Easy Practice — Set 1

Quick Recap — Matrices

• Order $m\times n$ = rows $\times$ columns; square if $m=n$.
• Special: identity $I$ (1's on the diagonal), diagonal, scalar, symmetric ($A^T=A$), skew-symmetric ($A^T=-A$, zero diagonal).
• Operations: add/subtract need the same order; $AB$ needs (cols of $A$) = (rows of $B$).
• $(A^T)^T=A$, $(AB)^T=B^TA^T$, $IA=A$; trace = sum of diagonal entries.

Beyond-NCERT JEE Formulae

Cayley-Hamilton Theorem

Every square matrix satisfies its own characteristic equation $\det(A-\lambda I)=0$. For a $2\times2$ matrix this becomes

$$A^2-(\operatorname{tr}A)\,A+|A|\,I=O.$$

Rearranging delivers the inverse with no separate adjoint computation:

$$A^{-1}=\dfrac{1}{|A|}\big[(\operatorname{tr}A)\,I-A\big],\qquad |A|\ne0.$$

Higher powers collapse the same way: multiply the identity by $A$ to get $A^3=(\operatorname{tr}A)\,A^2-|A|\,A$, and keep reducing so that every power stays in the shape $A^n=pA+qI$.

[JEE Tip] The moment a question wants $A^{-1}$, a high power $A^n$, or a relation such as $A^3=pA+qI$, reach for Cayley-Hamilton — it beats direct multiplication almost every time.

Characteristic-Equation Facts (2x2)

For a $2\times2$ matrix the characteristic equation is

$$\lambda^2-(\operatorname{tr}A)\,\lambda+|A|=0.$$

Its roots are the eigenvalues $\lambda_1$ and $\lambda_2$, and comparing coefficients gives

$$\lambda_1+\lambda_2=\operatorname{tr}A,\qquad \lambda_1\lambda_2=|A|.$$

So the sum of the eigenvalues equals the trace and their product equals the determinant — an instant check on any eigenvalue you compute.

Adjoint Identities Beyond the Basics

• Reversal law: $\operatorname{adj}(AB)=\operatorname{adj}B\,\operatorname{adj}A$ — the order flips, exactly like $(AB)^{-1}$ and $(AB)^T$.
• Inverse of the adjoint: $(\operatorname{adj}A)^{-1}=\operatorname{adj}(A^{-1})=\dfrac{A}{|A|}$.
• Determinant of a double adjoint: $|\operatorname{adj}(\operatorname{adj}A)|=|A|^{(n-1)^2}$ for an $n\times n$ matrix.
• The 2x2 shortcut: here $(n-1)^2=1$, and in fact $\operatorname{adj}(\operatorname{adj}A)=A$ itself — taking the adjoint twice returns the original matrix.

[JEE Tip] For general order $n$, $\operatorname{adj}(\operatorname{adj}A)=|A|^{\,n-2}A$; setting $n=2$ recovers $\operatorname{adj}(\operatorname{adj}A)=A$.

Solved Examples — Beyond-NCERT Formulae

Example 1 — Inverse of a 2x2 via Cayley-Hamilton

Find $A^{-1}$ for $A=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ without building the adjoint separately.

Step 1. Trace and determinant: $\operatorname{tr}A=2+2=4$ and $|A|=2\cdot2-1\cdot1=3$.

Step 2. Cayley-Hamilton gives $A^2-4A+3I=O$, so $3I=4A-A^2=A(4I-A)$.

Step 3. Therefore $A^{-1}=\dfrac{1}{3}\big(4I-A\big)=\dfrac{1}{3}\begin{pmatrix}2&-1\\-1&2\end{pmatrix}$.

Check. $A\,A^{-1}=\dfrac{1}{3}\begin{pmatrix}3&0\\0&3\end{pmatrix}=I$, as required.

Example 2 — Computing $A^3$ by reduction

Let $A=\begin{pmatrix}1&2\\1&3\end{pmatrix}$. Find $A^3$ without cubing directly.

Step 1. $\operatorname{tr}A=4$ and $|A|=1\cdot3-2\cdot1=1$, so Cayley-Hamilton gives $A^2=4A-I$.

Step 2. Multiply by $A$ and substitute: $A^3=4A^2-A=4(4A-I)-A=15A-4I$.

Step 3. Hence $A^3=15\begin{pmatrix}1&2\\1&3\end{pmatrix}-4I=\begin{pmatrix}11&30\\15&41\end{pmatrix}$.

The characteristic relation replaced two matrix multiplications with a single scalar reduction.

Example 3 — The zero-matrix verification

Confirm $A^2-(\operatorname{tr}A)\,A+|A|\,I=O$ for $A=\begin{pmatrix}2&3\\1&4\end{pmatrix}$.

Step 1. $\operatorname{tr}A=6$ and $|A|=8-3=5$.

Step 2. Compute $A^2=\begin{pmatrix}7&18\\6&19\end{pmatrix}$ and $6A=\begin{pmatrix}12&18\\6&24\end{pmatrix}$.

