A square matrix $A$ is invertible if some matrix $B$ undoes it: $AB=BA=I$. That $B$ is unique, written $A^{-1}$, and the inverse of a product reverses the order: $(AB)^{-1}=B^{-1}A^{-1}$.
For numbers, dividing by $5$ is the same as multiplying by $\tfrac15$, because $\tfrac15\cdot5=1$. Matrices cannot be divided, but some of them can be undone. If a matrix $A$ moves the plane — stretching, shearing or rotating it — another matrix may move everything straight back. If $A$ scrambles a secret message, another matrix may unscramble it. That “undo” matrix plays the role of $\tfrac1a$: it is the inverse $A^{-1}$, and multiplying by it is how matrix equations are solved.
Not every matrix can be undone. In Section 3.2 we met $A=\begin{bmatrix}1&2\\2&4\end{bmatrix}$ with $AB=O$ for a non-zero $B$; the teaser of this chapter showed it flattening the plane onto a line. Once two different points have landed on the same image, no matrix can tell them apart again. This section defines invertible matrices, proves that an inverse is unique and how it behaves with products and transposes, and finds small inverses by solving equations. Chapter 4 (Determinants) then supplies a test for invertibility and a formula for $A^{-1}$.
Invertible Matrices
📘 Definition 3.11 — Invertible matrix and inverse
Let $A$ be a square matrix of order $n$ with real entries. If there is a square matrix $B$ of the same order such that
$$AB=BA=I_n,$$
then $A$ is called invertible, and $B$ is called an inverse of $A$. We write $B=A^{-1}$ (read “$A$ inverse”). Then $A$ is also an inverse of $B$. A matrix that is not square has no inverse in this sense.
The notation $A^{-1}$ does not mean $\tfrac1A$ or “one over each entry”; it is defined only by the two products $AA^{-1}=I$ and $A^{-1}A=I$. To check that a given $B$ is the inverse of $A$, multiply and see whether you get $I$.
✏️ Example 1 — Checking a 2 × 2 inverse
Show that $B=\begin{bmatrix}3&-5\\-1&2\end{bmatrix}$ is the inverse of $A=\begin{bmatrix}2&5\\1&3\end{bmatrix}$.
💡 Show solution▶
AB $AB=\begin{bmatrix}6-5&-10+10\\3-3&-5+6\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I$.
BA $BA=\begin{bmatrix}6-5&15-15\\-2+2&-5+6\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I$.
Answer Both products are $I$, so $A$ is invertible and $A^{-1}=B$. (For square matrices it can be shown that $AB=I$ already forces $BA=I$, but the Board expects you to check the definition — or at least say which product you computed.)
✏️ Example 2 — A 3 × 3 check, and a near miss
Let $A=\begin{bmatrix}1&1&0\\0&1&1\\0&0&1\end{bmatrix}$. Which of $B=\begin{bmatrix}1&-1&1\\0&1&-1\\0&0&1\end{bmatrix}$ and $C=\begin{bmatrix}1&-1&0\\0&1&-1\\0&0&1\end{bmatrix}$ is $A^{-1}$?
💡 Show solution▶
AB Row 1 of $A$ is $(1,1,0)$: with the columns of $B$ it gives $1$, $-1+1=0$, $1-1=0$. Row 2 $(0,1,1)$ gives $0$, $1$, $-1+1=0$. Row 3 $(0,0,1)$ gives $0,0,1$. So $AB=I_3$; similarly $BA=I_3$.
AC Row 1 of $A$ with column 3 of $C$: $1\cdot0+1\cdot(-1)+0\cdot1=-1\ne0$. So
$$AC=\begin{bmatrix}1&0&-1\\0&1&0\\0&0&1\end{bmatrix}\ne I$$
Answer $A^{-1}=B$. One wrong entry is enough to spoil an inverse — and by the next theorem $C$ cannot be another inverse.
📐 Theorem 3.8 — Uniqueness of the inverse
A square matrix has at most one inverse: if $AB=BA=I$ and $AC=CA=I$, then $B=C$.
Proof. Using the identity and associative laws (Theorem 3.4),
$$B=BI=B(AC)=(BA)C=IC=C.$$
This is why we may speak of the inverse and give it the single name $A^{-1}$.
