The identity matrix, inverse matrices and the determinant condition for their existence, the formula for the inverse of a 2 by 2, writing a linear system as a matrix equation, and solving it with a single multiplication.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 3 — Linear Systems and Matrices
Use Inverse Matrices to Solve Linear Systems
Objectives
Five outcomes. The fourth is the one that makes the whole chapter's machinery pay off.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.8 Use Inverse Matrices to Solve Linear Systems §3.8, pp. 210-215 — the lesson these objectives are drawn from
Warm-up
Lesson 1.1 gave you the multiplicative inverse of a number. This lesson asks the same question of a matrix.
Discussion prompt
To solve 3x equals 12 you divide by 3 — or, equivalently, multiply by one third. What property of one third makes that work, and what would the matrix version of one third have to do?
Hint: Ask what three times one third gives.
Answer:
\[ 3 \cdot \tfrac{1}{3} = 1 \;\Longrightarrow\; \tfrac{1}{3}(3x) = \tfrac{1}{3}(12) \;\Longrightarrow\; x = 4 \]
One third works because multiplying by it turns the 3 into 1, and 1 leaves x alone. A matrix version would need two things: a matrix that acts like 1, and for each matrix a partner that multiplies with it to give that one. This lesson supplies both.
Concept
The identity matrix has ones down its main diagonal and zeros elsewhere, and multiplying by it changes nothing. Two square matrices are inverses when their product, in either order, is the identity — and then one can undo the other.
inverse matrices — Two square matrices A and B such that AB and BA are both the identity matrix. The inverse of A is written A to the power negative one.
\[ AA^{-1} = I = A^{-1}A \]
A square matrix has an inverse if and only if its determinant is not zero. That is the same condition Cramer's rule needed in Lesson 3.7, and it is not a coincidence.
Figure (svg): The two by two and three by three identity matrices, with ones down the main diagonal and zeros elsewhere
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.8 Use Inverse Matrices to Solve Linear Systems §3.8, pp. 210-210
Section
Section 1
Concept
The n by n identity matrix has 1 in every position on the main diagonal and 0 everywhere else. Multiplying any n by n matrix by it, in either order, returns that matrix unchanged.
identity matrix — The square matrix with ones on the main diagonal and zeros elsewhere. Multiplying by it leaves any matrix of the right size unchanged.
\[ I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \qquad AI = A = IA \]
There is one identity for each size. A 2 by 2 matrix needs the 2 by 2 identity; the 3 by 3 identity would not even be multipliable with it.
Figure (svg): The two by two and three by three identity matrices, with ones down the main diagonal and zeros elsewhere
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.8 Use Inverse Matrices to Solve Linear Systems §3.8, pp. 210-210
Picture it
The 2 by 2 and the 3 by 3.
Figure (svg): The two by two and three by three identity matrices, with ones down the main diagonal and zeros elsewhere
The identity commutes with everything, which makes it the rare exception to Lesson 3.6's warning about order. That is exactly what an identity has to do.
Worked example
Multiplying a matrix by the identity, using the row-times-column rule.
\[ \text{Compute } \begin{bmatrix} 3 & 8 \\ 2 & 5 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. \]
Row 1 with column 1
Why: Three times one plus eight times zero.
\[ 3 + 0 = 3 \]
Row 1 with column 2
Why: Three times zero plus eight times one.
\[ 0 + 8 = 8 \]
Row 2 with column 1
Why: Two times one plus five times zero.
\[ 2 + 0 = 2 \]
Row 2 with column 2
Why: Two times zero plus five times one.
\[ 0 + 5 = 5 \]
Figure (svg): The solution to Worked example verify the identity does nothing shown as a ladder of expressions, one row per algebraic move
\[ \begin{bmatrix} 3 & 8 \\ 2 & 5 \end{bmatrix} \]
Verify: see why each entry survives
Why: In every product, the identity contributes exactly one 1 and the rest zeros, so exactly one term of each sum survives and it is the original entry. The zeros do the work: they switch off every term except the one that should remain.
Sorting
Ones on the diagonal, zeros elsewhere, and square.
Sort into buckets
Sort each matrix.
The anti-diagonal case is worth testing: multiplying by it swaps the two rows, which is a real and useful operation but not an identity.
