If $A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, find $k$ so that $A^2 = kA - 2I$.

> [!TLDR] > $A$ satisfies its own characteristic equation $\lambda^2-\lambda+2=0$, i.e. $A^2 = A - 2I$; comparing with $A^2 = kA - 2I$ gives $k=1$. We want the number $k$ for which $A^2 = kA - 2I$. There are two clean routes: multiply $A$ by itself and match entries, or use the Cayley–Hamilton theorem. Both give the same $k$, and it is worth seeing why. ## Route 1 — direct multiplication (foolproof) $$A^2 = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}.$$ Entry by entry: - $(1,1):\ 3\cdot3+(-2)\cdot4 = 9-8 = 1$ - $(1,2):\ 3\cdot(-2)+(-2)\cdot(-2) = -6+4 = -2$ - $(2,1):\ 4\cdot3+(-2)\cdot4 = 12-8 = 4$ - $(2,2):\ 4\cdot(-2)+(-2)\cdot(-2) = -8+4 = -4$ So $A^2 = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix}$. Now $$kA - 2I = \begin{bmatrix} 3k-2 & -2k \\ 4k & -2k-2 \end{bmatrix}.$$ Setting this equal to $A^2$, the $(1,1)$ entry gives $3k-2 = 1 \Rightarrow k=1$, and every other entry ($-2k=-2$, $4k=4$, $-2k-2=-4$) also gives $k=1$. Consistent. ## Route 2 — Cayley–Hamilton (elegant) Every $2\times2$ matrix satisfies $\lambda^2-(\operatorname{tr}A)\lambda+(\det A)=0$. Here …

If A = 3 -2 \4 -2 and I = 1 0 \0 1 , find k so tha – Class 12 Mathematics

Exercise 3.2 · Q17

A = \begin{bmatrix} 3 & -2 \\ 4 & -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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$A$ satisfies its own characteristic equation $\lambda^2-\lambda+2=0$ , i.e. $A^2 = A - 2I$ ; comparing with $A^2 = kA - 2I$ gives $k=1$ .

We want the number $k$ for which $A^2 = kA - 2I$ . There are two clean routes: multiply $A$ by itself and match entries, or use the Cayley–Hamilton theorem. Both give the same $k$ , and it is worth seeing why.

Route 1 — direct multiplication (foolproof)

A^2 = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}.

Entry by entry:

$(1,1):\ 3\cdot3+(-2)\cdot4 = 9-8 = 1$
$(1,2):\ 3\cdot(-2)+(-2)\cdot(-2) = -6+4 = -2$
$(2,1):\ 4\cdot3+(-2)\cdot4 = 12-8 = 4$
$(2,2):\ 4\cdot(-2)+(-2)\cdot(-2) = -8+4 = -4$

So $A^2 = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix}$ . Now

kA - 2I = \begin{bmatrix} 3k-2 & -2k \\ 4k & -2k-2 \end{bmatrix}.

Setting this equal to $A^2$ , the $(1,1)$ entry gives $3k-2 = 1 \Rightarrow k=1$ , and every other entry ( $-2k=-2$ , $4k=4$ , $-2k-2=-4$ ) also gives $k=1$ . Consistent.

Route 2 — Cayley–Hamilton (elegant)

Every $2\times2$ matrix satisfies $\lambda^2-(\operatorname{tr}A)\lambda+(\det A)=0$ . Here …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AHSEC Higher Secondary (HS) Final Examination 2026Set ANNUAL4 marks
$Q. Answer both$
(a) and
(b) :
(a) [2 marks] If $A=\begin{pmatrix}3 & 1 \\ 7 & 5\end{pmatrix}$ , $I=\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$ , then find $x$ and $y$ such that $A^{2}+xI=yA$ .
(b) [2 marks] If $A$ and $B$ are invertible matrices of the same order, then prove that $(AB)^{-1}=B^{-1}A^{-1}$ .
›Reveal solutionSolution

(a) $A^2+xI=yA\Rightarrow x=8,y=8$ ; (b) $(AB)^{-1}=B^{-1}A^{-1}$ proved. (OR: (c) skew-symmetric diagonals are zero; (d) $x=5$ or $-3$ .)

