If $A = \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix}$, show that $(A - 2I)(A - 3I) = 0$.

If A = 4 2 \-1 1 , show that (A - 2I)(A - 3I) = 0. — CBSE Class 12 Mathematics 2019 PYQ

Question

A = \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix}

CBSECBSE Class XII Board 2019Subjective· 2mImportance★★★★★

✓ Free question

Concept understanding — Cayley-Hamilton Theorem

The Cayley–Hamilton Theorem

Matrices add, subtract, multiply and scale much like numbers (except $AB\neq BA$ in general), so we can substitute a matrix into a polynomial. For $p(x)=x^2-3x+2$ , replacing $x$ by a square matrix $A$ gives

p(A)=A^2-3A+2I,

where the constant $2$ becomes $2I$ so it can be added to matrices.

The Cayley–Hamilton theorem makes a striking claim: every square matrix satisfies its own characteristic equation.

The characteristic polynomial

Every $n\times n$ matrix $A$ has a characteristic polynomial

p(\lambda)=\det(\lambda I-A),

a degree- $n$ polynomial whose roots are the eigenvalues. For a $2\times2$ matrix it is $\lambda^2-(\operatorname{tr}A)\lambda+\det A$ . For $A=\begin{pmatrix}1&2\\3&4\end{pmatrix}$ this is $p(\lambda)=\lambda^2-5\lambda-2$ .

The statement

Important

If $p(\lambda)=\lambda^n+c_{n-1}\lambda^{n-1}+\cdots+c_1\lambda+c_0$ is the characteristic polynomial of $A$ , then

$p(A)=A^n+c_{n-1}A^{n-1}+\cdots+c_1A+c_0I=0,$

the $n\times n$ zero matrix.

For the example, $A^2-5A-2I=\begin{pmatrix}0&0\\0&0\end{pmatrix}$ .

Why it is surprising, and a quick check

The polynomial is built from $\det(\lambda I-A)$ , yet feeding $A$ back into it annihilates it — and this holds for any $A$ , invertible or not. You can verify it on the general $2\times2$ matrix $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ , where $p(\lambda)=\lambda^2-(a+d)\lambda+(ad-bc)$ : a short computation of $A^2-(a+d)A+(ad-bc)I$ gives the zero matrix.

Why it matters

Cayley–Hamilton lets you rewrite any high power $A^k$ (for $k\ge n$ ) as a combination of $I,A,\dots,A^{n-1}$ , which speeds up computing powers, exponentials and inverses.

Note

If $c_0\neq0$ (so $A$ is invertible), rearranging $p(A)=0$ expresses the inverse as a polynomial in $A$ :

$A^{-1}=-\frac{1}{c_0}\big(A^{n-1}+c_{n-1}A^{n-2}+\cdots+c_1I\big).$

One caution: the theorem is not proved by literally putting $\lambda=A$ into the scalar determinant — the honest proof uses the adjugate. For a first meeting, learn the statement and practise verifying it on small matrices.

The Cayley-Hamilton Theorem is not part of the core CBSE Class 12 Matrices syllabus but is an important topic for JEE Advanced and other engineering entrance exams that build on the NCERT Class 12 Mathematics curriculum on matrices and determinants. "Cayley Hamilton theorem proof and application to find inverse" is a frequently searched topic among students preparing for these higher-level linear algebra questions.

The Cayley-Hamilton theorem says every square matrix satisfies its own characteristic equation. For $A$ , the characteristic polynomial is $\lambda^2 - 5\lambda + 6 = (\lambda-2)(\lambda-3)$ , so plugging in $A$ gives $(A-2I)(A-3I)=0$ — we verify this directly by matrix multiplication.

The problem asks us to show that $(A-2I)(A-3I) = 0$ for the given $2\times 2$ matrix $A$ . This is a direct verification of the Cayley-Hamilton theorem, but let's understand why this product is zero before we compute.

The big idea: The characteristic polynomial of $A$ is found from $\det(A - \lambda I)$ . For a $2\times 2$ matrix, this is $\lambda^2 - (\text{trace})\lambda + \det A$ . Here trace $= 4+1 = 5$ and determinant $= (4)(1) - (2)(-1) = 4+2 = 6$ . So the characteristic polynomial is $\lambda^2 - 5\lambda + 6$ , which factors as $(\lambda-2)(\lambda-3)$ . The Cayley-Hamilton theorem states that $A$ satisfies its own characteristic equation: $A^2 - 5A + 6I = 0$ . But note that $A^2 - 5A + 6I = (A-2I)(A-3I)$ — because matrix multiplication distributes just like polynomial multiplication (since $A$ commutes with $I$ ). So showing $(A-2I)(A-3I)=0$ is exactly verifying Cayley-Hamilton for this $A$ .

Let's do it step by step.

Compute $A - 2I$ and $A - 3I$ . $I$ is the $2\times 2$ identity matrix $\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$ .

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

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

Multiply $(A-2I)(A-3I)$ . Multiply the two matrices:

(A-2I)(A-3I) = \begin{bmatrix}2 & 2 \\ -1 & -1\end{bmatrix} \begin{bmatrix}1 & 2 \\ -1 & -2\end{bmatrix}

Compute entry by entry:

Top-left: $(2)(1) + (2)(-1) = 2 - 2 = 0$
Top-right: $(2)(2) + (2)(-2) = 4 - 4 = 0$
Bottom-left: $(-1)(1) + (-1)(-1) = -1 + 1 = 0$
Bottom-right: $(-1)(2) + (-1)(-2) = -2 + 2 = 0$

So every entry is zero. Hence $(A-2I)(A-3I) = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix} = 0$ .

Watch out

A common mistake is to multiply in the wrong order: $(A-2I)(A-3I)$ is not the same as $(A-2I)(A-3I)$ ? Actually it is — but students sometimes forget that matrix multiplication is not commutative in general. Here, since both factors are polynomials in $A$ , they commute (because $A$ and $I$ commute), so order doesn't matter. But if you had two unrelated matrices, order would matter.

Tip

Notice that the two matrices $A-2I$ and $A-3I$ are each singular (their rows are multiples: first row is $-2$ times second row in the first matrix, and similarly in the second). Their product being zero is a special case: two nonzero matrices can multiply to zero — this is exactly what happens when they share eigenvectors in a certain way.

✓Final answer

(A-2I)(A-3I) = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}

Q.If A=[42−11]A = \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix}A=[4−1​21​], show that (A−2I)(A−3I)=0(A - 2I)(A - 3I) = 0(A−2I)(A−3I)=0.

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Q.If $A = \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix}$ , show that $(A - 2I)(A - 3I) = 0$ .

Unlock everything free for 14 days