Worked example
The identity is one of the few matrices that commutes with everything.
\[ \text{Compute } \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 3 & 8 \\ 2 & 5 \end{bmatrix}. \]
Row 1 of the identity with column 1
Why: One times three plus zero times two.
\[ 3 \]
Row 1 with column 2
Why: One times eight plus zero times five.
\[ 8 \]
Row 2 with column 1
Why: Zero times three plus one times two.
\[ 2 \]
Row 2 with column 2
Why: Zero times eight plus one times five.
\[ 5 \]
Figure (svg): The solution to Worked example the identity in the other order shown as a ladder of expressions, one row per algebraic move
\[ IA = A \]
Verify: compare with the previous product
Why: Both orders gave the same answer, the original matrix. This is the exception Lesson 3.6 promised: matrix multiplication is not commutative in general, but the identity commutes with every square matrix of its size, which is precisely what an identity is for.
Trap
\[ \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \quad \text{used as the identity} \]
Assume the identity is the matrix made entirely of ones
Why: The name suggests a matrix of ones rather than a matrix that acts like one.
\[ \begin{bmatrix} 3 & 8 \\ 2 & 5 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 11 & 11 \\ 7 & 7 \end{bmatrix} \]
That has changed the matrix completely, so it cannot be an identity for anything.
\[ I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]
Ones on the diagonal only, zeros elsewhere
Why: The zeros are as essential as the ones: they switch off every term that should not survive.
The name means the matrix that acts like the number one, not the matrix made of ones. Testing any candidate by multiplying settles it in four calculations.
Prediction
Commit before computing.
Predict first
You multiply a 3 by 3 matrix by the 3 by 3 identity. What do you get?
Correct: The original matrix, unchanged.
\[ AI = A = IA \quad \text{for every square } A \text{ of the right size} \]
Why: That is the defining property of an identity: multiplying by it changes nothing, in either order. In each entry's sum, the identity contributes exactly one 1 and the rest zeros, so exactly one term survives and it is the original entry. Getting anything else would mean the matrix used was not the identity.
Explain it to yourself
The zeros are doing as much work as the ones.
\[ \begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]
Discussion prompt
Explain what would go wrong if the off-diagonal entries of the identity were anything other than zero. Compute the top-left entry with a 1 in the bottom-left position instead and see.
Hint: Write out the sum for the top-left entry.
Answer:
\[ \text{top-left} = a(1) + b(0) = a \]
The zero switches off the b term, so only a survives. With a 1 there instead, the entry would be a plus b — the two entries would get mixed together and the matrix would not come back unchanged.
So the identity works because each of its columns picks out exactly one entry and suppresses the rest. That is what the pattern of ones and zeros is for.
Matching
Several ideas from Lesson 1.1 have matrix versions.
Match the pairs
Why: The last row is the sharpest analogy: exactly one number, zero, has no reciprocal, and exactly the matrices with zero determinant have no inverse. In both cases the failure is a division by zero — for numbers, dividing 1 by 0; for matrices, dividing by the determinant in the inverse formula.
Section
Section 2
Concept
Square matrices A and B are inverses when AB and BA are both the identity. The inverse of A is written A to the power negative one, and a matrix has one exactly when its determinant is not zero.
\[ AA^{-1} = I = A^{-1}A \]
Both orders must give the identity. For square matrices it turns out that one implies the other, but the definition asks for both and checking both is the safe habit.
Figure (svg): Two columns contrasting a matrix with an inverse against one without
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.8 Use Inverse Matrices to Solve Linear Systems §3.8, pp. 210-210
Picture it
The determinant decides, exactly as it decided Cramer's rule.
Figure (svg): Two columns contrasting a matrix with an inverse against one without
A matrix with zero determinant is called singular and has no inverse at all. The corresponding system has no unique solution, and those are two descriptions of one fact.
Worked example
Verifying the pair from Example 1 by multiplying in both orders.
\[ \text{Show that } \begin{bmatrix} 3 & 8 \\ 2 & 5 \end{bmatrix} \text{ and } \begin{bmatrix} -5 & 8 \\ 2 & -3 \end{bmatrix} \text{ are inverses.} \]
Multiply in the first order, row 1 with column 1
Why: Three times negative five plus eight times two.
\[ -15 + 16 = 1 \]
Row 1 with column 2
Why: Three times eight plus eight times negative three.
\[ 24 - 24 = 0 \]
Complete the first product
Why: Row 2 gives 2 times -5 plus 5 times 2, which is 0, and 2 times 8 plus 5 times -3, which is 1.
\[ 0\text{ and } 1 \]
Multiply in the other order and check again
Why: The second product also comes out as ones on the diagonal and zeros elsewhere.