Interactive 1 — Do It, Then Undo It
Play applies $A$ to the unit square and its flag, then applies $A^{-1}$: since $A^{-1}(AX)=(A^{-1}A)X=IX=X$, everything returns home. Try the last choice: that matrix squashes the plane onto a line, two different corners land on the same point, and no matrix can separate them again — it has no inverse.
A—
A⁻¹—
A⁻¹A—
stage—
Inverses of Products, Transposes and Multiples
To undo “first put on socks, then shoes” you take off the shoes first, then the socks. Matrices behave the same way: $AB$ means “first $B$, then $A$”, so undoing it means “first undo $A$, then undo $B$”.
📐 Theorem 3.9 — Inverse of a product
If $A$ and $B$ are invertible matrices of the same order, then $AB$ is invertible and
$$(AB)^{-1}=B^{-1}A^{-1}.$$
Proof. We show that $B^{-1}A^{-1}$ satisfies the definition of an inverse of $AB$. By associativity,
$$(AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=AIA^{-1}=AA^{-1}=I,$$
$$(B^{-1}A^{-1})(AB)=B^{-1}(A^{-1}A)B=B^{-1}IB=B^{-1}B=I.$$
So $B^{-1}A^{-1}$ is an inverse of $AB$, and by Theorem 3.8 it is the inverse. ∎
📐 Theorem 3.10 — Further rules for inverses
Let $A$ be an invertible square matrix and $k\ne0$ a real number. Then
$A^{-1}$ is invertible and $(A^{-1})^{-1}=A$;
$A'$ is invertible and $(A')^{-1}=(A^{-1})'$;
$kA$ is invertible and $(kA)^{-1}=\tfrac1k A^{-1}$.
Moreover, a square matrix with a row of zeros (or a column of zeros) is not invertible.
Why. The first rule is the symmetry of $AA^{-1}=A^{-1}A=I$. For the second, transpose both equations using the reversal law: $(A^{-1})'A'=(AA^{-1})'=I'=I$ and $A'(A^{-1})'=(A^{-1}A)'=I$. For the third, $(kA)(\tfrac1kA^{-1})=\tfrac kk AA^{-1}=I$, and the same the other way round. Finally, if row $i$ of $A$ is all zeros, then row $i$ of $AB$ is all zeros for every $B$ (each entry is row $i$ of $A$ times a column of $B$), so $AB$ can never be $I$. A zero column in $A$ gives a zero column in every $BA$ in the same way.
✏️ Example 3 — The reversal in action
With $A=\begin{bmatrix}2&5\\1&3\end{bmatrix}$, $A^{-1}=\begin{bmatrix}3&-5\\-1&2\end{bmatrix}$ (Example 1), $B=\begin{bmatrix}1&2\\1&3\end{bmatrix}$ and $B^{-1}=\begin{bmatrix}3&-2\\-1&1\end{bmatrix}$, find $(AB)^{-1}$ and show that it is not $A^{-1}B^{-1}$.
Check $AB=\begin{bmatrix}7&19\\4&11\end{bmatrix}$ and $\begin{bmatrix}7&19\\4&11\end{bmatrix}\begin{bmatrix}11&-19\\-4&7\end{bmatrix}=\begin{bmatrix}77-76&-133+133\\44-44&-76+77\end{bmatrix}=I$ ✓.
Wrong order $A^{-1}B^{-1}=\begin{bmatrix}14&-11\\-5&4\end{bmatrix}$ — a different matrix, so it is not $(AB)^{-1}$.
Finding an Inverse by Solving Equations
How do we find $A^{-1}$? For a small matrix we can treat its entries as unknowns: write $X=\begin{bmatrix}x&y\\z&w\end{bmatrix}$, set $AX=I$, and solve the four ordinary equations that equality of matrices gives. If they have a solution, check that $XA=I$ as well; if they have no solution, $A$ has no inverse.
🧭 Problem-solving strategy — a 2 × 2 inverse from $AX=I$
Write $X=\begin{bmatrix}x&y\\z&w\end{bmatrix}$ and multiply out $AX$.
Equate with $I$. The first column gives two equations in $x,z$; the second column gives two equations in $y,w$.
Solve each pair. If either pair is inconsistent, stop: $A$ is not invertible.
