Matrix A Satisfies A²⁰+αA¹⁹+βA=Given Matrix — Find β−α

JEE Main 2021. Previous Year Question.

Problem

If the matrix $A=\begin{bmatrix}1&0&0\\0&2&0\\3&0&-1\end{bmatrix}$ satisfies the equation $A^{20}+\alpha A^{19}+\beta A=\begin{bmatrix}1&0&0\\0&4&0\\0&0&1\end{bmatrix}$ for some real numbers $\alpha$ and $\beta$, then $\beta-\alpha$ is equal to?

Key insight. $A$ is a lower triangular matrix — every entry above the diagonal is zero — so its eigenvalues are simply its diagonal entries: $1$, $2$, $-1$. Any polynomial in $A$ (like $A^{20}+\alpha A^{19}+\beta A$) is also lower triangular, with diagonal entries equal to the same polynomial evaluated at each eigenvalue. That turns a matrix equation into three simple numerical equations.

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Approach

Identify $A$’s eigenvalues directly from its diagonal (since $A$ is lower triangular). Since $A^{20}+\alpha A^{19}+\beta A$ is a polynomial in $A$, it shares the same eigenvalue structure — its diagonal entries are the polynomial $p(\lambda)=\lambda^{20}+\alpha\lambda^{19}+\beta\lambda$ evaluated at each of $A$’s eigenvalues. Matching these against the diagonal of the given result matrix gives equations to solve for $\alpha$ and $\beta$.

Solution

Step 1 — Find the eigenvalues of A

Since $A$ is lower triangular (all entries above the diagonal are zero), its eigenvalues are exactly its diagonal entries:

$$\lambda_1=1, \quad \lambda_2=2, \quad \lambda_3=-1$$

Step 2 — Set up the polynomial equation for each eigenvalue

Let $p(\lambda)=\lambda^{20}+\alpha\lambda^{19}+\beta\lambda$. Since $A$’s diagonal entries in row/column order correspond to $\lambda_1,\lambda_2,\lambda_3$, the resulting matrix’s diagonal entries are $p(\lambda_1), p(\lambda_2), p(\lambda_3)$, matching the given matrix’s diagonal $1, 4, 1$:

$$p(1)=1, \qquad p(2)=4, \qquad p(-1)=1$$

Step 3 — Write out each equation

$$p(1) = 1+\alpha+\beta = 1 \implies \alpha+\beta=0 \quad \text{…(i)}$$

$$p(-1) = 1-\alpha-\beta = 1 \implies \alpha+\beta=0 \quad \text{(same as (i), no new information)}$$

$$p(2) = 2^{20}+\alpha\cdot2^{19}+2\beta = 4 \quad \text{…(ii)}$$

Step 4 — Substitute β = −α into equation (ii)

$$2^{20}+\alpha\cdot2^{19}-2\alpha = 4$$

$$2^{20} + \alpha(2^{19}-2) = 4$$

Step 5 — Solve for α

$$\alpha = \frac{4-2^{20}}{2^{19}-2} = \frac{4-1048576}{524288-2} = \frac{-1048572}{524286}$$

Dividing numerator and denominator by $2$:

$$\alpha = \frac{-524286}{262143} = -2$$

Step 6 — Find β

$$\beta = -\alpha = 2$$

Step 7 — Verify with equation (ii)

$$2^{20}+(-2)(2^{19})+2(2) = 1048576-1048576+4 = 4 \checkmark$$

Step 8 — Compute β − α

$$\beta-\alpha = 2-(-2) = 4$$

Answer

Common mistakes

• Assuming $A^{20}$ can be computed by directly raising each matrix entry to the 20th power. Matrix powers don’t work entry-wise — the eigenvalue approach sidesteps ever computing $A^{20}$ or $A^{19}$ explicitly.
• Forgetting that $A$’s eigenvalues match the diagonal only because $A$ is triangular. This shortcut doesn’t apply to a general (non-triangular) matrix, where eigenvalues would need to be found from the characteristic polynomial instead.

Practise next

• If $B=\begin{bmatrix}2&0&0\\0&3&0\\5&0&-2\end{bmatrix}$ satisfies $B^3+\alpha B^2+\beta B+12I=O$, find $\alpha+\beta$ using the same eigenvalue-matching method.