Introduces the identity matrix and the multiplicative inverse, computes inverses for two-by-two and larger matrices, and solves a whole system in one multiplication. Connects a vanishing determinant to the absence of an inverse and to the special cases of the earlier sections.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 9 — Systems of Equations and Inequalities
§9.7 Solving Systems with Inverses, pp. 1145-1160
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §9.7 Solving Systems with Inverses §9.7, pp. 1145-1160 — the pages these objectives are drawn from
Warm-up
For numbers the answer is familiar. For matrices it is not obvious.
Discussion prompt
To solve the equation that 3 times x equals 12, what do you do?
Hint: What do you multiply by, and what does the product equal?
Answer:
Multiply both sides by one third, which is 3's reciprocal. Three times one third is 1, and multiplying by 1 leaves the unknown alone.
So undoing a multiplication needs two things: something to multiply by, and something for the product to equal that acts like a do-nothing.
For matrices both pieces have to be built. The identity matrix plays the role of 1, and the inverse plays the role of the reciprocal — but not every matrix has one.
Concept
The inverse of a square matrix is the matrix that multiplies it, in either order, to give the identity — which is the matrix that leaves everything unchanged.
inverse matrix — the matrix which, multiplied by a given square matrix in either order, gives the identity matrix
\[ AA^{-1}=A^{-1}A=I \]
Requiring the product in both orders is a real condition, given that matrix multiplication is not commutative. For square matrices it turns out that one order implies the other, but that is a theorem rather than an assumption.
Figure (svg): The identity matrix shown with ones down its diagonal and zeros elsewhere, alongside the statement that multiplying by it changes nothing
OpenStax, Precalculus, §9.7 Solving Systems with Inverses §9.7, pp. 1145-1149
Section
Section 1
Concept
The identity matrix leaves any matrix unchanged when multiplied by it, in either order. It is the matrix analogue of the number one.
The identity is one of the few matrices that commutes with everything of the right size. That is exactly what makes it usable on either side of an equation, which the inverse method depends on.
Figure (svg): The identity matrix shown with ones down its diagonal and zeros elsewhere, alongside the statement that multiplying by it changes nothing
OpenStax, Precalculus, §9.7 Solving Systems with Inverses §9.7, pp. 1145-1150
Picture it
Ones on the diagonal and nothing else.
Figure (svg): The identity matrix shown with ones down its diagonal and zeros elsewhere, alongside the statement that multiplying by it changes nothing
The bottom line names why it matters here: without something for a product to equal, there would be no way to say what an inverse is.
Worked example
One product settles it.
\[ \text{Multiply } \begin{bmatrix}2&5\\3&1\end{bmatrix} \text{ by the two-by-two identity.} \]
Row 1 with column 1
Why: Two times one plus five times zero.
\[ 2 \]
Row 1 with column 2
Why: Two times zero plus five times one.
\[ 5 \]
Row 2 with both columns
Why: Same pattern.
\[ 3\text{ and } 1 \]
Compare
Why: The original matrix.
Figure (svg): The identity matrix shown with ones down its diagonal and zeros elsewhere, alongside the statement that multiplying by it changes nothing
\[ \begin{bmatrix}2&5\\3&1\end{bmatrix} \]
Verify: check the other order too
Why: Multiplying with the identity on the left gives the same result, which is unusual — most matrices do not commute. The identity commuting with everything is what allows it to be used on either side of an equation.
OpenStax, Precalculus, §9.7 Solving Systems with Inverses §9.7, pp. 1146-1147
Prediction
A matrix is multiplied by the identity of the right size.
Predict first
What is the result?
Correct: The original matrix, unchanged.
Why: Each column of the identity has a single one, which selects exactly one entry and leaves it as it was. This holds in either order, which makes the identity one of the few matrices that commutes with everything.
Worked example
Each entry picks out one column.
\[ \text{Explain why ones on the diagonal produce no change.} \]
Consider one entry of the product
Why: Row against column.
Look at the identity's column
Why: One nonzero entry.
\[ a\text{ single } 1 \]
See what survives
Why: Only the term against that 1.
Identify it
Why: The original entry in that position.
Figure (svg): The solution to Worked example why the diagonal of ones shown as a ladder of expressions, one row per legal move
\[ \text{the product reproduces }A \]
Verify: check what a different diagonal would do
Why: Replacing a diagonal one by a two would double every entry in that column of the product. The ones are what make the selection neutral rather than scaling, which is why any other value would fail.
