Chapter 4 · Determinants

4.2 Minors, Cofactors, Adjoint and Inverse

Delete a row and a column and a smaller determinant is left: a minor. Give it a sign and it becomes a cofactor. Arrange the cofactors, transpose, and you have the adjoint — and with it a formula for the inverse of any non-singular matrix.

In Section 3.4 you found the inverse of a $2\times2$ matrix by solving $AX=I$ for four unknowns, and noticed that the number $ad-bc$ kept turning up in the denominators. For a $3\times3$ matrix the same method means solving for nine unknowns — slow, and easy to get wrong. Is there a formula instead, one that says in advance whether the inverse exists and what it is?

There is, and it is built entirely from determinants. Each entry of a square matrix has a small determinant attached to it (its minor), which with the right sign becomes its cofactor. Arranging the cofactors in a matrix and transposing gives the adjoint $\operatorname{adj}A$, and one identity, $A(\operatorname{adj}A)=|A|\,I$, does all the work: when $|A|\ne0$, dividing by $|A|$ gives $A^{-1}$. When $|A|=0$ — a singular matrix — there is no inverse at all.

Minors and Cofactors

When you expanded a $3\times3$ determinant in Section 4.1, each entry $a_{ij}$ was multiplied by a sign $(-1)^{i+j}$ and by the $2\times2$ determinant left after deleting its row and column. These two ingredients get names.

📘 Definition 4.3 — Minor
Let $A=[a_{ij}]$ be a square matrix of order $n\ge2$. The minor $M_{ij}$ of the entry $a_{ij}$ is the determinant of the matrix of order $n-1$ obtained by deleting the $i$-th row and the $j$-th column of $A$. (For $n=2$ the minor is a single number; for $n=3$ it is a $2\times2$ determinant.)
📘 Definition 4.4 — Cofactor
The cofactor of $a_{ij}$ is $$A_{ij}=(-1)^{i+j}M_{ij}.$$ So $A_{ij}=M_{ij}$ when $i+j$ is even and $A_{ij}=-M_{ij}$ when $i+j$ is odd — the chessboard of signs from Section 4.1.

With this language the expansion rule of Theorem 4.1 becomes very short, and it has a surprising companion.

📐 Theorem 4.3 — Expanding with cofactors
Let $A=[a_{ij}]$ be a square matrix of order $3$.

(a) For any row $i$: $\ |A|=a_{i1}A_{i1}+a_{i2}A_{i2}+a_{i3}A_{i3}$; and for any column $j$: $\ |A|=a_{1j}A_{1j}+a_{2j}A_{2j}+a_{3j}A_{3j}$.

(b) If the entries of one row are multiplied by the cofactors of a different row, the sum is zero: for $i\ne k$, $$a_{i1}A_{k1}+a_{i2}A_{k2}+a_{i3}A_{k3}=0,$$ and the same holds for two different columns.

Why it is true. (a) is Theorem 4.1 written with cofactors. For (b), take $i=1$, $k=2$ (the other cases are the same). The cofactors $A_{21},A_{22},A_{23}$ do not use row 2 at all. So $a_{11}A_{21}+a_{12}A_{22}+a_{13}A_{23}$ is, by (a), the expansion along row 2 of the determinant $B$ obtained from $A$ by replacing row 2 with a copy of row 1. Now evaluate $B$ along its third row instead: each $2\times2$ minor consists of the shortened rows 1 and 2, which are equal, so each is of the form $\begin{vmatrix}p&q\\p&q\end{vmatrix}=pq-qp=0$. Hence $B=0$, which is (b). $\square$

