Chapter 4 · Determinants

Miscellaneous · Q8 (MCQ): Inverse of a diagonal matrix

Miscellaneous ExerciseInverse of a matrix · Section 4.5

The question

If $x, y, z$ are nonzero real numbers, then the inverse of matrix

A = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}

(A)

\begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix}

(B)

xyz\begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix}

(C)

\frac{1}{xyz}\begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}

(D)

\frac{1}{xyz}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

(The book introduces Questions 8 and 9 with "Choose the correct answer in Exercise 17 to 19." The numbers are left over from an older edition; the line applies to Questions 8 and 9.)

Exam answer

Expanding along $R_1$ ⁠:

|A| = x\begin{vmatrix} y & 0 \\ 0 & z \end{vmatrix} - 0 + 0 = xyz \ne 0

The cofactors are $A_{11} = yz$ ⁠, $A_{22} = xz$ ⁠, $A_{33} = xy$ ⁠, and all others are 0. So

\begin{aligned} A^{-1} &= \frac{1}{|A|}\operatorname{adj}A \\ &= \frac{1}{xyz}\begin{bmatrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{bmatrix} \\ &= \begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix} \end{aligned}

Answer: (A) $\begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix}$

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

A diagonal matrix only stretches each coordinate separately: $x$ times the first, $y$ times the second, $z$ times the third. To undo it, divide each coordinate by the same number, so the inverse is the diagonal matrix with entries $\frac1x$ ⁠, $\frac1y$ ⁠, $\frac1z$ ⁠. The adjoint method confirms this: $|A| = xyz$ and $\operatorname{adj}A$ has diagonal $yz, xz, xy$ ⁠, and dividing gives $\frac1x, \frac1y, \frac1z$ ⁠.

Given, to find, and the plan

Given: The $3\times3$ diagonal matrix A with diagonal $x, y, z$ ⁠, all non-zero.
To find: $A^{-1}$ ⁠, and the matching option.
Plan: (1) $|A|$ ⁠. (2) All nine cofactors and adj A. (3) $A^{-1} = \frac{1}{|A|}\operatorname{adj}A$ ⁠. (4) Check $AA^{-1} = I$ ⁠. (5) Examine every option.

Basics you need

1Determinant of a matrix with many zeros

Expand along a row with zeros: only the non-zero entries contribute. Lesson: expanding a 3 × 3 determinant

2Cofactors and adjoint

$A_{ij} = (-1)^{i+j}M_{ij}$ ⁠; adj A is the transpose of the cofactor matrix. Basics: the + − + sign pattern

3Inverse

If $|A| \ne 0$ ⁠, $A^{-1} = \frac{1}{|A|}\operatorname{adj}A$ ⁠, and it is the only matrix with $AA^{-1} = A^{-1}A = I$ ⁠. Basics: why the inverse divides by |A|

4Negative powers

$x^{-1}$ means $\frac1x$ ⁠, defined because $x \ne 0$ ⁠. Basics: algebra the chapter leans on

Solution

The determinant of A

Expand along $R_1 = (x, 0, 0)$ ⁠: only the first entry contributes.
$|A| = x\begin{vmatrix} y & 0 \\ 0 & z \end{vmatrix} - 0 + 0 = x(yz - 0) = xyz.$
Since $x$ ⁠, $y$ ⁠, $z$ are all non-zero, $|A| = xyz \ne 0$ ⁠, so A is invertible.

The two zeros in $R_1$ remove two terms (Remark (i) of Section 4.2.3). A product of non-zero real numbers is non-zero, which is why the question insists that $x$ ⁠, $y$ ⁠, $z$ are nonzero: if any were 0, $|A| = 0$ and A would have no inverse.