OpenStax, Precalculus, §9.7 Solving Systems with Inverses §9.7, pp. 1147-1150
Trap
\[ \text{multiply a }3\times 2\text{ matrix by the }3\times 3\text{ identity on the right} \]
Use whichever identity is at hand
Why: The size is not checked against the dimension rule.
The inner dimensions are 2 and 3, so the product does not exist.
The identity's size must fit the dimension rule. On the right of a 3 by 2 matrix, the 2 by 2 identity is needed.
On the left, the 3 by 3 identity is the one that fits. So a non-square matrix has different identities on each side.
For square matrices the two coincide, which is why the size question rarely comes up once inverses are in play — inverses exist only for square matrices.
Faded example
The three-by-three case.
Fill in the blanks
\textzeros3\text___3\times___
Why: The identity is always square, with ones down the main diagonal and zeros in every other position. There is one for each size, and the size must fit the dimension rule.
Sorting
The dimension rule decides.
Sort into buckets
Sort each situation for a 3 by 2 matrix.
Step zero
You want to define an inverse for matrices.
Discussion prompt
What has to exist first?
Hint: What does a product need to equal?
Answer:
Something for the product to equal — a matrix that plays the role the number 1 plays for ordinary multiplication.
Without it, saying that one matrix undoes another has no content, because there is nothing for 'undone' to mean.
So the identity comes first and the inverse is defined against it. A good grasp of this ordering explains why the identity gets its own definition rather than appearing as an afterthought.
Section
Section 2
Concept
For a two-by-two matrix the inverse has a formula: exchange the diagonal entries, negate the other two, and divide everything by the determinant.
The determinant appearing as a divisor is the whole story of when an inverse exists. A matrix with a zero determinant is the analogue of the number zero — the one value with no reciprocal.
Figure (svg): A card showing the two-by-two determinant as a difference of diagonal products, with the inverse formula beside it
OpenStax, Precalculus, §9.7 Solving Systems with Inverses §9.7, pp. 1150-1154
Picture it
The determinant is a difference of diagonal products.
Figure (svg): A card showing the two-by-two determinant as a difference of diagonal products, with the inverse formula beside it
The bottom line is the point. Because the determinant is a divisor, its vanishing is a division by zero and the inverse simply does not exist.
Worked example
Determinant, then the pattern.
\[ \text{Find the inverse of } \begin{bmatrix}2&5\\1&3\end{bmatrix}. \]
Compute the determinant
Why: Diagonal products, subtracted.
\[ 6 - 5 = 1 \]
Swap the diagonal entries
Why: Three and two.
\[ 3\text{ and } 2 \]
Negate the others
Why: Five and one.
\[ -5\text{ and } -1 \]
Divide by the determinant
Why: It is one, so no change.
Figure (svg): A card showing the two-by-two determinant as a difference of diagonal products, with the inverse formula beside it
\[ \begin{bmatrix}3&-5\\-1&2\end{bmatrix} \]
Verify: multiply to check
Why: Two times three plus five times negative one is 1, and two times negative five plus five times two is 0 — the first row of the identity. The second row checks the same way, confirming the inverse.
OpenStax, Precalculus, §9.7 Solving Systems with Inverses §9.7, pp. 1151-1152
Faded example
For a two-by-two matrix with entries 2, 5, 1 and 3.
Fill in the blanks
\det=2(3)-5(1)=6-5=1
Why: The determinant is the main-diagonal product minus the other diagonal product. It is computed first because it decides whether an inverse exists at all.
Worked example
The determinant vanishes.
\[ \text{Find the inverse of } \begin{bmatrix}2&4\\1&2\end{bmatrix}. \]
Compute the determinant
Why: Diagonal products.
\[ 4 - 4 = 0 \]
Attempt the division
Why: By zero.
Conclude
Why: No inverse exists.
Note the structure
Why: The second row is half the first.
Figure (svg): The solution to Worked example a matrix with no inverse shown as a ladder of expressions, one row per legal move
\[ \text{singular: no inverse} \]
Verify: connect to the rows
Why: The second row is exactly half the first, so as equations they describe the same line — a dependent system. The zero determinant and the dependent rows are the same fact seen two ways, which is why the determinant answers the classification question.
OpenStax, Precalculus, §9.7 Solving Systems with Inverses §9.7, pp. 1152-1154
Error analysis
A student computes an inverse.