Figure (svg): The solution to Worked example check a proposed inverse shown as a ladder of expressions, one row per algebraic move
\[ AA^{-1} = I = A^{-1}A \]
Verify: notice which entries had to cancel
Why: The off-diagonal entries came out as 24 minus 24 and negative 10 plus 10 — exact cancellations, not near misses. That is the signature of a genuine inverse: the diagonal entries survive as ones and everything else cancels to zero exactly.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.8 Use Inverse Matrices to Solve Linear Systems §3.8, pp. 210-210
Sorting
Compute the determinant.
Sort into buckets
Sort each matrix by whether it has an inverse.
Scanning for proportional rows spots a singular matrix instantly, without computing anything.
Worked example
The determinant test, applied before any formula.
\[ \text{Does } \begin{bmatrix} 2 & 3 \\ 4 & 6 \end{bmatrix} \text{ have an inverse?} \]
Compute the determinant
Why: Two times six minus four times three.
\[ 12 - 12 = 0 \]
Apply the condition
Why: A matrix has an inverse if and only if its determinant is not zero.
\[ \det = 0,\text{ so no inverse} \]
See why the formula fails
Why: The inverse formula divides by the determinant, and dividing by zero is undefined.
Connect it to the rows
Why: The second row is exactly twice the first, so the two rows carry the same information and nothing can undo that collapse.
Figure (svg): The solution to Worked example a matrix with no inverse shown as a ladder of expressions, one row per algebraic move
\[ \det = 0 \;\Longrightarrow\; \text{no inverse exists} \]
Verify: check that no inverse could exist
Why: Suppose some B satisfied AB equal to I. The second row of A is twice its first, so the second row of AB would be twice its first row — but the identity's second row, 0 and 1, is not twice its first row, 1 and 0. So no such B exists, which confirms the determinant test rather than merely restating it.
Trap
\[ AB = I \;\Longrightarrow\; \text{A and B are inverses} \]
Verify one product and stop
Why: The definition's second half is treated as a formality.
For square matrices the second product does follow, but that is a theorem rather than something obvious — and for non-square matrices it genuinely fails.
\[ AB = I \;\text{ and }\; BA = I \]
Check both orders, as the definition asks
Why: Matrix multiplication is not commutative, so AB equal to I and BA equal to I are different statements that happen to coincide here.
Checking both costs four more multiplications and confirms you have the inverse rather than something that merely works on one side.
Prediction
Commit before reasoning.
Predict first
Among 2 by 2 matrices, how common is it to have no inverse?
Correct: Only those whose determinant is exactly zero — a special condition.
It also mirrors Lesson 3.1: parallel lines are a special case, and two lines picked at random cross. Singular matrices correspond exactly to those special cases.
Why: The determinant ad minus cb is zero only when the two products happen to be exactly equal, which is a knife-edge condition: change any entry slightly and it becomes nonzero. So singular matrices are exceptional rather than typical. This mirrors numbers, where exactly one value out of infinitely many has no reciprocal.
Fill the middle
The off-diagonal entry that must vanish.
Fill in the blanks
3(8) + 8(-3) = 0
Why: Twenty-four minus twenty-four is zero, which is exactly what the top-right entry of the identity requires. The off-diagonal entries of an inverse check must cancel exactly — a value close to but not equal to zero would mean the proposed inverse is wrong, not merely imprecise.
Counterexample
A classmate reasons by analogy with numbers.
\[ \text{if } AB = AC \text{ then } B = C \]
Discussion prompt
For numbers this cancellation is valid whenever a is not zero. Find matrices where it fails, and say what property A must have for the cancellation to be legal.
Hint: Try a singular A, and recall from Lesson 3.6 that nonzero matrices can multiply to zero.
Answer:
\[ A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \; B = \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix}, \; C = \begin{bmatrix} 1 & 2 \\ 5 & 7 \end{bmatrix} \]
Both products equal the same matrix, yet B and C differ. The cancellation fails because A is singular.
If A has an inverse, multiplying both sides on the left by it gives B equals C directly. So invertibility is exactly the condition that makes cancellation legal — the matrix version of a being nonzero.
Section
Section 3
Concept
Swap the two entries on the main diagonal, change the sign of the other two, and divide every entry by the determinant. Three moves, and the third is the one that can fail.
\[ A^{-1} = \frac{1}{\lvert A \rvert}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]
Dividing by the determinant is scalar multiplication from Lesson 3.5, so it reaches every entry — including the two that were already negated.
Figure (svg): The formula for the inverse of a two by two matrix: swap the main diagonal, negate the other, and divide by the determinant
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.8 Use Inverse Matrices to Solve Linear Systems §3.8, pp. 210-210 — The Inverse of a 2 by 2 Matrix
Picture it
From a general 2 by 2 to its inverse.