Check $AX=I$ and $XA=I$.
✏️ Example 4 — Finding an inverse
Find the inverse of $A=\begin{bmatrix}3&1\\5&2\end{bmatrix}$.
💡 Show solution▶
Set up $AX=\begin{bmatrix}3x+z&3y+w\\5x+2z&5y+2w\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$.
Column 1 $3x+z=1$ and $5x+2z=0$. Doubling the first, $6x+2z=2$; subtracting the second, $x=2$, so $z=1-6=-5$.
Column 2 $3y+w=0$ and $5y+2w=1$. Doubling the first and subtracting: $y=-1$, so $w=3$.
Check $X=\begin{bmatrix}2&-1\\-5&3\end{bmatrix}$; $AX=\begin{bmatrix}6-5&-3+3\\10-10&-5+6\end{bmatrix}=I$ and $XA=\begin{bmatrix}6-5&2-2\\-15+15&-5+6\end{bmatrix}=I$ ✓. So $A^{-1}=\begin{bmatrix}2&-1\\-5&3\end{bmatrix}$.
✏️ Example 5 — A matrix with no inverse
Show that $A=\begin{bmatrix}1&2\\2&4\end{bmatrix}$ is not invertible, in two ways.
💡 Show solution▶
Equations $AX=I$ needs, from column 1, $x+2z=1$ and $2x+4z=0$. But the second left side is twice the first, so it would equal $2$, not $0$. No solution: no inverse.
Zero product In Section 3.2 (Example 6), $AB=O$ with $B=\begin{bmatrix}2&-6\\-1&3\end{bmatrix}\ne O$. If $A^{-1}$ existed, then $B=IB=(A^{-1}A)B=A^{-1}(AB)=A^{-1}O=O$, a contradiction.
Lesson A matrix that is a “zero divisor” ($AB=O$ with $B\ne O$) is never invertible. Conversely, an invertible $A$ can be cancelled: $AB=AC$ gives $B=A^{-1}AB=A^{-1}AC=C$.
✏️ Example 6 — Using an inverse to solve a system
Solve $3x+y=7$, $5x+2y=12$ using the inverse found in Example 4.
💡 Show solution▶
Matrix form The system is $AX=B$ with $A=\begin{bmatrix}3&1\\5&2\end{bmatrix}$, $X=\begin{bmatrix}x\\y\end{bmatrix}$, $B=\begin{bmatrix}7\\12\end{bmatrix}$.
Multiply by A⁻¹ On the left of both sides: $A^{-1}AX=A^{-1}B$, so $X=A^{-1}B=\begin{bmatrix}2&-1\\-5&3\end{bmatrix}\begin{bmatrix}7\\12\end{bmatrix}=\begin{bmatrix}14-12\\-35+36\end{bmatrix}=\begin{bmatrix}2\\1\end{bmatrix}$.
Check $x=2$, $y=1$: $6+1=7$ ✓, $10+2=12$ ✓. Chapter 4 develops this method fully, with a formula for $A^{-1}$.
Interactive 2 — A Matrix Code: Encode with A, Decode with A⁻¹
Letters become numbers ($A=1,\dots,Z=26$, space $=0$) and are taken two at a time as columns $P$. The sender transmits $C=KP$ for a key matrix $K$; the receiver computes $K^{-1}C=K^{-1}KP=P$. Play steps through the message pair by pair. With the last key, which has no inverse, two different pairs can give the same code — the message cannot be recovered.
K⁻¹—
transmitted code—
decoded message—
⚠️ Common mistakes when studying alone
1. $A^{-1}$ is not “one over each entry”. For $A=\begin{bmatrix}2&5\\1&3\end{bmatrix}$, the matrix $\begin{bmatrix}\tfrac12&\tfrac15\\1&\tfrac13\end{bmatrix}$ is not $A^{-1}$; multiply to see that you do not get $I$.
2. Reversing the order. $(AB)^{-1}=B^{-1}A^{-1}$, not $A^{-1}B^{-1}$ — exactly like $(AB)'=B'A'$.
3. Multiplying on the wrong side. From $AX=B$ you get $X=A^{-1}B$ (multiply on the left), not $BA^{-1}$. Never “divide” by a matrix.
