Exercise 4.4 · Q14: Find $a$ and $b$ with $A^2+aA+bI=O$

Basics you need

Solution

1. Compute $A^2$

   $A$ is $2\times2$⁠, so $A\cdot A$ is defined and is $2\times2$⁠.

   $$\begin{aligned} c_{11} &= 3\cdot3 + 2\cdot1 = 9 + 2 = 11\\ c_{12} &= 3\cdot2 + 2\cdot1 = 6 + 2 = 8\\ c_{21} &= 1\cdot3 + 1\cdot1 = 3 + 1 = 4\\ c_{22} &= 1\cdot2 + 1\cdot1 = 2 + 1 = 3 \end{aligned}$$ $$A^2 = \begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix}$$
   $c_{12}$ is row 1 of $A$ ($3, 2$⁠) times column 2 of $A$ ($2, 1$⁠). Every entry of $A^2$ is needed because the equation must hold in every position.
   Pair up, multiply, add. Basics: matrix facts the chapter uses

2. Write $A^2 + aA + bI$ as one matrix

   $$aA = \begin{bmatrix} 3a & 2a \\ a & a \end{bmatrix}, \quad bI = \begin{bmatrix} b & 0 \\ 0 & b \end{bmatrix}$$

   All three are $2\times2$⁠, so they can be added entry by entry:

   $$\begin{aligned} &A^2 + aA + bI\\ &= \begin{bmatrix} 11 + 3a + b & 8 + 2a \\ 4 + a & 3 + a + b \end{bmatrix} \end{aligned}$$
   $a$ and $b$ are numbers (scalars), so $aA$ and $bI$ are scalar multiples. The term $bI$ adds $b$ only on the diagonal, which is why $b$ appears in entries $(1,1)$ and $(2,2)$ but not in the other two.

3. Set every entry to zero and solve

   The matrix equals $O$ exactly when all four entries are $0$⁠:

   $$\begin{aligned} (1,2)&:\quad 8 + 2a = 0\\ (2,1)&:\quad 4 + a = 0\\ (1,1)&:\quad 11 + 3a + b = 0\\ (2,2)&:\quad 3 + a + b = 0 \end{aligned}$$

   The $(2,1)$ equation gives $a = -4$⁠. The $(1,2)$ equation gives $2a = -8$⁠, so $a = -4$ as well: the two agree.

   Put $a = -4$ into the $(2,2)$ equation: $3 - 4 + b = 0$⁠, so $b = 1$⁠. Put it into the $(1,1)$ equation: $11 - 12 + b = 0$⁠, so $b = 1$ again.

   All four equations hold with $a = -4$ and $b = 1$, and with no other values, because the $(2,1)$ equation forces $a = -4$ and then the $(2,2)$ equation forces $b = 1$⁠.

   Four equations in two unknowns could easily contradict each other, and then no such $a, b$ would exist. So it is not enough to solve two of them: we must check that the solution satisfies the other two as well. Starting with the off-diagonal entries is smart because $b$ does not appear there.
   Equal matrices give one equation per entry. Basics: rows, columns and positions · Basics: algebra

4. Check: $A^2 - 4A + I = O$

   $$4A = \begin{bmatrix} 12 & 8 \\ 4 & 4 \end{bmatrix}$$ $$\begin{aligned} (1,1)&:\ 11 - 12 + 1 = 0\\ (1,2)&:\ 8 - 8 + 0 = 0\\ (2,1)&:\ 4 - 4 + 0 = 0\\ (2,2)&:\ 3 - 4 + 1 = 0 \end{aligned}$$

   So $A^2 - 4A + I = O$ ✓.

   Substituting the answer into the original matrix equation is the final test that nothing was mis-copied between Steps 1–3.

5. Why $a = -(3+1)$ and $b = |A|$

   Notice $-4 = -(3 + 1)$⁠, minus the sum of the diagonal entries of $A$⁠, and $1 = 3\cdot1 - 2\cdot1 = |A|$⁠. This is no accident. For any $2\times2$ matrix $M$ with entries $p, q$ (row 1) and $r, s$ (row 2):

   $$M^2 = \begin{bmatrix} p^2 + qr & pq + qs \\ rp + sr & rq + s^2 \end{bmatrix}$$

   and $M^2 - (p+s)M + (ps - qr)I$ has entries

   $$\begin{aligned} (1,1)&:\ p^2 + qr - p^2 - sp + ps - qr = 0\\ (1,2)&:\ pq + qs - pq - sq = 0\\ (2,1)&:\ rp + sr - pr - sr = 0\\ (2,2)&:\ rq + s^2 - ps - s^2 + ps - qr = 0 \end{aligned}$$

   So every $2\times2$ matrix satisfies $M^2 - (p+s)M + |M|\,I = O$⁠. For our $A$ the diagonal sum is $3 + 1 = 4$ and $|A| = 1$⁠, so $a = -4$ and $b = 1$ again.

   The general calculation explains the answer and gives a quick way to check it: minus the diagonal sum, and the determinant. It also explains Example 15 and Exercise 13 of this set: there $2 + 2 = 4$⁠, $|A| = 1$ and $3 + 2 = 5$⁠, $|A| = 7$⁠, matching $A^2 - 4A + I$ and $A^2 - 5A + 7I$⁠.
   $|M| = ps - qr$⁠. Lesson: determinant of a square matrix

Every fact used, and where it comes from