Figure (svg): The formula for the inverse of a two by two matrix: swap the main diagonal, negate the other, and divide by the determinant
The determinant appears in the denominator, which is why a zero determinant means no inverse. The formula does not merely fail to help — it is undefined.
Worked example
Example 1. Three moves, in order.
\[ \text{Find the inverse of } A = \begin{bmatrix} 3 & 8 \\ 2 & 5 \end{bmatrix}. \]
Compute the determinant
Why: Three times five minus two times eight.
\[ 15 - 16 = -1 \]
Swap the main diagonal
Why: The 3 and the 5 change places.
Negate the other diagonal
Why: Eight becomes negative eight; two becomes negative two.
\[ -8\text{ and } -2 \]
Divide every entry by the determinant
Why: Dividing by negative one flips every sign.
\[ [[-5, 8], [2, -3]] \]
Figure (svg): The solution to Worked example find an inverse shown as a ladder of expressions, one row per algebraic move
\[ A^{-1} = \begin{bmatrix} -5 & 8 \\ 2 & -3 \end{bmatrix} \]
Verify: multiply A by the proposed inverse
Why: Three times negative five plus eight times two is negative fifteen plus sixteen, which is 1. Three times eight plus eight times negative three is 24 minus 24, which is 0. The other row gives 0 and 1. The product is the identity, so the inverse is correct.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.8 Use Inverse Matrices to Solve Linear Systems §3.8, pp. 210-210
Fill the middle
Guided Practice 2: the matrix with rows -1, 5 and -4, 8.
Fill in the blanks
\det = -8 + 20 = 12, \quad A^-5 = \tfrac______\begin___ 8 & ___ \\ 4 & -1 \end___
Why: The original's top-right entry is 5, and the formula negates it, giving negative 5. The bottom-left entry of the original is negative 4, and negating it gives positive 4, which is why that entry is positive in the answer. Only the main diagonal is swapped; the other two stay in place and change sign.
Worked example
Guided Practice 1. A determinant other than 1 produces fractional entries.
\[ \text{Find the inverse of } \begin{bmatrix} 6 & 1 \\ 2 & 4 \end{bmatrix}. \]
Compute the determinant
Why: Six times four minus two times one.
\[ 24 - 2 = 22 \]
Swap and negate
Why: The diagonal becomes 4 and 6; the other entries become negative 1 and negative 2.
\[ [[4, -1], [-2, 6]] \]
Divide every entry by 22
Why: Four over 22 is two elevenths; negative one over 22 stays as it is; negative two over 22 is negative one eleventh; six over 22 is three elevenths.
Write the answer
Why: Either as a scalar times a matrix, or with the fractions written out.
\[ (\frac{1}{22}) [[4, -1], [-2, 6]] \]
Figure (svg): The solution to Worked example an inverse with fractions shown as a ladder of expressions, one row per algebraic move
\[ A^{-1} = \tfrac{1}{22}\begin{bmatrix} 4 & -1 \\ -2 & 6 \end{bmatrix} \]
Verify: multiply out and check for the identity
Why: The top-left entry of the product is 6 times four twenty-seconds plus 1 times negative two twenty-seconds, which is 24 over 22 minus 2 over 22, or 22 over 22, which is 1. Leaving the answer as a scalar times a whole-number matrix makes this check far easier than writing every fraction out.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.8 Use Inverse Matrices to Solve Linear Systems §3.8, pp. 210-210
Trap
\[ A = \begin{bmatrix} 3 & 8 \\ 2 & 5 \end{bmatrix} \]
Swap the off-diagonal entries and negate the main diagonal
Why: The two operations are applied to the wrong pairs.
\[ \tfrac{1}{-1}\begin{bmatrix} -3 & 2 \\ 8 & -5 \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ -8 & 5 \end{bmatrix} \quad \text{(wrong)} \]
Multiplying A by this gives entries of -55 and 34, nothing like the identity.
\[ A^{-1} = \frac{1}{ad - cb}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]
SWAP the main diagonal; NEGATE the other one
Why: The a and d change places; the b and c stay put and change sign.
\[ A^{-1} = \begin{bmatrix} -5 & 8 \\ 2 & -3 \end{bmatrix} \]
The check is always available: multiply your answer by the original and look for the identity. Nothing else confirms an inverse.
Ranking
Finding the inverse of a 2 by 2 matrix.
Put in order
Why: The determinant comes first because a zero value means there is no inverse to find and the remaining steps would be wasted. The swap and the negation are independent and could be done in either order, but the division must come last since it applies to the finished array. The check is the only step that can confirm the answer.
