Show that $A = \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix}$ satisfies the equation $A^2 - 3A - 7I = O$ and hence find $A^{-1}$.

> [!TLDR] > The Cayley-Hamilton theorem says every square matrix satisfies its own characteristic equation. For $A$, the characteristic polynomial is $\lambda^2 - 3\lambda - 7 = 0$, so $A^2 - 3A - 7I = O$. Rearranging gives $A(A - 3I) = 7I$, so $A^{-1} = \frac{1}{7}(A - 3I) = \frac{1}{7}\begin{bmatrix}2 & 3 \\ -1 & -5\end{bmatrix}$. The problem asks two things: first, to verify that $A$ satisfies a given matrix equation, and second, to use that equation to find $A^{-1}$. The key idea is the **Cayley-Hamilton theorem** — a matrix obeys its own characteristic polynomial. But here, the equation $A^2 - 3A - 7I = O$ is handed to us; we just need to check it. Then, once verified, we can rearrange it to express $I$ in terms of $A$, which directly gives the inverse. Let’s go step by step. 1. **Compute $A^2$ directly.** $A = \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix}$. Multiply: $$A^2 = A \cdot A = \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix} \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix}$$ Top-left: $5 \cdot 5 + 3 \cdot (-1) = 25 - 3 = 22$ Top-right: $5 \cdot 3 + 3 \cdot (-2) = 15 - 6 = 9$ Bottom-left: $(-1) \cdot 5 + (-2) \cdot (-1) = -5 + 2 = -3$ Bottom-right: $(-1) \cdot 3 + (-2) \cdot (-2) = -3 + 4 = 1$ So $$A^2 = \begin{bmatrix} 22 & 9 \\ -3 & 1 \end{bmatrix}$$ 2. **Form $A^2 - 3A - 7I$.** First, $3A = \begin{bmatrix} 15 & 9 \\ -3 & -6 \end{bmatrix}$. And $7I = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$. Now subtract: $$A^2 - 3A - 7I = \begin{bmatrix} 22 & 9 \\ -3 & 1 \end{bmatrix} - \begin{bmatrix} 15 & 9 \\ -3 & -6 \end{bmatrix} - \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$$ Do the subtraction element-wise: - Top-left: $22 - 15 - 7 = 0$ - Top-right: $9 - 9 - 0 = 0$ - Bottom-left: $-3 - (-3) - 0 = -3 + 3 = 0$ - Bottom-right: $1 - (-6) - 7 = 1 + 6 - 7 = 0$ Every entry is zero, so indeed $A^2 - 3A - 7I = O$. The matrix satisfies the equation. > [!TIP] > You could also find

Show that A = 5 3 \-1 -2 satisfies the equation A² – Class 12 Mathematics

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A = \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix}

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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. …

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The Cayley-Hamilton theorem says every square matrix satisfies its own characteristic equation. For $A$ , the characteristic polynomial is $\lambda^2 - 3\lambda - 7 = 0$ , so $A^2 - 3A - 7I = O$ . Rearranging gives $A(A - 3I) = 7I$ , so $A^{-1} = \frac{1}{7}(A - 3I) = \frac{1}{7}\begin{bmatrix}2 & 3 \\ -1 & -5\end{bmatrix}$ .

The problem asks two things: first, to verify that $A$ satisfies a given matrix equation, and second, to use that equation to find $A^{-1}$ . The key idea is the Cayley-Hamilton theorem — a matrix obeys its own characteristic polynomial. But here, the equation $A^2 - 3A - 7I = O$ is handed to us; we just need to check it. Then, once verified, we can rearrange it to express $I$ in terms of $A$ , which directly gives the inverse.

Let’s go step by step.

Compute $A^2$ directly. $A = \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix}$ . Multiply:

A^2 = A \cdot A = \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix} \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix}

Top-left: $5 \cdot 5 + 3 \cdot (-1) = 25 - 3 = 22$

Top-right: $5 \cdot 3 + 3 \cdot (-2) = 15 - 6 = 9$

Bottom-left: $(-1) \cdot 5 + (-2) \cdot (-1) = -5 + 2 = -3$

Bottom-right: $(-1) \cdot 3 + (-2) \cdot (-2) = -3 + 4 = 1$

A^2 = \begin{bmatrix} 22 & 9 \\ -3 & 1 \end{bmatrix}

Form $A^2 - 3A - 7I$ . First, $3A = \begin{bmatrix} 15 & 9 \\ -3 & -6 \end{bmatrix}$ . And $7I = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$ . Now subtract:

A^2 - 3A - 7I = \begin{bmatrix} 22 & 9 \\ -3 & 1 \end{bmatrix} - \begin{bmatrix} 15 & 9 \\ -3 & -6 \end{bmatrix} - \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}

Do the subtraction element-wise:

Top-left: $22 - 15 - 7 = 0$
Top-right: $9 - 9 - 0 = 0$
Bottom-left: $-3 - (-3) - 0 = -3 + 3 = 0$
Bottom-right: $1 - (-6) - 7 = 1 + 6 - 7 = 0$ Every entry is zero, so indeed $A^2 - 3A - 7I = O$ . The matrix satisfies the equation.

Tip

You could also find the characteristic polynomial of $A$ : $\det(A - \lambda I) = (5-\lambda)(-2-\lambda) - (3)(-1) = \lambda^2 - 3\lambda - 7$ . Cayley-Hamilton then guarantees $A^2 - 3A - 7I = O$ without any multiplication. But the problem expects the direct verification.

Rearrange to find $A^{-1}$ . …

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Q.Show that $A = \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix}$ satisfies the equation $A^2 - 3A - 7I = O$ and hence find $A^{-1}$ .

Concept understanding — Cayley-Hamilton Theorem

The Cayley–Hamilton Theorem

The characteristic polynomial

The statement

Why it is surprising, and a quick check

Why it matters

Method: Finding $A^{-1}$ from a matrix's own polynomial equation

Steps

Common Mistakes

Q.Show that A=[53−1−2]A = \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix}A=[5−1​3−2​] satisfies the equation A2−3A−7I=OA^2 - 3A - 7I = OA2−3A−7I=O and hence find A−1A^{-1}A−1.

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Q.Show that $A = \begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix}$ satisfies the equation $A^2 - 3A - 7I = O$ and hence find $A^{-1}$ .

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