Practice: Matrix Inverses and Elementary Matrices

Direct application · Error diagnosis

A learner proposes $B = (\begin{matrix} 3 & - 1 \\ - 5 & 2 \end{matrix})$ as the inverse of $A = (\begin{matrix} 2 & 1 \\ 5 & 3 \end{matrix})$ .

Compute the entry in row 1, column 1 of the product $A B$ .

Your answer

Enter the value. It is checked against the answer and the precision this task asks for.

2 hints available, least help first.

Hint 1: Retrieval cue

Take row 1 of $A$ against column 1 of $B$ .

Hint 2: Concept cue

$2 \times 3 + 1 \times (- 5)$ .

Direct application

Let $A = (\begin{matrix} 2 & 1 \\ 5 & 3 \end{matrix})$ .

Compute $A^{- 1}$ and give the entry in row 1, column 1.

Your answer

Enter the value. It is checked against the answer and the precision this task asks for.

2 hints available, least help first.

Hint 1: Retrieval cue

Compute $det A = a d - b c$ first.

Hint 2: Concept cue

The $(1, 1)$ entry of the inverse is $d / det A$ .

Direct application

In a $2 \times 2$ system, the operation $R_{2} \to R_{2} - 5 R_{1}$ is performed by left multiplication by an elementary matrix $E$ , obtained by applying that operation to $I$ .

What is the entry of $E$ in row 2, column 1?

Your answer

Enter the value. It is checked against the answer and the precision this task asks for.

2 hints available, least help first.

Hint 1: Retrieval cue

Apply the operation to $I$ rather than to any other matrix.

Hint 2: Concept cue

What does row 2 of $I$ become after subtracting 5 times row 1?

Recognition · Error diagnosis

With $A = (\begin{matrix} 2 & 1 \\ 5 & 3 \end{matrix})$ and $B = (\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix})$ , a learner computes $A^{- 1} B^{- 1} = (\begin{matrix} 3 & - 7 \\ - 5 & 12 \end{matrix})$ and reports it as $(A B)^{- 1}$ . Is that right?

2 hints available, least help first.

Hint 1: Retrieval cue

Multiply $A B$ by each candidate and see which gives $I$ .

Hint 2: Concept cue

In $(A B) (B^{- 1} A^{- 1})$ , which two factors are adjacent in the middle?

Direct application

Factor $C = (\begin{matrix} 2 & 1 & 1 \\ 4 & - 6 & 0 \\ - 2 & 7 & 2 \end{matrix})$ as $L U$ by downward elimination.

The first step clears position $(2, 1)$ using $R_{2} \to R_{2} - 2 R_{1}$ . What is the entry $L_{21}$ ?

Your answer

Enter the value. It is checked against the answer and the precision this task asks for.

2 hints available, least help first.

Hint 1: Retrieval cue

What number multiplied row 1 before it was subtracted?

Hint 2: Concept cue

Check your answer by forming row 2 of $L U$ : it must give $(4, - 6, 0)$ .

Construction · Direct application · Explanation

(a) Compute $A^{- 1}$ for $A = (\begin{matrix} 1 & 2 \\ 3 & 7 \end{matrix})$ by reducing $[A ∣ I]$ , recording each row operation. Verify your answer.

(b) Write each of your row operations as an elementary matrix and confirm that their product, in the correct order, equals $A^{- 1}$ .

(c) Let $B = (\begin{matrix} 1 & 0 \\ 4 & 1 \end{matrix})$ . Compute $(A B)^{- 1}$ two ways: directly from $A B$ , and from $A^{- 1}$ and $B^{- 1}$ . State the rule your second route uses.

(d) Attempt to invert $S = (\begin{matrix} 2 & 6 \\ 1 & 3 \end{matrix})$ by the same reduction. Say what happens and what it proves, listing three other properties of $S$ that follow.

(e) Find the $L U$ factorisation of $A$ from part (a), and use it to compute $det A$ . Say why $L$ holds the multiplier with the sign it does.

Write your answer

Write your answer, then compare it with the worked solution.

3 hints available, least help first.

Hint 1: Retrieval cue

Record each row operation as you perform it; parts (b) and (e) both depend on having them written down.

Hint 2: Concept cue

An elementary matrix is the identity with the operation applied to it. The first operation performed stands rightmost in the product.

Hint 3: Strategy cue

In part (d), do not restart when a zero row appears, ask what a zero row on the left establishes about the rows of $S$ .

Compare with the worked solution

Comparing does not record a result. Judging your own written answer cannot show that you can do this without help.

(a) The inverse. Start from $[A ∣ I]$ :

[\begin{array}{cccc} 1 & 2 & 1 & 0 \\ 3 & 7 & 0 & 1 \end{array}] \overset{R_{2} \to R_{2} - 3 R_{1}}{\to} [\begin{array}{cccc} 1 & 2 & 1 & 0 \\ 0 & 1 & - 3 & 1 \end{array}] \overset{R_{1} \to R_{1} - 2 R_{2}}{\to} [\begin{array}{cccc} 1 & 0 & 7 & - 2 \\ 0 & 1 & - 3 & 1 \end{array}]

So $A^{- 1} = (\begin{matrix} 7 & - 2 \\ - 3 & 1 \end{matrix})$ . Verification. $A A^{- 1} = (\begin{matrix} 7 - 6 & - 2 + 2 \\ 21 - 21 & - 6 + 7 \end{matrix}) = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})$ ✓. Cross-checking by formula: $det A = 7 - 6 = 1$ , so the inverse is $\frac{1}{1} (\begin{matrix} 7 & - 2 \\ - 3 & 1 \end{matrix})$ ✓, and the determinant being 1 explains the integer entries. (b) As elementary matrices. The two operations were $R_{2} \to R_{2} - 3 R_{1}$ and then $R_{1} \to R_{1} - 2 R_{2}$ :

E_{1} = (\begin{matrix} 1 & 0 \\ - 3 & 1 \end{matrix}), E_{2} = (\begin{matrix} 1 & - 2 \\ 0 & 1 \end{matrix}) .