Discrimination
In the formula for the inverse of a 2 by 2.
Sort into buckets
Sort each entry of the original matrix by what happens to it.
Prediction
Commit before computing.
Predict first
What is the inverse of the 2 by 2 identity matrix?
Correct: The identity itself.
\[ I^{-1} = I \quad \text{since } II = I \]
Why: Its determinant is 1 times 1 minus 0 times 0, which is 1. Swapping the diagonal ones changes nothing, negating the zeros changes nothing, and dividing by 1 changes nothing — so the formula returns the identity. This mirrors the number 1, whose reciprocal is itself, and it is the reason the identity is the natural stopping point when you undo a matrix.
Section
Section 4
Concept
Any linear system can be written as a single matrix equation: the coefficient matrix times a column of variables equals a column of constants. Multiplying out recovers the original equations exactly.
matrix of variables — The column matrix whose entries are the variables of the system. The matrix of constants is the column of right-hand sides.
\[ \begin{bmatrix} 3 & 8 \\ 2 & 5 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ 1 \end{bmatrix} \]
This is what the row-times-column rule from Lesson 3.6 was built for: each row of the product reproduces one equation of the system.
Figure (svg): A system of two equations rewritten as a single matrix equation A times X equals B
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.8 Use Inverse Matrices to Solve Linear Systems §3.8, pp. 210-211
Picture it
The system 3x plus 8y equals 5 and 2x plus 5y equals 1.
Figure (svg): A system of two equations rewritten as a single matrix equation A times X equals B
Multiplying the first two matrices gives a column whose entries are 3x plus 8y and 2x plus 5y — precisely the left-hand sides of the two equations.
Worked example
Translating a system into the form A X equals B.
\[ \text{Write } \begin{cases} 3x + 8y = 5 \\ 2x + 5y = 1 \end{cases} \text{ as a matrix equation.} \]
Build the coefficient matrix
Why: One row per equation, one column per variable, keeping the variables in the same order in both rows.
\[ [[3, 8], [2, 5]] \]
Build the variable column
Why: A single column containing x and y, in that order.
\[ [[x], [y]] \]
Build the constant column
Why: The two right-hand sides, in the same order as the equations.
\[ [[5], [1]] \]
Check by multiplying out
Why: Row one gives 3x plus 8y and row two gives 2x plus 5y, which are the original left sides.
Figure (svg): The solution to Worked example write a system as a matrix equation shown as a ladder of expressions, one row per algebraic move
\[ AX = B \]
Verify: confirm the dimensions
Why: A 2 by 2 times a 2 by 1 gives a 2 by 1, which matches the constant column. Had the variable column been written as a row, the product would have been undefined — the variables must be a COLUMN for the multiplication to work.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.8 Use Inverse Matrices to Solve Linear Systems §3.8, pp. 211-211
Matching
Each system becomes one matrix equation.
Match the pairs
Why: In each case the coefficient matrix is exactly the array of numbers multiplying x and y, with signs included, and the constant column is the right-hand sides in the same order. Notice the negative entries survive into the matrix: in the second system the coefficient of y in the second equation is negative 5, not 5.
Worked example
The coefficient matrix requires the variables in the same order in every row.
\[ \text{Write } \begin{cases} 4y + 3x = 7 \\ 2x = 5 - y \end{cases} \text{ as a matrix equation.} \]
Put both equations in standard form
Why: Variables on the left in the order x then y, constants on the right.
\[ 3 x + 4 y = 7\text{ and } 2 x + y = 5 \]
Read the coefficients into the matrix
Why: Three and four in the first row, two and one in the second.
\[ [[3, 4], [2, 1]] \]
Write the variable and constant columns
Why: x over y, and 7 over 5.
\[ [[x], [y]]\text{ and } [[7], [5]] \]
Note why the rearrangement mattered
Why: Reading coefficients off the unrearranged equations would have put 4 and 3 in the first row, describing a different system entirely.
Figure (svg): The solution to Worked example a system needing rearrangement first shown as a ladder of expressions, one row per algebraic move
\[ \begin{bmatrix} 3 & 4 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ 5 \end{bmatrix} \]
Verify: multiply out and compare with the originals
Why: Row one gives 3x plus 4y equals 7, which rearranges to the given 4y plus 3x equals 7. Row two gives 2x plus y equals 5, which rearranges to 2x equals 5 minus y. Both original equations are recovered, so the rearrangement preserved the system.
Error analysis
A student writes a matrix equation without first rearranging.
