Q.If $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$, prove that $A^3 - 6A^2 + 7A + 2I = 0$.

Concept

To evaluate $f(A)=A^2-2A-3I$ for a matrix $A$, the scalar constant $-3$ becomes $-3I$. After computing the matrix, classify it: symmetric ($M^T=M$), skew-symmetric ($M^T=-M$, zero diagonal), null, or identity.

Solution

1. $A^2$. $\begin{pmatrix}1&2\\2&3\end{pmatrix}\begin{pmatrix}1&2\\2&3\end{pmatrix}=\begin{pmatrix}5&8\\8&13\end{pmatrix}.$
2. $2A$ and $3I$. $2A=\begin{pmatrix}2&4\\4&6\end{pmatrix},\ 3I=\begin{pmatrix}3&0\\0&3\end{pmatrix}.$
3. Combine. $f(A)=\begin{pmatrix}5&8\\8&13\end{pmatrix}-\begin{pmatrix}2&4\\4&6\end{pmatrix}-\begin{pmatrix}3&0\\0&3\end{pmatrix}=\begin{pmatrix}0&4\\4&4\end{pmatrix}.$ …

The key insight here is that multiplying a matrix by the identity matrix leaves it unchanged: $A I = A$. So the expression $(A I)^2$ is just $A^2$. The problem then reduces to squaring the given $2 \times 2$ matrix and reading off the entries.

1. Simplify the expression

   Since $I$ is the identity matrix, $A I = A$. Therefore $(A I)^2 = A^2$. No matrix multiplication with $I$ changes anything — it’s like multiplying a number by 1.

2. Square the matrix $A$

   We have

Compute $A^2$ by standard matrix multiplication:

3. Simplify each entry
   • Top-left: $a^2 + b^2$
   • Top-right: $ab + ba = 2ab$
   • Bottom-left: $ba + ab = 2ab$
   • Bottom-right: $b^2 + a^2 = a^2 + b^2$ So

4. Match to the given form …

Concept

To evaluate a polynomial $f(t)=t^{2}-3t+7$ at a square matrix $A$, replace $t$ by $A$ and the constant $7$ by $7I$ (the identity of the same order):

Then add the given matrix entrywise.

Solution

1. $A^{2}$:

2. $-3A$ and $7I$:

3. $f(A)$:

4. Add the given matrix: …

Concept & Intuition

When a function $f(t)=t^2-3t+7$ is applied to a square matrix $A$, we replace the scalar variable $t$ by the matrix $A$ and the constant term $7$ by $7I$ (the identity matrix of the same size). This is the standard definition of a matrix polynomial. After computing $A^2$ and combining terms, we add the extra matrix. The whole expression simplifies dramatically — a sign that the given matrix might satisfy its own characteristic equation (Cayley‑Hamilton), but here we just do the arithmetic.

Step‑by‑step solution

1. Compute $A^2$

2. Compute $f(A) = A^2 - 3A + 7I$

   First, $3A = 3\begin{pmatrix}1 & -2 \\ 4 & 5\end{pmatrix} = \begin{pmatrix}3 & -6 \\ 12 & 15\end{pmatrix}$.

   The identity matrix of size 2 is $I = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$, so $7I = \begin{pmatrix}7 & 0 \\ 0 & 7\end{pmatrix}$.

   Now combine:

Do it entry‑wise:

• Top‑left: $-7 - 3 + 7 = -3$ …

Concept & Intuition

When a matrix satisfies a quadratic equation like $P^2 = I - P$, its higher powers can be expressed as linear combinations of $I$ and $P$ only. This is because any polynomial in $P$ can be reduced using the relation, much like reducing powers of a number that satisfies $x^2 = 1 - x$. The coefficients follow a recurrence, and we can solve for the exponent $n$ by matching the given form.

Step-by-step solution

1. Set up the recurrence From $P^2 = I - P$, multiply both sides by $P^{n-2}$ (for $n \ge 2$):

This is a linear recurrence for powers of $P$.

2. Assume a linear form Since the recurrence is linear and the initial terms are $P^0 = I$ and $P^1 = P$, every power can be written as

where $a_n, b_n$ are integers.

3. Find recurrence for coefficients Substitute the form into $P^n = P^{n-2} - P^{n-1}$:

Equate coefficients of $I$ and $P$:

4. Initial conditions

   For $n=0$: $P^0 = I = 1\cdot I + 0\cdot P$ → $a_0 = 1, b_0 = 0$.

   For $n=1$: $P^1 = P = 0\cdot I + 1\cdot P$ → $a_1 = 0, b_1 = 1$.

5. Compute coefficients until match

   Use $a_n = a_{n-2} - a_{n-1}$ (same recurrence for $b_n$):

   $n$	$a_n$	$b_n$
   0	1	0
   1	0	1
   2	$a_0 - a_1 = 1 - 0 = 1$	$b_0 - b_1 = 0 - 1 = -1$

The key insight is that $A$ is not just any $2\times2$ matrix — it is a rank‑1 symmetric matrix. That means it can be expressed as an outer product of a vector with itself. When you multiply such a matrix by itself, the result is just a scalar multiple of the original matrix. This is because the dot product of the vector with itself reappears as a scalar factor.

Let’s see why.

1. Write $A$ as an outer product. Let $\mathbf{v} = \begin{pmatrix}1 \\ -1\end{pmatrix}$. Then

This representation is the key: every column is a multiple of $\mathbf{v}$, and every row is a multiple of $\mathbf{v}^T$.

2. Compute $A^2$ using the outer product.

The middle term $\mathbf{v}^T\mathbf{v}$ is just the dot product of $\mathbf{v}$ with itself:

So

3. General pattern by induction. Suppose $A^k = 2^{k-1} A$ for some $k\ge 1$. Then

The base case $k=1$ gives $A^1 = 2^{0}A = A$, which is true. So by induction, for every integer $n\ge 0$,

4. Apply to the given problem. …

Concept: Direct matrix multiplication.

A = [[2, -2], [-2, 2]]

A^2 = A x A:

Entry (1,1) = (2)(2) + (-2)(-2) = 4 + 4 = 8

Entry (1,2) = (2)(-2) + (-2)(2) = -4 - 4 = -8

Entry (2,1) = (-2)(2) + (2)(-2) = -4 - 4 = -8

Entry (2,2) = (-2)(-2) + (2)(2) = 4 + 4 = 8 …

For $A=\begin{bmatrix}1&1\\1&1\end{bmatrix}$:

Therefore

$A^4=(A^2)^2=(2A)^2=4A^2=4(2A)=8A=2^3A.$ …