Since $E_{2} E_{1} A = I$ , the product $E_{2} E_{1}$ is $A^{- 1}$ . Multiplying, with $E_{1}$ applied first and therefore rightmost:

E_{2} E_{1} = (\begin{matrix} 1 & - 2 \\ 0 & 1 \end{matrix}) (\begin{matrix} 1 & 0 \\ - 3 & 1 \end{matrix}) = (\begin{matrix} 1 + 6 & - 2 \\ - 3 & 1 \end{matrix}) = (\begin{matrix} 7 & - 2 \\ - 3 & 1 \end{matrix}) = A^{- 1} ✓

Reversing the order would give $E_{1} E_{2} = (\begin{matrix} 1 & - 2 \\ - 3 & 7 \end{matrix})$ , which is not the inverse. The order records which operation came first. (c) The inverse of a product.

A B = (\begin{matrix} 1 & 2 \\ 3 & 7 \end{matrix}) (\begin{matrix} 1 & 0 \\ 4 & 1 \end{matrix}) = (\begin{matrix} 9 & 2 \\ 31 & 7 \end{matrix}) .

Directly. $det (A B) = 63 - 62 = 1$ , so $(A B)^{- 1} = (\begin{matrix} 7 & - 2 \\ - 31 & 9 \end{matrix})$ . From the factors. $B$ is elementary, it adds $4 \times$ row 1 to row 2, so $B^{- 1} = (\begin{matrix} 1 & 0 \\ - 4 & 1 \end{matrix})$ . Then

B^{- 1} A^{- 1} = (\begin{matrix} 1 & 0 \\ - 4 & 1 \end{matrix}) (\begin{matrix} 7 & - 2 \\ - 3 & 1 \end{matrix}) = (\begin{matrix} 7 & - 2 \\ - 28 - 3 & 8 + 1 \end{matrix}) = (\begin{matrix} 7 & - 2 \\ - 31 & 9 \end{matrix}) ✓

The rule is $(A B)^{- 1} = B^{- 1} A^{- 1}$ : the order reverses. Checking the wrong order, $A^{- 1} B^{- 1} = (\begin{matrix} 15 & - 2 \\ - 7 & 1 \end{matrix})$ , which is not the inverse of $A B$ . (d) A matrix with no inverse.

[\begin{array}{cccc} 2 & 6 & 1 & 0 \\ 1 & 3 & 0 & 1 \end{array}] \overset{R_{1} \to \frac{1}{2} R_{1}}{\to} [\begin{array}{cccc} 1 & 3 & \frac{1}{2} & 0 \\ 1 & 3 & 0 & 1 \end{array}] \overset{R_{2} \to R_{2} - R_{1}}{\to} [\begin{array}{cccc} 1 & 3 & \frac{1}{2} & 0 \\ 0 & 0 & - \frac{1}{2} & 1 \end{array}]

The left block has a zero row, so it can never be reduced to $I$ . The reduction has not failed; it has proved that $S$ has no inverse. The rows were dependent from the start: row 1 is twice row 2. Three consequences, each an entry of the equivalence list: $det S = 6 - 6 = 0$ ; $rank S = 1 < 2$ , so the columns are dependent, as $(6, 3) = 3 (2, 1)$ shows; and the kernel is nontrivial, since $S (3, - 1)^{T} = (6 - 6, 3 - 3)^{T} = 0$ , which also says $0$ is an eigenvalue. (e) $L U$ for $A$ . Only one elimination step is needed, $R_{2} \to R_{2} - 3 R_{1}$ , with multiplier 3:

L = (\begin{matrix} 1 & 0 \\ 3 & 1 \end{matrix}), U = (\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}) .

Check. $L U = (\begin{matrix} 1 & 2 \\ 3 & 6 + 1 \end{matrix}) = (\begin{matrix} 1 & 2 \\ 3 & 7 \end{matrix}) = A$ ✓. $det A = det L \cdot det U = 1 \times (1) (1) = 1$ , agreeing with part (a). Why the sign. $L$ is built so that $L U = A$ . It rebuilds $A$ from the reduced form, undoing the elimination. The step subtracted $3 R_{1}$ from row 2, so restoring row 2 requires adding $3 R_{1}$ back, and $L_{21} = + 3$ . The elementary matrix $E_{1} = (\begin{matrix} 1 & 0 \\ - 3 & 1 \end{matrix})$ carries $- 3$ because it performs the elimination; $L = E_{1}^{- 1}$ carries $+ 3$ because it reverses it. The two signs are consistent, and confusing them is caught immediately by multiplying $L U$ out.

A complete answer does each of these:

computes the inverse
verifies the inverse
identifies elementary matrices
applies order reversal
produces lu factorisation
reads a stalled reduction

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Practice: Matrix Inverses and Elementary Matrices

Direct application · Error diagnosis

Direct application

Direct application

Recognition · Error diagnosis

Direct application

Construction · Direct application · Explanation

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