CEGEP Linear Algebra — The Inverse of a Matrix Worksheet

All 6 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

1. Q1The Inverse of a Square Matrix

   An $n\times n$ matrix $A$ is invertible when there is an $n\times n$ matrix $A^{-1}$ with $A{A}^{-1}=A^{-1}A=I_n$. For a $2\times 2$ matrix $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ there is a rule: if the number $ad-bc$ is not zero, then $$A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right],$$ and if $ad-bc=0$ the rule produces nothing — part (b) asks you to settle that case yourself.

   1. Find $A^{-1}$ for $A=\left[\begin{array}{cc}3& 5\\ 1& 2\end{array}\right]$, and confirm your answer by computing $A{A}^{-1}$.
   2. Apply the rule to $B=\left[\begin{array}{cc}4& 6\\ 2& 3\end{array}\right]$. Then, without using the rule at all, give a non-zero column $\overrightarrow{x}$ with $B\overrightarrow{x}=\overrightarrow{0}$ and say in one sentence why the existence of such an $\overrightarrow{x}$ already rules out an inverse for $B$.
   3. A matrix $M$ satisfies $M^{-1}=\left[\begin{array}{cc}3& -4\\ -2& 3\end{array}\right]$. Recover $M$.
2. Q2Finding an Inverse by Row Reduction

   Find $A^{-1}$ for $A=\left[\begin{array}{ccc}1& 2& 1\\ 1& 3& 4\\ 2& 5& 6\end{array}\right]$ by reducing the augmented array $[A\mid I_3]$ to $[I_3\mid A^{-1}]$. Name each row operation you use, and check one column of your answer by multiplying it back into $A$.

3. Q3Elementary Matrices

   An elementary matrix is what you get by performing a single row operation on the identity matrix. Work throughout with $3\times 3$ matrices, and take $A=\left[\begin{array}{ccc}1& 3& 2\\ 0& 2& 5\\ 4& 1& 0\end{array}\right]$.

   1. Write down the elementary matrices $E_1$, $E_2$ and $E_3$ corresponding to $R_3\to R_3-4R_1$, to $R_2\to \frac{1}{2}{R}_2$, and to $R_1\leftrightarrow R_2$.
   2. Compute $E_{1}A$, and confirm that it is $A$ with that row operation carried out.
   3. Write down $E_1^{-1}$, $E_2^{-1}$ and $E_3^{-1}$, naming the row operation each one performs, then explain in two sentences why every elementary matrix is invertible and why its inverse is again elementary.
4. Q4Solving a System with the Inverse Matrix

   A soap workshop makes three blends, $X$, $Y$ and $Z$. One batch of each blend uses the kilograms of olive oil, coconut oil and shea butter recorded in the corresponding column of $$A=\left[\begin{array}{ccc}1& 1& 2\\ 2& 3& 5\\ 1& 0& 2\end{array}\right],\text{rows: olive oil, coconut oil, shea butter.}$$ If $\overrightarrow{x}$ lists the numbers of batches made, then $A\overrightarrow{x}$ lists the kilograms consumed, so “a delivery $\overrightarrow{b}$ is used up exactly” means $A\overrightarrow{x}=\overrightarrow{b}$.

   1. Verify that $A^{-1}=\left[\begin{array}{ccc}6& -2& -1\\ 1& 0& -1\\ -3& 1& 1\end{array}\right]$ by computing $A{A}^{-1}$.
   2. This week's delivery is $9$ kg of olive oil, $22$ kg of coconut oil and $8$ kg of shea butter. Use $\overrightarrow{x}=A^{-1}\overrightarrow{b}$ to find the batch numbers.
   3. Next week's delivery is $7$, $18$ and $5$ kg. Find those batch numbers too.
   4. In one sentence, say why computing $A^{-1}$ once is the efficient method here, whereas it would not be for a single delivery.
5. Q5Conditions Equivalent to Invertibility