Section 4.5, Theorem 4: invertible ⇔ non-singular. Lesson: inverse of a matrix
All nine cofactors, and adj A
$\begin{aligned} A_{11} &= +\begin{vmatrix} y & 0 \\ 0 & z \end{vmatrix} = yz \\ A_{12} &= -\begin{vmatrix} 0 & 0 \\ 0 & z \end{vmatrix} = 0 \\ A_{13} &= +\begin{vmatrix} 0 & y \\ 0 & 0 \end{vmatrix} = 0 \\ A_{21} &= -\begin{vmatrix} 0 & 0 \\ 0 & z \end{vmatrix} = 0 \\ A_{22} &= +\begin{vmatrix} x & 0 \\ 0 & z \end{vmatrix} = xz \\ A_{23} &= -\begin{vmatrix} x & 0 \\ 0 & 0 \end{vmatrix} = 0 \\ A_{31} &= +\begin{vmatrix} 0 & 0 \\ y & 0 \end{vmatrix} = 0 \\ A_{32} &= -\begin{vmatrix} x & 0 \\ 0 & 0 \end{vmatrix} = 0 \\ A_{33} &= +\begin{vmatrix} x & 0 \\ 0 & y \end{vmatrix} = xy \end{aligned}$
The cofactor matrix is diagonal, so it equals its transpose:
$\operatorname{adj}A = \begin{bmatrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{bmatrix}.$
Each off-diagonal minor contains a whole row or column of zeros (for example $M_{12}$ has first column $(0, 0)$ ⁠), so it is 0 and the sign does not matter. Each diagonal minor is the product of the other two diagonal entries: $M_{11} = yz$ ⁠, $M_{22} = xz$ ⁠, $M_{33} = xy$ ⁠.

Section 4.4, Definitions 1, 2; Section 4.5.1, Definition 3. Lesson: the adjoint
$A^{-1} = \frac{1}{|A|}\operatorname{adj}A$
$A^{-1} = \frac{1}{xyz}\begin{bmatrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{bmatrix} = \begin{bmatrix} \frac1x & 0 & 0 \\ 0 & \frac1y & 0 \\ 0 & 0 & \frac1z \end{bmatrix}.$
In the notation of the options, this is the matrix with diagonal $x^{-1}, y^{-1}, z^{-1}$ ⁠.

Each diagonal entry simplifies: $\frac{yz}{xyz} = \frac1x$ ⁠, $\frac{xz}{xyz} = \frac1y$ ⁠, $\frac{xy}{xyz} = \frac1z$ ⁠. The cancellations are allowed because $x, y, z \ne 0$ ⁠.

Section 4.5, Theorem 4. Basics: why the inverse divides by |A|
Check $AA^{-1} = I$

Rows of A: $(x, 0, 0)$ ⁠, $(0, y, 0)$ ⁠, $(0, 0, z)$ ⁠. Columns of $A^{-1}$ ⁠: $(\frac1x, 0, 0)$ ⁠, $(0, \frac1y, 0)$ ⁠, $(0, 0, \frac1z)$ ⁠. Each diagonal entry is $x\cdot\frac1x = 1$ ⁠, $y\cdot\frac1y = 1$ ⁠, $z\cdot\frac1z = 1$ ⁠; each off-diagonal entry is a sum of three products each containing a 0, so it is 0. Hence $AA^{-1} = I$ ✓, and in the same way $A^{-1}A = I$ ⁠.

The definition of the inverse requires both products to be I. For diagonal matrices the product is again diagonal, with the diagonal entries multiplied in pairs, so this check is immediate.

Product entry formula. Basics: matrix facts the chapter uses

Examine every option

Multiply A by each option and see whether the product is I. Take $x = 1$ ⁠, $y = 2$ ⁠, $z = 3$ (so $xyz = 6$ ⁠) as a test case.

Option	A times the option	Verdict
(A)	Diagonal $x\cdot x^{-1}, y\cdot y^{-1}, z\cdot z^{-1}$ ⁠, that is $1, 1, 1$ ⁠: the product is I	Correct for all non-zero $x, y, z$ (Steps 3, 4).
(B)	(B) is $xyz\,A^{-1} = \operatorname{adj}A$ ⁠, so the product is $xyz\,I$ ⁠; for the test case, $6I$	Wrong: it is I only when $xyz = 1$ ⁠. (B) is adj A, not $A^{-1}$ ⁠.
(C)	$\frac{1}{xyz}A^2$ ⁠, diagonal $\frac{x^2}{xyz}, \frac{y^2}{xyz}, \frac{z^2}{xyz}$ ⁠; for the test case $\frac16, \frac46, \frac96$	Wrong: not I in general.
(D)	$\frac{1}{xyz}A$ ⁠, diagonal $\frac{1}{yz}, \frac{1}{xz}, \frac{1}{xy}$ ⁠; for the test case $\frac16, \frac13, \frac12$	Wrong: not I in general.