Annotate
On: \( \begin{cases} 4y + 3x = 7 \\ 2x + y = 5 \end{cases} \;\Longrightarrow\; \begin{bmatrix} 4 & 3 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ 5 \end{bmatrix} \)
Put every equation in the same standard form before reading any coefficients. The matrix has no way of knowing which column you meant.
Fill the middle
For the system 2x plus 5y equals negative 5, and x plus 3y equals 3.
Fill in the blanks
A = \begin1 & 3 2 & 5 \\ ___ \end___
Why: The second equation is x plus 3y equals 3, so the coefficient of x is 1 — written explicitly even though it is invisible in the equation — and the coefficient of y is 3. An unwritten coefficient of 1 is the commonest thing to miss when building a coefficient matrix.
Prediction
Commit before reasoning.
Predict first
Why is the variable matrix written as a column rather than a row?
Correct: So that the dimensions allow the multiplication.
\[ (2 \times 2)(2 \times 1) = 2 \times 1 \quad \text{but} \quad (2 \times 2)(1 \times 2) \text{ is undefined} \]
Why: The coefficient matrix is 2 by 2, so to multiply it on the right the other matrix must have 2 rows — which means a 2 by 1 column. A 1 by 2 row would give inner numbers 2 and 1, which do not match, and the product would not exist. The column form is forced by Lesson 3.6's dimension rule rather than chosen for readability.
Explain it to yourself
The connection is worth stating rather than assuming.
\[ \begin{bmatrix} 3 & 8 \\ 2 & 5 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} \]
Discussion prompt
Multiply this out using the row-times-column rule and explain why the result reproduces the left-hand sides of the two equations. What does each row of the product correspond to?
Hint: Compute the top entry of the product.
Answer:
\[ \begin{bmatrix} 3x + 8y \\ 2x + 5y \end{bmatrix} \]
The top entry pairs row 1 of the coefficient matrix with the variable column, giving 3 times x plus 8 times y — exactly the left side of the first equation. Each ROW of the coefficient matrix is one equation.
Setting that column equal to the constant column then says both equations at once, which is why a whole system fits into a single matrix equation.
Section
Section 5
Concept
From A X equals B, multiplying on the left by the inverse of A gives X equals the inverse of A times B. One matrix multiplication produces every variable at once.
\[ AX = B \;\Longrightarrow\; X = A^{-1}B \]
On the LEFT, on both sides. Because matrix multiplication is not commutative, multiplying on the right would give a different and useless equation.
Figure (svg): The chain from A X equals B to X equals A inverse B, with the identity cancelling in the middle
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.8 Use Inverse Matrices to Solve Linear Systems §3.8, pp. 211-211
Picture it
Each step uses a property matrices actually have.
Figure (svg): The chain from A X equals B to X equals A inverse B, with the identity cancelling in the middle
Associativity lets the brackets move, the inverse collapses to the identity, and the identity disappears. Every step was established earlier in this chapter.
Worked example
The system whose coefficient matrix was inverted in Example 1.
\[ \begin{cases} 3x + 8y = 5 \\ 2x + 5y = 1 \end{cases} \]
Write it as a matrix equation
Why: Coefficient matrix, variable column, constant column.
\[ AX = B \]
Find the inverse of the coefficient matrix
Why: Determinant negative one, so the inverse is the swapped and negated matrix with every sign flipped.
\[ [[-5, 8], [2, -3]] \]
Multiply the inverse by the constant column, on the left
Why: Negative five times five plus eight times one, and two times five plus negative three times one.
\[ -25 + 8\text{ and } 10 - 3 \]
Read off the solution
Why: The variable column comes out as negative seventeen over seven.
\[ x = -17, y = 7 \]
Figure (svg): The solution to Worked example solve a system with an inverse shown as a ladder of expressions, one row per algebraic move
\[ X = A^{-1}B = \begin{bmatrix} -17 \\ 7 \end{bmatrix} \]
Verify: substitute into both original equations
Why: First: 3 times -17 plus 8 times 7 is -51 plus 56, which is 5 — correct. Second: 2 times -17 plus 5 times 7 is -34 plus 35, which is 1 — correct. Both equations hold, and the whole solution came from a single matrix multiplication.
Ranking
Solving a system with an inverse matrix.
Put in order
Why: The matrix equation must exist before its coefficient matrix can be inverted. The determinant check comes before the inverse because a zero value means no inverse exists and the method cannot be used. And the final substitution uses the original equations, which is the only check that tests the whole chain including the rearrangement.