Main (a). With $A=\begin{pmatrix}3&1\\7&5\end{pmatrix}$ ,

$A^{2}=\begin{pmatrix}3&1\\7&5\end{pmatrix}\begin{pmatrix}3&1\\7&5\end{pmatrix}=\begin{pmatrix}16&8\\56&32\end{pmatrix}.$

Then $A^{2}+xI=\begin{pmatrix}16+x&8\\56&32+x\end{pmatrix}$ and $yA=\begin{pmatrix}3y&y\\7y&5y\end{pmatrix}$ .

Comparing the $(1,2)$ entries: $8=y$ . Comparing $(1,1)$ : $16+x=3y=24\Rightarrow x=8$ . (Check $(2,2)$ : $32+8=40=5\cdot 8$ .) So $x=8,\ y=8$ .

Main (b). To prove $(AB)^{-1}=B^{-1}A^{-1}$ , verify it is the inverse of $AB$ :

$(AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=A\,I\,A^{-1}=AA^{-1}=I,$

$(B^{-1}A^{-1})(AB)=B^{-1}(A^{-1}A)B=B^{-1}IB=I.$

Since $B^{-1}A^{-1}$ multiplies $AB$ to $I$ on both sides, $(AB)^{-1}=B^{-1}A^{-1}$ .

…

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AHSEC Higher Secondary (HS) Final Examination 2022Set ANNUAL4 marks
$A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$
›Reveal solutionSolution

Computing $A^2$ and matching it against $-\lambda A-\mu I$ entrywise gives $\lambda=-4,\ \mu=1$ . (OR: the coefficient determinant of the 3-equation system equals $a$ , so the system is consistent — with a unique solution — for every $a\ne0$ , and inconsistent at $a=0$ .)

Find $\lambda,\mu$ for $A^2+\lambda A+\mu I=0$ , $A=\begin{bmatrix}2&3\\1&2\end{bmatrix}$ .

$A^2 = \begin{bmatrix}2&3\\1&2\end{bmatrix}\begin{bmatrix}2&3\\1&2\end{bmatrix} = \begin{bmatrix}2(2)+3(1) & 2(3)+3(2)\\1(2)+2(1)&1(3)+2(2)\end{bmatrix} = \begin{bmatrix}7&12\\4&7\end{bmatrix}.$

$A^2+\lambda A+\mu I = \begin{bmatrix}7+2\lambda+\mu & 12+3\lambda\\4+\lambda & 7+2\lambda+\mu\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix}.$

From the $(1,2)$ entry: $12+3\lambda=0 \Rightarrow \lambda=-4$ .

Check $(2,1)$ entry: $4+\lambda=0 \Rightarrow \lambda=-4$ ✓ (consistent).

From the $(1,1)$ entry: $7+2\lambda+\mu=0 \Rightarrow 7-8+\mu=0 \Rightarrow \mu=1$ .

$\boxed{\lambda=-4,\ \mu=1}$

OR: Determine $a$ for which the system $x+y+z=1,\ 2x+3y+2z=2,\ ax+ay+2az=4$ is consistent.

The coefficient matrix is $\begin{bmatrix}1&1&1\\2&3&2\\a&a&2a\end{bmatrix}$ . Its determinant:

$\Delta = 1(3\cdot2a-2\cdot a) - 1(2\cdot2a-2\cdot a) + 1(2a-3a) = (6a-2a)-(4a-2a)+(-a) = 4a-2a-a = a.$

…

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Q.If $A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ , find $k$ so that $A^2 = kA - 2I$ .

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

Route 1 — direct multiplication (foolproof)

Route 2 — Cayley–Hamilton (elegant)

Method: Finding an unknown scalar in a matrix identity

Steps

Common Mistakes

Q.If A=[3−24−2]A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}A=[34​−2−2​] and I=[1001]I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}I=[10​01​], find kkk so that A2=kA−2IA^2 = kA - 2IA2=kA−2I.

Route 1 — direct multiplication (foolproof)

Route 2 — Cayley–Hamilton (elegant)

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Q.If $A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ , find $k$ so that $A^2 = kA - 2I$ .

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