Let α and β Be Real Numbers and Consider a 3×3 Matrix A Such That A² = 3A + αI. If A⁴ = 21A + βI, Find the Values of α and β

A standard matrix algebra problem, not tied to a specific exam paper.

Problem

Let $\alpha$ and $\beta$ be real numbers. Consider a $3\times3$ matrix $A$ such that $A^2=3A+\alpha I$. If $A^4=21A+\beta I$, find the values of $\alpha$ and $\beta$.

Key insight. $A^4$ is just $(A^2)^2$ — squaring the given expression for $A^2$, and using the fact that $A$ and $I$ commute with everything, turns $A^4$ into another expression purely in terms of $A$ and $I$, which can then be compared directly against $21A+\beta I$.

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Approach

Since $A^4=(A^2)^2$, substitute the given expression $A^2=3A+\alpha I$ and expand the square. Because $A$ commutes with $I$ (and with itself), this expansion behaves exactly like squaring an ordinary binomial. The result is an expression of the form $(\text{something})A + (\text{something})I$; matching its coefficients against $21A+\beta I$ gives two ordinary equations — one for $\alpha$, then one for $\beta$.

Solution

Step 1 — Write A⁴ as (A²)²

$$A^4 = (A^2)^2 = (3A+\alpha I)^2$$

Step 2 — Expand the square

Since matrices commute with $I$ and with themselves, this expands just like $(3x+\alpha)^2$:

$$(3A+\alpha I)^2 = 9A^2+6\alpha A+\alpha^2 I$$

Step 3 — Substitute A² again

The term $9A^2$ still contains $A^2$, so substitute $A^2=3A+\alpha I$ once more:

$$9A^2 = 9(3A+\alpha I) = 27A+9\alpha I$$

So:

$$A^4 = 27A+9\alpha I+6\alpha A+\alpha^2 I = (27+6\alpha)A + (9\alpha+\alpha^2)I$$

Step 4 — Compare with the given A⁴ = 21A + βI

Matching the coefficient of $A$:

$$27+6\alpha = 21 \implies 6\alpha=-6 \implies \alpha=-1$$

Step 5 — Find β

Matching the coefficient of $I$, using $\alpha=-1$:

$$\beta = 9\alpha+\alpha^2 = 9(-1)+(-1)^2 = -9+1 = -8$$

Answer

Common mistakes

• Trying to find the actual matrix A. This problem never requires knowing what $A$ looks like — every step works purely with the algebraic relation $A^2=3A+\alpha I$, treating $A$ like a single unknown quantity that happens to commute with $I$.
• Forgetting to substitute $A^2$ a second time. After expanding $(3A+\alpha I)^2=9A^2+6\alpha A+\alpha^2I$, the $9A^2$ term still isn’t in the form “coefficient of $A$ plus coefficient of $I$” — it needs one more substitution using the original relation before coefficients can be compared.
• Matching coefficients in the wrong order. Solving for $\beta$ before $\alpha$ is possible in principle but circular here, since $\beta$’s expression depends on $\alpha$ — finding $\alpha$ first from the $A$-coefficient equation is what makes $\beta$ a simple substitution afterward.

Practise next

• Let $A^2=2A+\alpha I$ and $A^4=14A+\beta I$ for a $3\times3$ matrix $A$. Find $\alpha$ and $\beta$.