Annotate
On: \( \begin{bmatrix}3&1\\2&4\end{bmatrix}^{-1}=\begin{bmatrix}4&-1\\-2&3\end{bmatrix} \)
The division is what makes the product come out as the identity rather than a multiple of it. Multiplying the candidate by the original matrix is a one-line check that catches this and every other slip.
Prediction
You compute a determinant of zero.
Predict first
What follows?
Correct: No inverse exists.
Why: The determinant appears as a divisor in the formula, so a zero determinant is a division by zero. This is the matrix analogue of zero being the one number with no reciprocal.
Sorting
Compute the determinant.
Sort into buckets
Sort each two-by-two matrix by its entries in reading order.
Explain it to yourself
It appears as a divisor in the formula.
Discussion prompt
Explain what that implies.
Hint: What is the one number with no reciprocal?
Answer:
Dividing by the determinant is the last step, so a determinant of zero makes the formula undefined — a division by zero.
That makes the determinant the matrix analogue of the number itself in ordinary arithmetic: zero is the one number with no reciprocal, and a zero determinant is the one case with no inverse.
And it is not an accident of the formula. A zero determinant means the rows are dependent, which means the system has no unique solution — so the algebra and the geometry agree, as a good explanation should point out.
Section
Section 3
Concept
For matrices larger than two by two there is no short formula. Writing the matrix beside the identity and row reducing until the left side becomes the identity turns the right side into the inverse.
The method works because the row operations that turn the matrix into the identity, applied to the identity, build exactly the matrix that reverses them. That is why the two halves must be reduced together with identical operations.
Figure (svg): A contrast between a matrix with an inverse and one without, showing what each says about the corresponding system
OpenStax, Precalculus, §9.7 Solving Systems with Inverses §9.7, pp. 1154-1157
Picture it
The left column is the invertible case.
Figure (svg): A contrast between a matrix with an inverse and one without, showing what each says about the corresponding system
The third row on each side is what the reduction reveals. A full staircase means the identity is reachable; a row of zeros means it is not.
Worked example
Matrix and identity side by side.
\[ \text{Set up the inverse computation for } \begin{bmatrix}1&2\\3&4\end{bmatrix}. \]
Write the matrix
Why: On the left.
\[ 1, 2, 3, 4 \]
Append the identity
Why: Same size, on the right.
\[ 1, 0, 0, 1 \]
Note the goal
Why: Left side becomes the identity.
Note the rule
Why: Every operation spans both halves.
Figure (svg): A contrast between a matrix with an inverse and one without, showing what each says about the corresponding system
\[ \left[\begin{array}{cc|cc}1&2&1&0\\3&4&0&1\end{array}\right] \]
Verify: check the shape
Why: The array is 2 by 4 — the matrix and the identity side by side. Every row operation applies across all four entries of a row, exactly as the augmented matrices of §9.6 included their constant column.
OpenStax, Precalculus, §9.7 Solving Systems with Inverses §9.7, pp. 1155-1156
Prediction
The left side has been reduced to the identity.
Predict first
What is the right side now?
Correct: The inverse of the original matrix.
Why: The row operations that turned the matrix into the identity, applied to the identity, assemble into the matrix that reverses them. That is exactly the inverse.
Worked example
The left side cannot become the identity.
\[ \text{What happens reducing } \begin{bmatrix}1&2\\2&4\end{bmatrix} \text{ beside the identity?} \]
Clear below the pivot
Why: Subtract twice row 1.
\[ 0, 0\text{ on the left} \]
Look at the left side
Why: A row of zeros.
Conclude
Why: The identity is unreachable.
Check the determinant
Why: Four minus four.
Figure (svg): The solution to Worked example when the reduction fails shown as a ladder of expressions, one row per legal move
\[ \text{no inverse exists} \]
Verify: connect the two tests
Why: The determinant is zero and the reduction produces a zero row — two different computations giving the same verdict. That agreement is expected, since both are detecting the same dependence among the rows.
OpenStax, Precalculus, §9.7 Solving Systems with Inverses §9.7, pp. 1156-1157
Trap
\[ \text{reduce the left half, then work on the right separately} \]
Treat the two halves as independent
Why: Operations are applied to one side at a time.
The right side no longer records the operations that transformed the left, so it is not the inverse.
Every operation spans the whole row, both halves at once. The right side is a record of what was done to the left.
That record is precisely what builds the inverse: the operations reversing the matrix, assembled into a matrix themselves.
The bar is a reading aid, exactly as in §9.6, and never a boundary for operations.
Sorting
The left half's final form decides.