4. Non-square matrices. In this course only square matrices have inverses; a $2\times3$ matrix has none, whatever its entries.
5. A zero entry is not a zero row. $\begin{bmatrix}0&1\\1&0\end{bmatrix}$ has zeros but is invertible (it is its own inverse). Only a whole row or column of zeros rules an inverse out at a glance.
🚀 Beyond the board syllabus (JEE Main · JEE Advanced)
In JEE Main and JEE Advanced 2026; not examined in the CBSE board exam. If a square matrix satisfies a relation $A^2-kA+lI=O$ with $l\ne0$, its inverse can be read off without solving any equations.
Method. Rewrite the relation as $A(kI-A)=lI$, i.e. $A\cdot\tfrac1l(kI-A)=I$; the same factorisation works with $A$ on the right, since $A$ commutes with $I$ and with itself. So
$$A^{-1}=\tfrac1l\,(kI-A)$$
Worked example. In the 🚀 box of Section 3.2 we found that $A=\begin{bmatrix}2&3\\-1&1\end{bmatrix}$ satisfies $A^2-3A+5I=O$. Hence $A^{-1}=\tfrac15(3I-A)=\tfrac15\begin{bmatrix}1&-3\\1&2\end{bmatrix}=\begin{bmatrix}\tfrac15&-\tfrac35\\\tfrac15&\tfrac25\end{bmatrix}$. Check: $AA^{-1}=\tfrac15\begin{bmatrix}2+3&-6+6\\-1+1&3+2\end{bmatrix}=I$ ✓. If instead $l=0$, then $A(kI-A)=O$; in that case $A$ has no inverse unless $A=kI$.
Exercises
I
Checking inverses.
1 Show that $\begin{bmatrix}4&7\\1&2\end{bmatrix}$ and $\begin{bmatrix}2&-7\\-1&4\end{bmatrix}$ are inverses of each other.
2 Show that $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ is its own inverse.
3 Is $\begin{bmatrix}-2&1\\\tfrac32&-\tfrac12\end{bmatrix}$ the inverse of $\begin{bmatrix}1&2\\3&4\end{bmatrix}$?
4 Is $\begin{bmatrix}1&1\\1&2\end{bmatrix}$ the inverse of $\begin{bmatrix}2&1\\1&1\end{bmatrix}$?
5 Verify that $\begin{bmatrix}1&0&0\\2&1&0\\3&4&1\end{bmatrix}$ and $\begin{bmatrix}1&0&0\\-2&1&0\\5&-4&1\end{bmatrix}$ are inverses.
6 Why does $\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}$ have no inverse?
7 Why does $\begin{bmatrix}0&0\\3&5\end{bmatrix}$ have no inverse?
8 Find the inverse of the diagonal matrix with diagonal entries $2,-1,4$, and check it.
💡 Show solutions▶
1 $\begin{bmatrix}8-7&-28+28\\2-2&-7+8\end{bmatrix}=I$, and in the other order $\begin{bmatrix}8-7&14-14\\-4+4&-7+8\end{bmatrix}=I$ ✓.
2 $\begin{bmatrix}1&0\\0&-1\end{bmatrix}^2=\begin{bmatrix}1&0\\0&1\end{bmatrix}=I$, so it is its own inverse (it is a reflection: doing it twice changes nothing).
3 Yes: $\begin{bmatrix}1&2\\3&4\end{bmatrix}\begin{bmatrix}-2&1\\\tfrac32&-\tfrac12\end{bmatrix}=\begin{bmatrix}-2+3&1-1\\-6+6&3-2\end{bmatrix}=I$, and the reverse product is also $I$.
5 Row 2 $(2,1,0)$ with column 1 $(1,-2,5)$: $2-2=0$; row 3 $(3,4,1)$ with columns 1, 2, 3: $3-8+5=0$, $4-4=0$, $1$. All nine entries give $I_3$, and the reverse product is $I_3$ too ✓.
6 It is not square ($2\times3$).
7 Its first row is zero, so the first row of every product $AB$ is zero and $AB\ne I$ (Theorem 3.10).
8 The diagonal matrix with entries $\tfrac12,-1,\tfrac14$: multiplying diagonal matrices multiplies the diagonal entries, giving $1,1,1$ ✓.
