Chapter 3 · Matrices

Exercise 3.2 · Q17: Find $k$ with $A^2=kA-2I$

Exercise 3.2Multiplication of matrices · Section 3.4

The question

A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}, \quad I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},

find $k$ so that $A^2 = kA - 2I$ ⁠.

Exam answer

\begin{aligned} A^2 &= \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \\ &= \begin{bmatrix} 9-8 & -6+4 \\ 12-8 & -8+4 \end{bmatrix} \\ &= \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} \end{aligned}

\begin{aligned} kA - 2I &= \begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \\ &= \begin{bmatrix} 3k - 2 & -2k \\ 4k & -2k - 2 \end{bmatrix} \end{aligned}

Since $A^2 = kA - 2I$ ⁠, equating corresponding entries,

\begin{aligned} 3k - 2 = 1 &\Rightarrow k = 1 \\ -2k = -2 &\Rightarrow k = 1 \\ 4k = 4 &\Rightarrow k = 1 \\ -2k - 2 = -4 &\Rightarrow k = 1 \end{aligned}

$\therefore\ k = 1$ ⁠.

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

The one idea this problem is built on

A matrix equation between $2\times2$ matrices is really four ordinary equations, one for each entry. Compute both sides as matrices, set corresponding entries equal, and solve. Because there is one unknown but four equations, a value of $k$ exists only if all four equations give the same $k$ ⁠; checking that is part of the answer.

Given, to find, and the plan

Given: The $2\times2$ matrix $A$ and the identity $I$ of order 2.
To find: The number $k$ for which $A^2 = kA - 2I$ ⁠.
Plan: (1) Check orders. (2) Compute $A^2$ entry by entry. (3) Write $kA - 2I$ entry by entry in terms of $k$ ⁠. (4) Equate the four pairs of entries and solve each. (5) Confirm that one $k$ satisfies all four, and substitute back.

Basics you need

1Row times column

The $(i,k)$ entry of $PQ$ is row $i$ of $P$ times column $k$ of $Q$ ⁠: multiply matching entries and add. Basics: pair up, multiply, add

2Scalar multiples and differences

$kA$ multiplies every entry by $k$ ⁠; $P - Q$ subtracts entry by entry (same order needed). Basics: adding only like with like

3Equal matrices, equal entries

Two matrices of the same order are equal exactly when every pair of corresponding entries is equal. Lesson: equality of matrices

4Solving a linear equation

Collect the unknown on one side, then divide by its coefficient. Basics: solving equations

Solution

Check the orders

$A$ is $2\times2$ ⁠, so $A^2 = AA$ is $(2\times2)(2\times2)$ ⁠, defined, of order $2\times2$ ⁠. $kA$ and $2I$ are both $2\times2$ ⁠, so $kA - 2I$ is defined and $2\times2$ ⁠. Both sides of the equation are $2\times2$ ⁠, so comparing them makes sense.

Equality of matrices first requires the same order; only then can entries be compared one to one.
Compute $A^2$
$A^2 = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$
The rows of $A$ are $(3, -2)$ and $(4, -2)$ ⁠; its columns are $(3, 4)$ and $(-2, -2)$ read downwards.
$\begin{aligned} c_{11} &= 3\cdot3 + (-2)\cdot4\\ &= 9 - 8 = 1\\ c_{12} &= 3\cdot(-2) + (-2)(-2)\\ &= -6 + 4 = -2\\ c_{21} &= 4\cdot3 + (-2)\cdot4\\ &= 12 - 8 = 4\\ c_{22} &= 4\cdot(-2) + (-2)(-2)\\ &= -8 + 4 = -4 \end{aligned}$ $A^2 = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix}$
Watch the signs: $(-2)(-2) = +4$ ⁠. This is where most slips in this question happen.

A negative times a negative is positive. Basics: pair up, multiply, add
Write $kA - 2I$ entry by entry
$kA = \begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix}, \quad 2I = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$
Subtracting entry by entry:
$\begin{aligned} (1,1)&: 3k - 2\\ (1,2)&: -2k - 0 = -2k\\ (2,1)&: 4k - 0 = 4k\\ (2,2)&: -2k - 2 \end{aligned}$ $kA - 2I = \begin{bmatrix} 3k - 2 & -2k \\ 4k & -2k - 2 \end{bmatrix}$
$2I$ has 2 only on the diagonal, so it changes only the $(1,1)$ and $(2,2)$ entries. The off-diagonal entries of $kA - 2I$ are just those of $kA$ ⁠.