Sort into buckets
Sort each outcome.
Faded example
For a three-by-three matrix.
Fill in the blanks
\text33\times6\text___3\times___\text___
Why: The identity has the same size as the matrix, so the combined array has twice as many columns. Every row operation runs across all six entries of a row.
Explain it
Reducing beside the identity produces the inverse.
Discussion prompt
Explain to a classmate why that happens.
Hint: What is the right side recording?
Answer:
The row operations turn the original matrix into the identity — that is, they undo whatever the matrix does.
Applying those same operations to the identity records them: the right side accumulates the effect of every step.
So the right side ends up being the matrix that performs all those undoing operations at once — which is the inverse. A good explanation stresses that this is why both halves must be reduced together, since the record is only accurate if it captures every operation.
Section
Section 4
Concept
A system becomes a single equation with a coefficient matrix, a column of unknowns and a column of constants. Multiplying by the inverse on the left isolates the unknowns.
Multiplying on the left is essential, not a convention. Since matrix multiplication is not commutative, multiplying on the right would give a product that does not simplify and, for a column of unknowns, would not even be defined.
Figure (svg): A diagram showing a system written as a matrix equation and solved in one multiplication by the inverse
OpenStax, Precalculus, §9.7 Solving Systems with Inverses §9.7, pp. 1157-1159
Picture it
One multiplication replaces the whole reduction.
Figure (svg): A diagram showing a system written as a matrix equation and solved in one multiplication by the inverse
The bottom line is when this is worth doing. One inverse serves every right-hand side, so several systems sharing a coefficient matrix are solved almost free after the first.
Worked example
One multiplication.
\[ \text{Solve } 2x+5y=11, \; x+3y=6 \text{ using the inverse found earlier.} \]
Write the matrix equation
Why: Coefficients, unknowns, constants.
\[ AX = B \]
Recall the inverse
Why: From the earlier example.
\[ [[3, -5], [-1, 2]] \]
Multiply the inverse by the constants
Why: Row against column.
\[ 33 - 30\text{ and } -11 + 12 \]
Read the answers
Why: The column of unknowns.
\[ x = 3, y = 1 \]
Figure (svg): A diagram showing a system written as a matrix equation and solved in one multiplication by the inverse
\[ (x,y)=(3,1) \]
Verify: substitute into both equations
Why: Six plus five is 11 — correct. Three plus three is 6 — also correct. One matrix multiplication produced both values at once, where row reduction would have taken several operations plus a back-substitution.
OpenStax, Precalculus, §9.7 Solving Systems with Inverses §9.7, pp. 1158-1159
Prediction
The equation reads coefficient matrix times unknowns equals constants.
Predict first
Where does the inverse go?
Correct: On the left of both sides.
Why: Only the left puts the inverse adjacent to the coefficient matrix, where the two combine into the identity. On the right it would sit next to the unknowns and nothing would cancel.
Worked example
Order matters here.
\[ \text{Why multiply } AX=B \text{ by the inverse on the left?} \]
Consider the left product
Why: The inverse meets the coefficient matrix.
Note what remains
Why: The identity times the unknowns.
Consider the right instead
Why: The inverse would meet the unknowns.
Check the dimensions
Why: A column times a square matrix.
Figure (svg): The solution to Worked example why the left shown as a ladder of expressions, one row per legal move
\[ A^{-1}(AX)=(A^{-1}A)X=X \]
Verify: check the dimension argument
Why: The column of unknowns is n by 1, and multiplying it on the right by an n by n matrix pairs 1 with n — undefined. So the wrong order fails twice over, algebraically and dimensionally.
OpenStax, Precalculus, §9.7 Solving Systems with Inverses §9.7, pp. 1159-1160
Error analysis
A student solves a matrix equation.
Annotate
On: \( AX=B \;\Longrightarrow\; X=BA^{-1} \)
Matrix multiplication is not commutative, so which side an operation is applied to is part of the operation. The left is the side that puts the inverse adjacent to the matrix it cancels.
Faded example
Multiplying the inverse by the constant column.
Fill in the blanks
3(11)+(-5)(6)=33-30=3, \quad (-1)(11)+2(6)=1
Why: Each row of the inverse pairs with the constant column to give one unknown. A single matrix multiplication produces the whole solution at once.
Sorting
It costs more to set up and less to reuse.
Sort into buckets
Sort each situation.
Explain it
Row reduction also solves the system.
Discussion prompt
Explain to a classmate when the inverse is the better choice.