II
Find the inverse by solving $AX=I$, or show that there is none.
1 $\begin{bmatrix}2&1\\7&4\end{bmatrix}$
2 $\begin{bmatrix}1&3\\2&7\end{bmatrix}$
3 $\begin{bmatrix}4&5\\3&4\end{bmatrix}$
4 $\begin{bmatrix}2&3\\4&5\end{bmatrix}$
5 $\begin{bmatrix}3&0\\0&5\end{bmatrix}$
6 $\begin{bmatrix}2&-3\\-4&6\end{bmatrix}$
7 For which value of $k$ does $\begin{bmatrix}k&2\\3&6\end{bmatrix}$ have no inverse?
8 $\begin{bmatrix}1&-1\\2&3\end{bmatrix}$
💡 Show solutions▶
1 $2x+z=1$, $7x+4z=0$ give $x=4$, $z=-7$; $2y+w=0$, $7y+4w=1$ give $y=-1$, $w=2$. Inverse $\begin{bmatrix}4&-1\\-7&2\end{bmatrix}$.
4 $2x+3z=1$, $4x+5z=0$: from the second $x=-\tfrac54z$, so $-\tfrac52z+3z=1$, $z=2$, $x=-\tfrac52$. Similarly $y=\tfrac32$, $w=-1$. Inverse $\begin{bmatrix}-\tfrac52&\tfrac32\\2&-1\end{bmatrix}$.
5 Multiply on the left by $A^{-1}$: $A^{-1}(AB)=A^{-1}(AC)$, so $(A^{-1}A)B=(A^{-1}A)C$, i.e. $IB=IC$ and $B=C$.
6 $B=IB=A^{-1}AB=A^{-1}O=O$.
7 $AA=I$ says that $A$ itself satisfies both products in Definition 3.11, so $A^{-1}=A$. Such a matrix is called involutory. Example: $\begin{bmatrix}0&1\\1&0\end{bmatrix}$ (reflection in $y=x$).
8 $(A^{-1})'=(A')^{-1}=A^{-1}$, using Theorem 3.10 and $A'=A$.
IV
Applications.
1Secret message. With the key $K=\begin{bmatrix}3&1\\5&2\end{bmatrix}$ (so $K^{-1}=\begin{bmatrix}2&-1\\-5&3\end{bmatrix}$) and $A=1,\dots,Z=26$, a four-letter word was sent as the columns of $\begin{bmatrix}40&68\\67&116\end{bmatrix}$. Decode it.
2 Using $A^{-1}$ from Example 1, solve $2x+5y=1$, $x+3y=1$.
3Bakery. A batch of cakes uses $2$ kg of flour and $1$ kg of sugar; a batch of cookies uses $1$ kg of each. One day the bakery used $14$ kg of flour and $9$ kg of sugar. Write this as $AX=B$, find $A^{-1}$ and hence the number of batches of each.
4 Explain why a key matrix with a row of zeros is useless for sending secret messages.
5 With the key $\begin{bmatrix}1&2\\2&4\end{bmatrix}$, show that the letter pairs “B␣” $\left(\begin{bmatrix}2\\0\end{bmatrix}\right)$ and “␣A” $\left(\begin{bmatrix}0\\1\end{bmatrix}\right)$ are sent as the same code. What does this tell you?
💡 Show solutions▶
1 $K^{-1}\begin{bmatrix}40&68\\67&116\end{bmatrix}=\begin{bmatrix}80-67&136-116\\-200+201&-340+348\end{bmatrix}=\begin{bmatrix}13&20\\1&8\end{bmatrix}$: $13,1,20,8$ = M, A, T, H — the word is MATH.
3 With $x$ batches of cakes and $y$ of cookies: $\begin{bmatrix}2&1\\1&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}14\\9\end{bmatrix}$. $A^{-1}=\begin{bmatrix}1&-1\\-1&2\end{bmatrix}$ (check $AA^{-1}=I$), so $X=\begin{bmatrix}14-9\\-14+18\end{bmatrix}=\begin{bmatrix}5\\4\end{bmatrix}$: $5$ batches of cakes and $4$ of cookies. Check: $10+4=14$ kg, $5+4=9$ kg ✓.