Since the inverse is unique, at most one option can equal

A^{-1}

for every choice of

x, y, z

⁠, and (A) does. A single test case is enough to rule out (B), (C) and (D) as general answers. Note that with

x = y = z = 1

⁠, A is I and all four options reduce to I, which is why that choice would not separate them.

Uniqueness of the inverse (Chapter 3, Section 3.7, Theorem 3). Lesson: inverse of a matrix

Result

The correct answer is (A): $|A| = xyz$ ⁠, $\operatorname{adj}A$ has diagonal $yz, xz, xy$ ⁠, and so $A^{-1}$ is the diagonal matrix with entries $x^{-1}, y^{-1}, z^{-1}$ ⁠. (Textbook answer: A.)

Check it with numbers

For $x = 1$ ⁠, $y = 2$ ⁠, $z = 3$ ⁠: A has diagonal $1, 2, 3$ and option (A) has diagonal $1, \frac12, \frac13$ ⁠. The product has diagonal $1\cdot1 = 1$ ⁠, $2\cdot\frac12 = 1$ ⁠, $3\cdot\frac13 = 1$ ⁠: it is I ✓. With negative entries, $x = -2$ ⁠, $y = 5$ ⁠, $z = -\frac12$ ⁠: option (A) has diagonal $-\frac12, \frac15, -2$ ⁠, and $(-2)(-\frac12) = 1$ ⁠, $5\cdot\frac15 = 1$ ⁠, $(-\frac12)(-2) = 1$ ✓.

Common mistakes

Choosing (B) because it contains $xyz$ ⁠: (B) is the adjoint, which still has to be divided by $|A| = xyz$ ⁠.
Inverting the whole matrix entry by entry, including the zeros. Only the diagonal entries are inverted, and that works only because A is diagonal; for a general matrix, $A^{-1}$ is not the matrix of reciprocals.
Testing with $x = y = z = 1$ , which makes every option look right.

What to take away

The inverse of a diagonal matrix with non-zero diagonal entries is the diagonal matrix of their reciprocals.
For a diagonal matrix, $|A|$ is the product of the diagonal entries, and adj A is diagonal with "the product of the other entries" in each place.
To check an MCQ about an inverse, multiply A by each option and look for I, using test values that are not all equal to 1.

Every fact used, and where it comes from

Fact	Used in	Source
Expansion along a row with zeros	Step 1	Section 4.2.3, Remark (i) (Lesson: expanding a 3 × 3 determinant)
Minors, cofactors, adjoint	Step 2	Section 4.4, Definitions 1, 2; Section 4.5.1, Definition 3 (Lesson: the adjoint)
Invertible ⇔ non-singular; $A^{-1} = \frac{1}{\|A\|}\operatorname{adj}A$	Steps 1, 3	Section 4.5, Theorem 4 (Lesson: inverse of a matrix)
The inverse is unique	Step 5	Chapter 3, Section 3.7, Theorem 3
Every option's product with A, symbolically and for the test values	Steps 3–5, Check	Recomputed with sympy in the site's verification script

Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →

Miscellaneous · Q8 (MCQ): Inverse of a diagonal matrix

Basics you need

Solution

The determinant of A

All nine cofactors, and adj A

A−1=1∣A∣adj⁡AA^{-1} = \frac{1}{|A|}\operatorname{adj}AA−1=∣A∣1​adjA

Check AA−1=IAA^{-1} = IAA−1=I

Examine every option

Every fact used, and where it comes from

$A^{-1} = \frac{1}{|A|}\operatorname{adj}A$

Check $AA^{-1} = I$