Hint: How many systems are there?
Answer:
For one system, row reduction is quicker — computing the inverse is more work than reducing the augmented matrix, and a multiplication is still needed afterwards.
For several systems sharing the same coefficient matrix, the inverse is computed once and each solve becomes a single multiplication.
So the choice is about reuse. A good explanation adds that this pattern recurs throughout computing: pay a setup cost once when the result will be used many times, and skip it when it will not.
Section
Section 5
Concept
A nonzero determinant means the system has exactly one solution. A zero determinant means it has none or infinitely many, and which is decided by the constants.
This connects three things that looked separate: the §9.1 classification, the §9.6 special rows, and the existence of an inverse. All three are detecting whether the rows of the coefficient matrix are independent.
| determinant | the matrix | the system |
|---|---|---|
| not zero | has an inverse | exactly one solution |
| zero | has no inverse | none or infinitely many |
| zero, constants consistent | singular | infinitely many |
| zero, constants inconsistent | singular | no solution |
Figure (svg): A contrast between a matrix with an inverse and one without, showing what each says about the corresponding system
OpenStax, Precalculus, §9.7 Solving Systems with Inverses §9.7, pp. 1153-1160
Picture it
The determinant decides which column applies.
Figure (svg): A contrast between a matrix with an inverse and one without, showing what each says about the corresponding system
The caption is the payoff: a single number, computed before any solving, answers the question the whole of §9.1 was about.
Worked example
One determinant.
\[ \text{Does } 3x+6y=9, \; x+2y=4 \text{ have a unique solution?} \]
Write the coefficient matrix
Why: Coefficients only.
\[ 3, 6, 1, 2 \]
Compute the determinant
Why: Diagonal products.
\[ 6 - 6 = 0 \]
Conclude on uniqueness
Why: Zero determinant.
Check the constants
Why: Three times the second is 12, not 9.
Figure (svg): A contrast between a matrix with an inverse and one without, showing what each says about the corresponding system
\[ \text{singular and inconsistent: no solution} \]
Verify: confirm by looking at the rows
Why: The first equation is three times the second on the left but its constant is 9 rather than 12, so the two describe parallel lines. The determinant detected the dependence and the constants settled which of the two singular cases it was.
OpenStax, Precalculus, §9.7 Solving Systems with Inverses §9.7, pp. 1154-1157
Prediction
The coefficient matrix has determinant zero.
Predict first
What can you conclude?
Correct: There is not exactly one solution.
Why: A zero determinant means the coefficient rows are dependent, so uniqueness fails. Whether the system has none or infinitely many depends on the constants, which the determinant does not involve.
Worked example
Same determinant, different constants.
\[ \text{Classify } 3x+6y=12, \; x+2y=4. \]
Compute the determinant
Why: Same coefficients as before.
\[ 0 \]
Conclude on uniqueness
Why: Not unique.
Check the constants
Why: Three times four is twelve.
Conclude
Why: The equations agree.
Figure (svg): The solution to Worked example the other singular case shown as a ladder of expressions, one row per legal move
\[ \text{dependent: infinitely many} \]
Verify: compare the two examples
Why: Both have the same coefficient matrix and the same zero determinant, but different constants — and different outcomes. The determinant alone cannot distinguish them, which is why it answers the uniqueness question and not the existence one.
OpenStax, Precalculus, §9.7 Solving Systems with Inverses §9.7, pp. 1157-1160
Trap
\[ \det=0 \;\Longrightarrow\; \text{no solution} \]
Read a zero determinant as inconsistency
Why: The two singular cases are not distinguished.
A system with infinitely many solutions is reported as having none.
A zero determinant rules out uniqueness and nothing more. It says the coefficient rows are dependent.
Whether the system has none or infinitely many depends on the constants, which the determinant never sees.
Check the constants for consistency when the determinant vanishes. Row reduction does this automatically, which is one reason it remains useful.
Sorting
Coefficients and constants answer different questions.
Sort into buckets
Sort each piece of information.
Faded example
A coefficient matrix with entries 3, 6, 1, 2.
Fill in the blanks
\det=3(2)-6(1)=6-6=0 \;\Longrightarrow\; \text___
Why: The determinant vanishes because the first row is three times the second. That dependence rules out a unique solution, and the constants then decide between none and infinitely many.
Explain it to yourself
The determinant, the inverse and the §9.1 cases are linked.
Discussion prompt
Explain what all three are detecting.
Hint: What property of the rows?