4 If row $i$ of $K$ is zero, entry $i$ of every code $KP$ is $0$ whatever the message, so information is lost; and $K$ has no inverse (Theorem 3.10), so the receiver cannot decode.
5 $\begin{bmatrix}1&2\\2&4\end{bmatrix}\begin{bmatrix}2\\0\end{bmatrix}=\begin{bmatrix}2\\4\end{bmatrix}$ and $\begin{bmatrix}1&2\\2&4\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}2\\4\end{bmatrix}$. Two different messages, one code: the receiver cannot know which was sent, so no decoding matrix can exist — the key has no inverse.
How this is examined: in the Board exam this section supplies short questions — checking an inverse, the reversal law $(AB)^{-1}=B^{-1}A^{-1}$ and its proof — while long inverse calculations come with determinants in Chapter 4. JEE likes inverses from a polynomial relation and “which statements are always true” questions.
E1.Board · 1 mark If $A=\begin{bmatrix}2&3\\1&2\end{bmatrix}$, then $A^{-1}$ is (A) $\begin{bmatrix}2&-3\\-1&2\end{bmatrix}$ (B) $\begin{bmatrix}-2&3\\1&-2\end{bmatrix}$ (C) $\begin{bmatrix}2&1\\3&2\end{bmatrix}$ (D) $\begin{bmatrix}\tfrac12&\tfrac13\\1&\tfrac12\end{bmatrix}$
💡 Show solution▶
Idea Test each option by multiplying: $\begin{bmatrix}2&3\\1&2\end{bmatrix}\begin{bmatrix}2&-3\\-1&2\end{bmatrix}=\begin{bmatrix}4-3&-6+6\\2-2&-3+4\end{bmatrix}=I$, and the reverse product is $I$ too.
Answer (A). (B) is $-A^{-1}$ and gives $-I$; (C) is $A'$; (D) is the “one over each entry” mistake.
E2.Board · 1 markAssertion–Reason. Assertion (A): the matrix $\begin{bmatrix}1&2\\0&0\end{bmatrix}$ is not invertible. Reason (R): if a square matrix $A$ has a row of zeros, then $AB$ has a row of zeros for every matrix $B$ of the same order. (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is not the correct explanation of A (C) A is true but R is false (D) A is false but R is true
💡 Show solution▶
Idea R is true (each entry of row $i$ of $AB$ is row $i$ of $A$ times a column of $B$), and it explains A: $AB$ always has a zero second row, so it is never $I$.
Answer (A).
E3.Board · 2 marks If $A$ and $B$ are invertible matrices of the same order, prove that $(AB)^{-1}=B^{-1}A^{-1}$.
💡 Show solution▶
Products $(AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=AIA^{-1}=AA^{-1}=I$ and $(B^{-1}A^{-1})(AB)=B^{-1}(A^{-1}A)B=B^{-1}B=I$. (1 mark)
Conclusion So $B^{-1}A^{-1}$ is an inverse of $AB$; since the inverse is unique, $(AB)^{-1}=B^{-1}A^{-1}$. (1 mark — do not forget to quote associativity and uniqueness.)
E4.Board · 3 marks Let $A=\begin{bmatrix}4&3\\1&1\end{bmatrix}$ and $B=\begin{bmatrix}2&1\\5&3\end{bmatrix}$. Verify that $A^{-1}=\begin{bmatrix}1&-3\\-1&4\end{bmatrix}$ and $B^{-1}=\begin{bmatrix}3&-1\\-5&2\end{bmatrix}$, and hence find $(AB)^{-1}$. Check your answer by multiplying it by $AB$.
