Q.For the matrix $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}$. Show that $A^3 - 6A^2 + 5A + 11 I = O$. Hence, find $A^{-1}$.

Concept: Cayley-Hamilton Theorem — every square matrix satisfies its own characteristic equation.

Step 1: Find the characteristic polynomial of $A$.

Expanding along the first row:

Simplify each term:

• First term: $(1-\lambda)(\lambda^2 - 5\lambda + 3)$
• Second term: $-(9 - \lambda)$
• Third term: $1 \cdot (2\lambda - 5)$

Adding:

So the characteristic polynomial is $-\lambda^3 + 6\lambda^2 - 5\lambda - 11 = 0$, or

Step 2: Apply Cayley-Hamilton Theorem.

Step 3: Find $A^{-1}$ from the equation.

Multiply the equation by $A^{-1}$:

Compute $A^2$: …

The problem asks two things: first, to verify a matrix polynomial identity, and second, to use it to find the inverse. The key idea is the Cayley-Hamilton theorem — a matrix satisfies its own characteristic equation. So if we find the characteristic polynomial of $A$, plugging $A$ into it must give the zero matrix. That gives us the identity we need to show. Then, once we have $A^3 - 6A^2 + 5A + 11I = O$, we can factor out an $A$ (carefully, since matrix multiplication isn't commutative, but $A$ commutes with itself and with $I$) to get an expression for $A^{-1}$.

Let's work through it.

1. Find the characteristic polynomial of $A$. The characteristic polynomial is $p(\lambda) = \det(\lambda I - A)$. Compute:

Expand the determinant. Using the first row:

Compute each $2\times 2$ determinant:

• First: $(\lambda - 2)(\lambda - 3) - (3)(1) = \lambda^2 - 5\lambda + 6 - 3 = \lambda^2 - 5\lambda + 3$
• Second: $(-1)(\lambda - 3) - (3)(-2) = -\lambda + 3 + 6 = -\lambda + 9$
• Third: $(-1)(1) - (\lambda - 2)(-2) = -1 + 2\lambda - 4 = 2\lambda - 5$

So:

Expand $(\lambda - 1)(\lambda^2 - 5\lambda + 3) = \lambda^3 - 5\lambda^2 + 3\lambda - \lambda^2 + 5\lambda - 3 = \lambda^3 - 6\lambda^2 + 8\lambda - 3$.

Then add $(-\lambda + 9)$ and subtract $(2\lambda - 5)$:

Simplify: $8 - 1 - 2 = 5$, and $-3 + 9 + 5 = 11$. So:

2. Apply Cayley-Hamilton theorem. The theorem states that $p(A) = O$. That is:

This is exactly what we needed to show. So the first part is done.

3. Find $A^{-1}$ from the polynomial identity. We have $A^3 - 6A^2 + 5A + 11I = O$. Rearrange to isolate $A$:

Factor $A$ on the left (since $A$ commutes with itself and with $I$):

Multiply both sides by $-\frac{1}{11}$:

This shows that $A^{-1} = -\frac{1}{11}(A^2 - 6A + 5I)$.

4. Compute $A^2$ and then $A^{-1}$ explicitly (optional but good for verification). First, $A^2$:

Compute each entry: …

Method: Cayley-Hamilton for a 3×3 Matrix, Then Solving for the Inverse

This method both derives a cubic matrix identity and uses it to find $A^{-1}$, for a $3\times3$ matrix, without computing the adjoint.

Steps

Step 1: Find the characteristic polynomial of $A$

Either expand $\det(A-\lambda I)$ directly, or use the shortcut for a $3\times3$ matrix:

The "principal minors" are the three $2\times2$ determinants obtained by deleting the same row and column number (delete row 1 & column 1, then row 2 & column 2, then row 3 & column 3).

Step 2: Apply Cayley-Hamilton to get the cubic identity

This is the identity the question asks you to verify — the coefficients in the given equation should match $\operatorname{tr}A$, the sum of principal minors, and $\det A$ exactly. …

Common Mistakes

Mistake 1: An arithmetic error computing $A^2$ for a $3\times3$ matrix

Why it's wrong: nine dot-product entries, several with negative numbers, give many chances for a sign or multiplication slip — and every later step (the cubic identity, and then $A^{-1}$) depends on $A^2$ being exactly right. Correct approach: compute each entry of $A^2$ as a labelled row·column dot product, and where possible sanity-check one or two entries a second time before moving on.

Mistake 2: Mismatching the sign of the sum-of-principal-minors term in the characteristic polynomial

Why it's wrong: the $3\times3$ characteristic polynomial is $\lambda^3-(\operatorname{tr}A)\lambda^2+(\text{sum of principal minors})\lambda-\det A$ — the middle coefficient is added, not subtracted, which is easy to get backwards by analogy with the $-\det A$ term at the end. Correct approach: derive the coefficients from $\det(A-\lambda I)$ directly, rather than guessing a sign pattern. …