   For an $n\times n$ matrix $A$ the following statements are either all true or all false: $A$ is invertible; the reduced row echelon form of $A$ is $I_n$; $A$ has $n$ pivots; $A\overrightarrow{x}=\overrightarrow{0}$ has only the solution $\overrightarrow{x}=\overrightarrow{0}$; $A\overrightarrow{x}=\overrightarrow{b}$ has exactly one solution for every $\overrightarrow{b}$; the columns of $A$ are linearly independent. Apply that list to $$P=\left[\begin{array}{ccc}1& 2& 3\\ 2& 4& 6\\ 0& 1& 1\end{array}\right],Q=\left[\begin{array}{ccc}1& 0& 2\\ 0& 1& -1\\ 2& 1& 3\end{array}\right],R=\left[\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 1& 0& 2\end{array}\right].$$ For each matrix decide whether it is invertible, justifying the decision by counting pivots. For each one that is not invertible, also give a non-zero $\overrightarrow{x}$ sent to $\overrightarrow{0}$ by that matrix, and write one of its columns as a combination of the other two, so that two further statements on the list are seen to fail as well. Compute no inverses.

6. Q6Synthesis — drawing on several topics in this set

   Let $A=\left[\begin{array}{cc}1& 2\\ 3& 5\end{array}\right]$.

   1. Reduce $A$ to $I_2$ using exactly three row operations, and write down the elementary matrices $E_1$, $E_2$, $E_3$ of those operations, in the order you used them.
   2. From $E_3{E}_2{E}_{1}A=I_2$, read off $A^{-1}$ as a product of elementary matrices and multiply it out. Confirm the result against the $2\times 2$ rule.
   3. Write $A$ itself as a product of elementary matrices, and explain what this computation illustrates about the statement “$A$ is a product of elementary matrices” as an addition to the invertibility list.

The 4 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

An inverse has to undo the matrix from both sides

Q1 opens with the definition rather than a formula, and the definition is a pair of equations: A⁻¹ is the matrix with AA⁻¹ = I and A⁻¹A = I. Both. That looks like pedantry until you remember that matrix multiplication does not commute, so a matrix that undoes A from the left is not automatically the same object as one that undoes it from the right — for square matrices they do coincide, but that is a theorem, not something the notation gives you. It is why part (a) asks you to confirm your answer by multiplying the product out, and why the notion is only defined for a square matrix at all.

For a 2x2 matrix Q1 hands you the closed rule: swap the two diagonal entries, negate the other two, and divide the whole thing by the number ad − bc. Part (b) applies it to a matrix where that number is 0 and the rule collapses. Part (c) runs it backwards — you are given M⁻¹ and asked for M — which is worth doing slowly, because the fact it rests on is that inverting is its own undoing: the inverse of A⁻¹ is A.

Why this set characterises invertibility without ever mentioning determinants

The number ad − bc that appears in the 2x2 rule is a determinant, and you may already have heard "a matrix is invertible exactly when its determinant is non-zero". That statement is true and it is the next worksheet's business. On this one it is unavailable on purpose, and the ordering is what makes the ideas mean something rather than being a slogan you repeat.

A determinant compresses all of that into a single number, which is convenient and which hides the mechanism. Meeting the mechanism first is what lets you answer the questions the number cannot: which vector is sent to zero, which column is redundant, how many pivots were lost. Q1(b) and Q5 both make that concrete by asking for a non-zero x with Ax = 0 — a single such vector is a complete proof that no inverse exists, since multiplying Ax = 0 on the left by a hypothetical A⁻¹ would force x = 0.

[A | I] to [I | A⁻¹], and why that is not a trick

Q2 is the general method: augment A with an identity of the same size, row reduce until the left block is the identity, and read the inverse off the right block. The question asks you to name each row operation as you use it, which is the habit that makes the method checkable — and the point of Q3 is that each of those named operations is itself a matrix.

The reason it works is the whole content of Q3. An elementary matrix is what you get by doing one row operation to the identity, and left-multiplying by it performs that same operation — part (b) asks you to verify exactly this on one product. So a reduction of A to I is a chain of left multiplications, and applying that same chain to the I sitting on the right assembles the matrix that undoes A. The array is not two calculations side by side; it is one calculation, recorded twice.