Answer:
All three detect whether the coefficient matrix's rows are independent — whether each equation says something the others do not.
If they are, the determinant is nonzero, the inverse exists, row reduction gives a full staircase, and the system has exactly one solution.
If they are not, all four fail together: zero determinant, no inverse, a row of zeros, and no unique solution. A good explanation stresses that these are not four facts but one, seen through four different computations.
Comparison
Fill the blanks from memory. The analogy holds with one important gap.
Comparison matrix
| numbers | matrices | |
|---|---|---|
| the do-nothing element | one | the identity matrix |
| undoing multiplication | the reciprocal | the inverse matrix |
| when it fails | only for zero | whenever the determinant is zero |
| does order matter | no | yes, except with the identity |
The third row is where the analogy is loosest. Among numbers only zero lacks a reciprocal; among matrices a great many lack an inverse.
Pattern
Five steps, and the second decides whether the rest is possible.
Step 2 is not optional. Attempting an inverse for a singular matrix wastes the whole computation, and the determinant costs one subtraction to check.
OpenStax Algebra and Trigonometry 2e, §11.7 Solving Systems with Inverses §11.7
Check
The identity.
Check your understanding
What does multiplying a matrix by the identity do?
Answer: A
Why: Each column of the identity has a single one, selecting exactly one entry and leaving it as it was. This holds in either order, which makes the identity one of the few matrices that commutes with everything.
Check
Existence of an inverse.
Check your understanding
A two-by-two matrix has determinant zero. What follows?
Answer: A
Why: The determinant appears as a divisor in the inverse formula, so a zero determinant makes the formula a division by zero. This mirrors zero being the one number with no reciprocal.
Check
Solving.
Check your understanding
To solve the matrix equation AX equals B, what do you do?
Answer: A
Why: Only the left puts the inverse adjacent to A, where the two combine into the identity and leave the unknowns alone. On the right the inverse would sit next to the unknowns and nothing would cancel.
Real world
Cryptography uses matrix inverses to encode and decode messages.
Discussion prompt
A message is encoded by multiplying blocks of it by a matrix. What must be true of that matrix?
Hint: How is the message recovered?
Answer:
It must have an inverse, since decoding means multiplying by the matrix that undoes the encoding.
A matrix with a zero determinant would destroy information — different messages could encode to the same result, making decoding impossible in principle rather than merely difficult.
So the sender must check the determinant before choosing a key. The abstract question of whether an inverse exists becomes the concrete question of whether the message can be recovered, which is the clearest illustration of why singular matrices matter.
Commit first
State your confidence along with your answer.
Predict first
Why does a zero determinant mean no inverse exists?
Correct: The determinant is a divisor in the inverse formula.
Why: Dividing every entry by the determinant is the last step, so a zero value makes it a division by zero. Underlying that, a zero determinant means the rows are dependent, so the matrix destroys information and cannot be undone.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
A classmate asks why some matrices have no inverse when almost every number has a reciprocal.
Hint: What does a singular matrix do to information?
Answer:
Among numbers only zero has no reciprocal, because only zero destroys information — multiplying by it collapses everything to the same result.
A matrix with a zero determinant does the same thing: different inputs can produce the same output, so there is no way to work backwards.
The difference is that many matrices do this, not just one. Any matrix whose rows are dependent collapses some directions, and a good explanation notes that this is the same dependence that makes a system fail to have a unique solution.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The fourth ties together three sections' worth of material into one number, and the second is the computation most worth having automatic before the next section.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Write the identity matrix and state what it does. Beside it, give the two-by-two inverse formula with the determinant marked as the divisor. Underneath, write a system as a matrix equation and show the one multiplication that solves it. Finish with a table connecting the determinant to the number of solutions.
If your table shows that a zero determinant rules out uniqueness without deciding between none and infinitely many, the section's subtlest point is on the page.
Recap
Five things, and the last connects them to everything before.
| if you remember one thing | it should be this |
|---|---|
| about the identity | it is the 1 of matrix multiplication, and it commutes |
| about the inverse formula | the determinant is the divisor, so zero means none |
| about solving | multiply on the left; order is part of the operation |
| about the determinant | it settles uniqueness, and the constants settle existence |
Section 9.8 closes the chapter with Cramer's rule, which expresses each unknown as a ratio of determinants — and makes the zero-determinant case visible as a denominator that vanishes.
OpenStax, Precalculus, §9.7 Solving Systems with Inverses §9.7, pp. 1145-1160 — everything on these slides traces back here
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