
  3. \displaystyle \begin{bmatrix} -2 & -\frac{5}{2} \\ -1 & -\frac{3}{2} \end{bmatrix}. The determinant is 12 - 10 = 2, and \displaystyle \frac{1}{2}\begin{bmatrix} -4 & -5 \\ -2 & -3 \end{bmatrix} simplifies to this.
  4. \displaystyle \begin{bmatrix} \frac{1}{6} & -\frac{1}{2} \\ \frac{1}{3} & 0 \end{bmatrix}. The determinant is 0 - (3)(-2) = 6, and \displaystyle \frac{1}{6}\begin{bmatrix} 1 & -3 \\ 2 & 0 \end{bmatrix} simplifies to this.
  5. \begin{bmatrix} 6 & -2 \\ -6 & 3 \end{bmatrix}. The determinant is \displaystyle \frac{1}{2} - \frac{1}{3} = \frac{1}{6}, and dividing by \displaystyle \frac{1}{6} multiplies by 6: \displaystyle 6\begin{bmatrix} 1 & -\frac{1}{3} \\ -1 & \frac{1}{2} \end{bmatrix}.
  6. The first has no inverse: 36 - 36 = 0. The second has determinant -5 + 6 = 1 and inverse \begin{bmatrix} -1 & 2 \\ -3 & 5 \end{bmatrix}.
  7. \displaystyle \begin{bmatrix} \frac{1}{3} & 0 \\ 0 & -\frac{1}{5} \end{bmatrix}. The determinant is -15, and \displaystyle \frac{1}{-15}\begin{bmatrix} -5 & 0 \\ 0 & 3 \end{bmatrix} simplifies to this. A diagonal matrix inverts entry by entry on its diagonal.
  8. k = 3 or k = -4. The determinant is k(k + 1) - 12 = k^2 + k - 12 = (k - 3)(k + 4), which is 0 exactly there.
  9. \begin{bmatrix} \cos t & \sin t \\ -\sin t & \cos t \end{bmatrix}. The determinant is \cos^2 t + \sin^2 t = 1, by the Pythagorean identity, so it is never zero.
  10. A = \begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}. A is the inverse of A^{-1}; the determinant of A^{-1} is 8 - 7 = 1, so swap and negate.
  11. (x, y) = (2, -3). \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}\begin{bmatrix} 4 \\ 5 \end{bmatrix} = \begin{bmatrix} 12 - 10 \\ -28 + 25 \end{bmatrix}.
  12. (x, y) = (2, -5). \displaystyle \begin{bmatrix} 1 & -\frac{1}{2} \\ -3 & 2 \end{bmatrix}\begin{bmatrix} 3 \\ 2 \end{bmatrix} = \begin{bmatrix} 3 - 1 \\ -9 + 4 \end{bmatrix}.
  13. (x, y) = (2, 1). The determinant is 2 + 12 = 14, so \displaystyle X = \frac{1}{14}\begin{bmatrix} 1 & 3 \\ -4 & 2 \end{bmatrix}\begin{bmatrix} 1 \\ 9 \end{bmatrix} = \frac{1}{14}\begin{bmatrix} 28 \\ 14 \end{bmatrix}.
  14. \displaystyle (x, y) = \left(3, -\frac{7}{2}\right). The determinant is 12 - 10 = 2, so \displaystyle X = \frac{1}{2}\begin{bmatrix} 4 & -2 \\ -5 & 3 \end{bmatrix}\begin{bmatrix} 2 \\ 1 \end{bmatrix} = \frac{1}{2}\begin{bmatrix} 6 \\ -7 \end{bmatrix}.
  15. A^{-1} = \begin{bmatrix} 0 & -2 & 1 \\ 1 & 1 & -1 \\ -1 & 0 & 1 \end{bmatrix}. R_3 - R_1 \to R_3 gives row 3 as 0, 0, 1 \mid -1, 0, 1; then R_2 - R_3 \to R_2 and R_1 - R_3 \to R_1 clear the third column; then R_1 - 2R_2 \to R_1 clears the second.
  16. (x, y, z) = (3, -1, 2). A^{-1}\begin{bmatrix} 3 \\ 1 \\ 5 \end{bmatrix} = \begin{bmatrix} 0 - 2 + 5 \\ 3 + 1 - 5 \\ -3 + 0 + 5 \end{bmatrix}.
  17. RAIN. The key has determinant 6 - 5 = 1 and inverse \begin{bmatrix} 2 & -5 \\ -1 & 3 \end{bmatrix}, and \begin{bmatrix} 2 & -5 \\ -1 & 3 \end{bmatrix}\begin{bmatrix} 59 & 97 \\ 20 & 37 \end{bmatrix} = \begin{bmatrix} 18 & 9 \\ 1 & 14 \end{bmatrix}, which reads 18, 1, 9, 14.
  18. 59, 37. G and O are 7 and 15, and \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} 7 \\ 15 \end{bmatrix} = \begin{bmatrix} 14 + 45 \\ 7 + 30 \end{bmatrix}.
  19. \begin{bmatrix} 5 & -3 \\ -8 & 5 \end{bmatrix} both ways. AB = \begin{bmatrix} 5 & 3 \\ 8 & 5 \end{bmatrix} with determinant 1. Also A^{-1} = \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix}, B^{-1} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}, and B^{-1}A^{-1} = \begin{bmatrix} 4 + 1 & -2 - 1 \\ -6 - 2 & 3 + 2 \end{bmatrix}.
  20. X = \begin{bmatrix} 3 & -1 \\ 2 & 0 \end{bmatrix}. Multiply both sides on the right by A^{-1} = \begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}: X = CA^{-1}. Check: XA = \begin{bmatrix} 6 - 5 & 3 - 3 \\ 4 & 2 \end{bmatrix} = C.
  21. For AB to be defined B must be 3 \times m, and for BA to be defined as well, m must be 2. Then AB is 2 \times 2 while BA is 3 \times 3. They cannot both equal one identity matrix, so the definition cannot be met.
  22. With D = ad - bc, \displaystyle \det(A^{-1}) = \frac{d}{D} \cdot \frac{a}{D} - \frac{-b}{D} \cdot \frac{-c}{D} = \frac{ad - bc}{D^2} = \frac{1}{D}. The equations AA^{-1} = A^{-1}A = I also say that A is an inverse of A^{-1}, and inverses are unique, so (A^{-1})^{-1} = A.
  23. BA^{-1} is (2 \times 1)(2 \times 2); the inner sizes 1 and 2 do not match, so it is undefined. Multiplying AX = B by A^{-1} must happen on the left of both sides, giving A^{-1}B, because A^{-1} has to sit next to A to cancel it.
  24. A\begin{bmatrix} d \\ -c \end{bmatrix} = \begin{bmatrix} ad - bc \\ cd - dc \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} and A\begin{bmatrix} -b \\ a \end{bmatrix} = \begin{bmatrix} 0 \\ -bc + ad \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}, and at least one of these columns is not zero. If A^{-1} existed, every solution of AX = 0 would equal A^{-1}0 = 0, a contradiction.
  25. C, D is 3, 4 and F, B is 6, 2. Both give \begin{bmatrix} 4(3) + 6(4) \\ 2(3) + 3(4) \end{bmatrix} = \begin{bmatrix} 4(6) + 6(2) \\ 2(6) + 3(2) \end{bmatrix} = \begin{bmatrix} 36 \\ 18 \end{bmatrix}. The determinant is 12 - 12 = 0, so the key has no inverse: Chinedu receiving 36, 18 cannot tell which pair was meant.
kg1128 · updated 2026-10-20 · CC-BY-SA-4.0