Worked example
The system from Lesson 3.7's Example 3, solved a third way.
\[ \begin{cases} 9x + 4y = -6 \\ 3x - 5y = -21 \end{cases} \]
Find the determinant
Why: Nine times negative five minus three times four.
\[ -45 - 12 = -57 \]
Build the inverse
Why: Swap the diagonal to give negative 5 and 9; negate the others to give negative 4 and negative 3; divide by negative 57.
\[ (1 / - 57) [[-5, -4], [-3, 9]] \]
Multiply by the constant column
Why: Negative five times negative six plus negative four times negative twenty-one is 30 plus 84, which is 114. Negative three times negative six plus nine times negative twenty-one is 18 minus 189, which is negative 171.
\[ 114\text{ and } -171 \]
Divide by the determinant
Why: One hundred and fourteen over negative 57 is negative 2; negative 171 over negative 57 is 3.
\[ x = -2, y = 3 \]
Figure (svg): The solution to Worked example a second system, same method shown as a ladder of expressions, one row per algebraic move
\[ X = \begin{bmatrix} -2 \\ 3 \end{bmatrix} \]
Verify: compare with the Cramer's rule answer
Why: Lesson 3.7 solved this same system by Cramer's rule and got (-2, 3). The numerators 114 and negative 171 that appeared here are exactly the determinants Cramer's rule computed — which is not a coincidence, since Cramer's rule is what you get by writing out the inverse-matrix method entry by entry.
Error analysis
A student solves a matrix equation by multiplying on the right.
Annotate
On: \( AX = B \;\Longrightarrow\; AXA^{-1} = BA^{-1} \;\Longrightarrow\; X = BA^{-1} \)
The side matters. Left-multiply to cancel a matrix on the left, and check the dimensions of every product you write down.
Comparison
Fill the blanks. All three give the same exact answer.
Comparison matrix
| Method | What you compute | Best when |
|---|---|---|
| Elimination (3.2) | a sequence of equations | one system, done once |
| Cramer's rule (3.7) | three determinants | you need only one variable |
| Inverse matrix (3.8) | one inverse and one product | several systems share a coefficient matrix |
| All three | the same exact solution | always - they cannot disagree |
The inverse method's real advantage is the third row: once A-inverse is known, any number of constant columns can be solved in one multiplication each.
Real world
A factory's output is governed by the same two constraints every week, but the weekly targets change.
Discussion prompt
The coefficient matrix is [[3,8],[2,5]] every week, while the constant column changes. Explain why the inverse-matrix method is the right choice here, and how much work each new week costs once the inverse is known.
Hint: Count what has to be recomputed.
Answer:
\[ A^{-1} = \begin{bmatrix} -5 & 8 \\ 2 & -3 \end{bmatrix} \quad \text{computed once} \]
Each new week is a single matrix multiplication: A-inverse times the new constant column, which is four multiplications and two additions. Elimination or Cramer's rule would restart from scratch every week.
This is the practical reason inverses matter. A system solved once is quicker by elimination; a coefficient matrix reused fifty times makes the inverse overwhelmingly the better investment.
Commit first
Answer, then rate your confidence honestly.
Predict first
A system's coefficient matrix has determinant zero. What happens when you try the inverse method?
Correct: It fails — A has no inverse, and correspondingly the system has no unique solution.
Elimination remains available and will tell you which degenerate case applies — no solution or infinitely many — which neither of the matrix methods can distinguish.
Why: The inverse formula divides by the determinant, so a zero determinant makes it undefined. That is not a shortcoming of the method: a system with a zero coefficient determinant genuinely has no unique solution, so there is no single answer for any method to find. The determinant test is the same one that governed Cramer's rule in Lesson 3.7, which is why all three matrix methods agree about exactly which systems they can handle.
Comparison
Fill the blanks. Solving a matrix equation copies solving a number equation.
Comparison matrix
| Step | For numbers: 3x = 12 | For matrices: AX = B |
|---|---|---|
| The undoing object | the reciprocal, one third | the inverse matrix |
| What it produces | the number 1 | the identity matrix |
| The move | multiply both sides by it | left-multiply both sides by it |
| When it fails | when the coefficient is 0 | when the determinant is 0 |
| The answer | x = 4 | X = A inverse times B |
Only the third row needs a qualification, and it needs one because matrix multiplication is not commutative.
Pattern
One routine solves a system with matrices.
Step two is doing two jobs at once: it warns you about the division in step three and it tells you whether the system has a unique solution at all.
OpenStax Algebra and Trigonometry 2e, §11.7 Solving Systems with Inverses §11.7
Check
The inverse formula. Swap one diagonal, negate the other.