$kA = [k\,a_{ij}]$ and $P - Q = [p_{ij} - q_{ij}]$ ⁠. Lesson: addition and scalar multiplication
Equate corresponding entries and solve each

$A^2 = kA - 2I$ means
$\begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3k - 2 & -2k \\ 4k & -2k - 2 \end{bmatrix}$
so the four entry equations are:
$\begin{aligned} (1,1)&: 3k - 2 = 1\\ &\;\Rightarrow 3k = 3 \Rightarrow k = 1\\ (1,2)&: -2k = -2 \Rightarrow k = 1\\ (2,1)&: 4k = 4 \Rightarrow k = 1\\ (2,2)&: -2k - 2 = -4\\ &\;\Rightarrow -2k = -2 \Rightarrow k = 1 \end{aligned}$
Each equation is linear in $k$ with a non-zero coefficient of $k$ ⁠, so each has exactly one solution, and all four give $k = 1$ ⁠. Hence $k = 1$ , and it is the only value that works.

Any one equation would already suggest $k=1$ ⁠, but the matrix equation demands all four at once. If two entries had given different values of $k$ ⁠, no $k$ would exist at all. Here they agree, so $k=1$ is a genuine solution; and since the $(1,2)$ equation alone forces $k=1$ ⁠, no other value can work.

Divide both sides by the coefficient of $k$ ⁠: from $3k = 3$ we get $k = 1$ ⁠. Basics: one unknown at a time
Substitute back

With $k = 1$ ⁠:
$A - 2I = \begin{bmatrix} 3 - 2 & -2 - 0 \\ 4 - 0 & -2 - 2 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix}$
which is exactly $A^2$ from Step 2. So $A^2 = A - 2I$ ⁠.

Substituting the answer into the original equation catches any arithmetic slip in Steps 2 to 4.

Result

$k = 1$ , that is, $A^2 = A - 2I$ ⁠. (Textbook answer: $k = 1$ ⁠.)

Check it with numbers

A test on a column vector: take $v$ with entries $1, 0$ ⁠. Then $Av$ is the first column of $A$ ⁠: $3, 4$ ⁠. Applying $A$ again: $3\cdot3 + (-2)\cdot4 = 1$ and $4\cdot3 + (-2)\cdot4 = 4$ ⁠, so $A^2v$ has entries $1, 4$ ⁠. On the other side, $Av - 2v$ has entries $3 - 2 = 1$ and $4 - 0 = 4$ ✓. With $v$ having entries $0, 1$ ⁠: $Av$ is $-2, -2$ ⁠; $A(Av)$ is $3(-2) + (-2)(-2) = -2$ and $4(-2) + (-2)(-2) = -4$ ⁠; and $Av - 2v$ is $-2 - 0 = -2$ ⁠, $-2 - 2 = -4$ ✓.

Common mistakes

Squaring the entries to get $A^2$ with rows $(9, 4)$ ⁠, $(16, 4)$ ⁠. That is not matrix multiplication.
Treating $-2I$ as $-2$ subtracted from every entry. It changes only the diagonal; the off-diagonal entries of $kA - 2I$ are $-2k$ and $4k$ ⁠.
Using only one entry. One entry suggests a value; all four must agree for that value to be an answer.
Sign slip in $(-2)(-2)$ ⁠, which gives $+4$ ⁠.

What to take away

A matrix equation of order $m\times n$ is $mn$ scalar equations, one per entry.
With fewer unknowns than entries, the system is overdetermined: a solution exists only if every entry equation agrees, so check them all.
$A^2 = A - 2I$ is another polynomial identity of a square matrix, like the one in Exercise 3.2 Q16; it lets you replace $A^2$ by $A - 2I$ in any longer calculation.

Every fact used, and where it comes from

Fact	Used in	Source
Product defined when columns of the first = rows of the second; entry = row times column	Steps 1, 2	Section 3.4.5 (Lesson: multiplication)
Scalar multiple and difference of matrices	Steps 3, 5	Section 3.4.2
Identity matrix $I$ of order 2	Step 3	Section 3.3 (vi)
Equal matrices have equal corresponding entries	Step 4	Definition 2, Section 3.3.1
All products and the value $k = 1$	Steps 2–5, Check	Recomputed in the site's verification script

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Exercise 3.2 · Q17: Find kkk with A2=kA−2IA^2=kA-2IA2=kA−2I

Basics you need

Solution

Check the orders

Compute A2A^2A2

Write kA−2IkA - 2IkA−2I entry by entry

Equate corresponding entries and solve each

Substitute back

Every fact used, and where it comes from

Exercise 3.2 · Q17: Find $k$ with $A^2=kA-2I$

Compute $A^2$

Write $kA - 2I$ entry by entry