Step 3. Then $A^2-6A+5I=\begin{pmatrix}7-12+5&18-18\\6-6&19-24+5\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}=O$, so the theorem checks out.

Example 4 — A 3x3 characteristic equation

For $A=\begin{pmatrix}2&1&1\\0&1&0\\1&1&2\end{pmatrix}$, write its characteristic equation and use it to find $A^{-1}$.

Step 1. For order $3$ the characteristic equation is $\lambda^3-(\operatorname{tr}A)\,\lambda^2+S\,\lambda-|A|=0$, where $S$ is the sum of the three principal $2\times2$ minors.

Step 2. Here $\operatorname{tr}A=5$, then $S=2+3+2=7$ and $|A|=3$, so $\lambda^3-5\lambda^2+7\lambda-3=0$.

Step 3. By Cayley-Hamilton $A^3-5A^2+7A-3I=O$, so $3I=A\big(A^2-5A+7I\big)$ and $A^{-1}=\dfrac{1}{3}\big(A^2-5A+7I\big)$.

Step 4. With $A^2=\begin{pmatrix}5&4&4\\0&1&0\\4&4&5\end{pmatrix}$ this yields $A^{-1}=\dfrac{1}{3}\begin{pmatrix}2&-1&-1\\0&3&0\\-1&-1&2\end{pmatrix}$.

Example 5 — The adjoint reversal law

For $A=\begin{pmatrix}1&2\\3&4\end{pmatrix}$ and $B=\begin{pmatrix}2&0\\1&3\end{pmatrix}$, verify $\operatorname{adj}(AB)=\operatorname{adj}B\,\operatorname{adj}A$.

Step 1. $AB=\begin{pmatrix}4&6\\10&12\end{pmatrix}$, so $\operatorname{adj}(AB)=\begin{pmatrix}12&-6\\-10&4\end{pmatrix}$.

Step 2. Separately, $\operatorname{adj}A=\begin{pmatrix}4&-2\\-3&1\end{pmatrix}$ and $\operatorname{adj}B=\begin{pmatrix}3&0\\-1&2\end{pmatrix}$.

Step 3. Then $\operatorname{adj}B\,\operatorname{adj}A=\begin{pmatrix}3&0\\-1&2\end{pmatrix}\begin{pmatrix}4&-2\\-3&1\end{pmatrix}=\begin{pmatrix}12&-6\\-10&4\end{pmatrix}$, matching Step 1.

Order matters: the reversed product $\operatorname{adj}A\,\operatorname{adj}B=\begin{pmatrix}14&-4\\-10&2\end{pmatrix}$ is different, so the flip in the law is essential.

Example 6 — Eigenvalues from the characteristic equation

Find the eigenvalues of $A=\begin{pmatrix}4&1\\2&3\end{pmatrix}$ and check them against the trace and determinant.

Step 1. $\operatorname{tr}A=7$ and $|A|=12-2=10$, so the characteristic equation is $\lambda^2-7\lambda+10=0$.

Step 2. Factoring, $(\lambda-2)(\lambda-5)=0$, so $\lambda_1=2$ and $\lambda_2=5$.

Step 3. Cross-check: $\lambda_1+\lambda_2=7=\operatorname{tr}A$ and $\lambda_1\lambda_2=10=|A|$, confirming the roots.

Example 7 — Taking the adjoint twice

For $A=\begin{pmatrix}3&1\\2&4\end{pmatrix}$, show that $\operatorname{adj}(\operatorname{adj}A)=A$ and find $(\operatorname{adj}A)^{-1}$.

Step 1. $\operatorname{adj}A=\begin{pmatrix}4&-1\\-2&3\end{pmatrix}$.

Step 2. Taking the adjoint again, $\operatorname{adj}(\operatorname{adj}A)=\begin{pmatrix}3&1\\2&4\end{pmatrix}=A$, exactly as the 2x2 rule predicts.

Step 3. Since $|A|=12-2=10$, the identity $(\operatorname{adj}A)^{-1}=\dfrac{A}{|A|}$ gives $(\operatorname{adj}A)^{-1}=\dfrac{1}{10}\begin{pmatrix}3&1\\2&4\end{pmatrix}$.

Example 8 — A high power through eigenvalues

Compute $A^5$ for $A=\begin{pmatrix}0&-2\\1&3\end{pmatrix}$ using the characteristic equation.

Step 1. $\operatorname{tr}A=3$ and $|A|=0\cdot3-(-2)\cdot1=2$, so $\lambda^2-3\lambda+2=0$ with roots $\lambda=1$ and $\lambda=2$.

Step 2. Write $A^5=pA+qI$. The same scalar relation holds at each eigenvalue, so $1^5=p+q$ and $2^5=2p+q$.

Step 3. Subtracting the first equation from the second, $p=32-1=31$, then $q=1-31=-30$, giving $A^5=31A-30I$.

Step 4. Substitute: $A^5=31\begin{pmatrix}0&-2\\1&3\end{pmatrix}-30I=\begin{pmatrix}-30&-62\\31&63\end{pmatrix}$.