Check your understanding
What is the inverse of [[3, 8], [2, 5]]?
Answer: A
Why: The determinant is 15 minus 16, which is -1. Swapping the diagonal gives 5 and 3, negating the others gives -8 and -2, and dividing by -1 flips every sign to give [[-5, 8], [2, -3]].
Check
The existence condition.
Check your understanding
Which matrix has NO inverse?
Answer: A
Why: Its determinant is 2 times 6 minus 4 times 3, which is 0, so the inverse formula would divide by zero. The second row is exactly twice the first, which is the visible signature.
Check
Solving with the inverse. Watch which side.
Check your understanding
For AX = B, what is X?
Answer: A
Why: Left-multiplying both sides by A-inverse gives A-inverse A X on the left, which collapses to the identity times X, which is X. The right side becomes A-inverse times B.
Real world
A simple cipher encodes a message by grouping letters into pairs, treating each pair as a column, and multiplying by a fixed 2 by 2 matrix.
Discussion prompt
Explain how the recipient decodes the message, what condition the encoding matrix must satisfy, and what would go wrong if a careless sender chose a matrix with determinant zero.
Hint: Decoding is undoing a multiplication.
Answer:
The recipient multiplies each encoded column on the left by the INVERSE of the encoding matrix, which returns the original pair — exactly the AX equals B argument with X as the plaintext.
The encoding matrix must therefore have a nonzero determinant, or no inverse exists and there is no way back.
With a zero determinant the encoding would be irreversible: different plaintext pairs would encode to the same ciphertext, so even knowing the matrix the recipient could not tell which message was sent. That is the same information collapse a zero determinant always signals.
Commit first
Answer, then rate your confidence honestly.
Predict first
Is the inverse of the inverse of A equal to A?
Correct: Yes — undoing the undoing returns the original.
\[ (A^{-1})^{-1} = A \]
Why: By definition A times A-inverse is the identity, and that same statement says A is an inverse of A-inverse. Since inverses are unique, A-inverse-inverse is A. It is the same fact as the reciprocal of the reciprocal of a number being the number itself, and it can be checked directly: inverting the matrix with rows -5, 8 and 2, -3 returns rows 3, 8 and 2, 5.
Explain it
They can solve 3x equals 12 and are unsure what a matrix inverse is for.
Discussion prompt
In four sentences or fewer, explain what an inverse matrix does by analogy with the reciprocal, and describe how it solves a whole system at once.
Hint: Start from the number case they already know.
Answer:
To solve 3x equals 12 you multiply both sides by one third, because three times one third is one and one leaves x alone. A matrix inverse does the same job: it is the matrix that multiplies with A to give the identity, which is the matrix version of the number one.
Write the system as a coefficient matrix times a column of variables equals a column of constants, then multiply both sides on the left by the inverse. The coefficient matrix cancels and the column of variables is left equal to the inverse times the constants — every variable at once.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the formula, say swap the main, negate the other — and check by multiplying. For the division, treat it as scalar multiplication so it reaches all four entries. For AX equals B, rearrange every equation into standard form first. For the side, remember that only an adjacent pair can cancel, so the inverse must go next to the A. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top of a page write the 2 by 2 identity matrix and verify, in full, that multiplying a matrix of your own by it changes nothing. Below, write a general 2 by 2 matrix, draw an arrow showing which diagonal is swapped and which is negated, and write the formula for the inverse with the determinant in the denominator. Then take a specific matrix with a nonzero determinant, find its inverse, and multiply the two together to confirm you get the identity. In the lower half, write a system of your own, convert it to AX equals B, solve it by X equals A-inverse B, and substitute the answer into both original equations. Finally, in a margin, write the one condition that must hold and what it means when it fails.
The margin condition should be that the determinant is nonzero, and its failure should be described as both no inverse and no unique solution. Those are the same fact, and this chapter has now shown it three times.
Recap
Five things, and together they close the chapter's arc from graphing to a single multiplication.
| If you see | Then |
|---|---|
| Ones on the diagonal, zeros elsewhere | The identity; it changes nothing |
| A 2 by 2 to invert | Swap the main, negate the other, divide |
| A determinant of zero | No inverse, and no unique solution |
| A system to solve | Write AX = B first |
| AX = B | Left-multiply by A inverse |
That completes Chapter 3. Chapter 4 leaves lines behind entirely and takes up quadratic functions, where a single equation can have two solutions, one, or none — and where the graph is a curve rather than anything straight.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.8 Use Inverse Matrices to Solve Linear Systems §3.8, pp. 210-215 — everything on these slides traces back here
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