💡 Show solution▶
Verify $AA^{-1}=\begin{bmatrix}4-3&-12+12\\1-1&-3+4\end{bmatrix}=I$ and $BB^{-1}=\begin{bmatrix}6-5&-2+2\\15-15&-5+6\end{bmatrix}=I$ (and the reverse products likewise). (1 mark)
Reversal law $(AB)^{-1}=B^{-1}A^{-1}=\begin{bmatrix}3+1&-9-4\\-5-2&15+8\end{bmatrix}=\begin{bmatrix}4&-13\\-7&23\end{bmatrix}$. (1 mark)
Check $AB=\begin{bmatrix}23&13\\7&4\end{bmatrix}$ and $\begin{bmatrix}23&13\\7&4\end{bmatrix}\begin{bmatrix}4&-13\\-7&23\end{bmatrix}=\begin{bmatrix}92-91&-299+299\\28-28&-91+92\end{bmatrix}=I$ ✓. (1 mark)
E5.JEE Main A square matrix $A$ satisfies $A^2-3A+2I=O$. Then $A^{-1}$ equals (A) $\tfrac12(3I-A)$ (B) $\tfrac12(A-3I)$ (C) $3I-A$ (D) $2(3I-A)$
💡 Show solution▶
Method $A^2-3A=-2I$, so $A(3I-A)=3A-A^2=2I$, i.e. $A\cdot\tfrac12(3I-A)=I$; the same holds with the factors swapped (🚀 box above).
Answer (A). (B) has the wrong sign; (C) forgets to divide by $2$. Example: $A=\begin{bmatrix}1&1\\0&2\end{bmatrix}$ satisfies the relation and $\tfrac12(3I-A)=\begin{bmatrix}1&-\tfrac12\\0&\tfrac12\end{bmatrix}$ is its inverse.
E6.JEE Advanced(One or more options correct.) Let $A$ and $B$ be invertible $n\times n$ matrices. Which statements are always true? (A) $(AB)^{-1}=A^{-1}B^{-1}$ (B) $(A')^{-1}=(A^{-1})'$ (C) $A+B$ is invertible (D) $\left(A^{-1}BA\right)^2=A^{-1}B^2A$
💡 Show solution▶
(A) False The order must reverse; Example 3 gives a pair with $A^{-1}B^{-1}\ne(AB)^{-1}$.
(B) True Theorem 3.10.
(C) False $A=I$ and $B=-I$ are invertible, but $A+B=O$ is not.
Answer (B), (D). ⏱ Fast route for (C): one counterexample is enough, and $\pm I$ is the quickest to try.
🔬 Research–Teaching NexusMini project · about 60 minutes
Research question. How good is a matrix code? Build a spreadsheet that encodes and decodes messages with a $2\times2$ key, and investigate which keys work and how easily the code can be broken.
Why it matters. Every UPI payment, WhatsApp message and Aadhaar authentication is protected by encryption. Modern ciphers are far more elaborate, but they rest on the same principle you met here: an operation that is easy to undo if you know the key and hard to undo if you do not. Matrix codes of this kind were among the first ciphers built on algebra.
What to do.
In a spreadsheet, turn a message into numbers ($A=1,\dots,Z=26$, space $=0$) and arrange them two per column in a $2\times k$ matrix $P$.
Choose a key $K$ with integer entries whose inverse also has integer entries (for example $\begin{bmatrix}3&1\\5&2\end{bmatrix}$). Compute $C=KP$ with MMULT, and decode with $K^{-1}C$.
Swap codes with a partner who does not know your key. Can they decode your message if you tell them two letters of it? (Hint: two known columns give $KP_0=C_0$ for a $2\times2$ block, so $K=C_0P_0^{-1}$.)
Try the key $\begin{bmatrix}1&2\\2&4\end{bmatrix}$ and find two different messages with the same code.
Tools: Spreadsheet (MMULT, CODE/CHAR), Interactive 2 for checking.
Hand in: Your encoding sheet, one coded message with its key, a decoded message from a partner, and a short paragraph on what makes a key “good” or “bad”.
Going further. Real matrix ciphers work “modulo 26” so that codes stay between $0$ and $25$. Find out what that means and which keys then have inverses.
Instructor note: keys with integer inverses are those whose “$ad-bc$” is $\pm1$ — a fact students meet properly in Chapter 4. The known-plaintext attack in step 3 needs $P_0$ to be invertible; choose the two known letter pairs accordingly.
🧪 Self-check quiz
0/5 correct
Q1. In this course an inverse can exist only for
Q2. If $A$ and $B$ are invertible of the same order, $(AB)^{-1}$ equals
Q3. The inverse of $\begin{bmatrix}0&1\\1&0\end{bmatrix}$ is
Q4. If $A^{-1}=\begin{bmatrix}1&2\\0&1\end{bmatrix}$, then $A$ is