Q3(c) then asks why every elementary matrix is invertible. The answer is that every allowed row operation can be undone by a row operation of the same kind — add back what you subtracted, multiply back by the reciprocal, or repeat the swap — so the inverse is elementary too. Notice where "allowed" is doing the work: scaling a row by 0 is not a legal operation, and that is precisely the case that would have no reciprocal to multiply back by.

The one detail this set is most often lost on: the order reverses

Everything above builds products of matrices, and every product has to be inverted at some point. There is exactly one rule for that, and it is not the one most people write first.

Q6 is where you watch it happen on a real product. Part (a) reduces a 2x2 matrix to the identity in three operations, so E₃E₂E₁A = I and therefore A⁻¹ = E₃E₂E₁ — the operations in the order you performed them. Part (c) asks for A itself, and it comes out as E₁⁻¹E₂⁻¹E₃⁻¹: the same three operations, each undone, and the whole list read backwards. That is the reversal rule in the only form you are likely to remember, because it also says something obvious about undoing a sequence of actions in the real world.

Solving a system by inversion, and knowing when not to

Q4 is a workshop making three blends of soap from three raw materials, with the recipe of each blend down a column of A, so that Ax lists the kilograms consumed. Part (a) verifies a supplied inverse; parts (b) and (c) then solve for two different weekly deliveries by the same product x = A⁻¹b. Part (d) asks the question the sheet is really about.

The other half of the judgement is that this method assumes an answer of a shape it may not have. x = A⁻¹b delivers exactly one solution, so it can only ever be used once A is known to be invertible — which, on this worksheet, means once the pivots have been counted. A system whose coefficient matrix is not invertible still has a perfectly good solution set; it is just empty or infinite, and only row reduction of the augmented array will report which.

The list, and how to use it as a list

Q5 gives you three 3x3 matrices and forbids you to compute a single inverse. You decide invertibility by counting pivots, and then, for each matrix that fails, you produce two more symptoms of the same failure: a non-zero vector sent to 0, and one column written as a combination of the other two.

Step 4 is the one worth practising, because it is one idea in two notations rather than two facts. Ax is a combination of the columns of A with the entries of x as coefficients, so a non-zero x with Ax = 0 is a statement that some combination of the columns, not all coefficients zero, gives the zero vector — and rearranging it writes one column in terms of the others. Q5 asks for the vector and for the relation between columns separately so that you see them as the same finding twice.

Q6(c) then adds one more entry to the list: A is a product of elementary matrices. It comes free from the reduction, since inverting E₃E₂E₁A = I in the other direction exhibits A as exactly such a product, and nothing in that argument depended on the particular matrix. A square matrix reduces to I precisely when it can be built out of row operations, and those are two readings of one calculation.

Getting the most out of it

Count pivots before you compute anything

Almost every question here is settled by the downward half of a row reduction. Get the matrix to echelon form, count the leading entries, and only then decide whether an inverse is worth chasing. Starting with the arithmetic of [A | I] and discovering a zero row three operations later is the expensive route to the same answer.

Apply every operation across the full row

In an augmented array the bar is punctuation, not a wall. A row operation acts on the whole row, both blocks, and a reduction that quietly leaves the right-hand block behind still looks tidy and produces a confident wrong answer. This is the one error the check in Q2 is designed to catch.

Say which side you multiplied on

Matrix multiplication does not commute, so "multiply by A⁻¹" is not an instruction until you have said left or right. Write x = A⁻¹b, not bA⁻¹; write (AB)⁻¹ = B⁻¹A⁻¹, not A⁻¹B⁻¹. On a CEGEP final the justification clause — which inverse, applied on which side, and why the equation forces it — is where the reasoning marks live.

Check by multiplying back

An inverse is the one answer in the course that verifies itself in a few seconds: multiply it back into the original and look for the identity. Q1, Q2, Q4 and Q6 all ask for some form of this, and one column is usually enough to catch a slip. Do it even when the question does not ask.

Common questions