✏️ Example 1 — Minors and cofactors
(a) Find all minors and cofactors of $\begin{bmatrix}3&-2\\5&4\end{bmatrix}$.  (b) For $A=\begin{bmatrix}1&2&-1\\0&3&4\\-2&1&5\end{bmatrix}$, find the minors and cofactors of the entries of the first row, and use them to find $|A|$.
(a) Minors Deleting row 1 and column 1 leaves $4$: $M_{11}=4$. Likewise $M_{12}=5$, $M_{21}=-2$, $M_{22}=3$ (for order 2 each minor is just the “opposite” entry).
(a) Cofactors Signs $\begin{smallmatrix}+&-\\-&+\end{smallmatrix}$: $A_{11}=4$, $A_{12}=-5$, $A_{21}=-(-2)=2$, $A_{22}=3$.
(b) Row 1 $M_{11}=\begin{vmatrix}3&4\\1&5\end{vmatrix}=11$, $M_{12}=\begin{vmatrix}0&4\\-2&5\end{vmatrix}=0+8=8$, $M_{13}=\begin{vmatrix}0&3\\-2&1\end{vmatrix}=0+6=6$. Cofactors: $A_{11}=11$, $A_{12}=-8$, $A_{13}=6$.
(b) $|A|$ By Theorem 4.3(a): $|A|=1(11)+2(-8)+(-1)(6)=11-16-6=-11$.
Check Along column 1 (entries $1,0,-2$): $A_{31}=+\begin{vmatrix}2&-1\\3&4\end{vmatrix}=11$, so $|A|=1(11)+0+(-2)(11)=-11$ ✓.
⚠️ Minor or cofactor?

A minor never has a sign attached; a cofactor always does. They differ only in the positions where $i+j$ is odd: $(1,2)$, $(2,1)$, $(2,3)$, $(3,2)$. Read the question carefully — “find $M_{23}$” and “find $A_{23}$” have opposite answers unless the value is $0$. Note also the notation: $A_{ij}$ (capital) is a cofactor, $a_{ij}$ (small) is an entry.

The Adjoint and the Identity $A(\operatorname{adj}A)=|A|\,I$

Theorem 4.3 says: a row of entries “against” its own cofactors gives $|A|$, and “against” the cofactors of another row gives $0$. That is exactly the pattern of a matrix product whose answer is $|A|$ on the diagonal and $0$ elsewhere — provided we put the cofactors of row $k$ into column $k$. That is why the cofactor matrix is transposed.

📘 Definition 4.5 — Adjoint of a matrix
Let $A=[a_{ij}]$ be a square matrix of order $n$ and let $A_{ij}$ be the cofactor of $a_{ij}$. The adjoint of $A$ is the transpose of the matrix of cofactors: $$\operatorname{adj}A=[A_{ij}]'=[A_{ji}].$$ For order $3$: $$\operatorname{adj}A=\begin{bmatrix}A_{11}&A_{21}&A_{31}\\A_{12}&A_{22}&A_{32}\\A_{13}&A_{23}&A_{33}\end{bmatrix}.$$ The cofactors of row $i$ of $A$ fill column $i$ of $\operatorname{adj}A$.

For a $2\times2$ matrix $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ the cofactors are $A_{11}=d$, $A_{12}=-c$, $A_{21}=-b$, $A_{22}=a$, so $$\operatorname{adj}\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}:$$ swap the diagonal entries and change the signs of the other two. This shortcut is worth memorising — but it works for order 2 only.

📐 Theorem 4.4 — The adjoint identity
For every square matrix $A$ of order $n$, $$A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|\,I_n,$$ where $I_n$ is the identity matrix of order $n$.

Proof (order 3). The entry in row $i$, column $k$ of $A(\operatorname{adj}A)$ is row $i$ of $A$ times column $k$ of $\operatorname{adj}A$. Column $k$ of $\operatorname{adj}A$ holds the cofactors of row $k$ of $A$, so the entry is $$a_{i1}A_{k1}+a_{i2}A_{k2}+a_{i3}A_{k3}.$$ By Theorem 4.3 this is $|A|$ when $i=k$ and $0$ when $i\ne k$. So $A(\operatorname{adj}A)$ has $|A|$ in every diagonal position and $0$ elsewhere: it is $|A|\,I_3$. The product $(\operatorname{adj}A)A$ pairs cofactors of a column with entries of a column, and the column versions of Theorem 4.3 give $|A|\,I_3$ again. The $2\times2$ case can be checked directly (Exercise II). $\square$

✏️ Example 2 — Building the adjoint and checking Theorem 4.4
For $A=\begin{bmatrix}1&2&-1\\0&3&4\\-2&1&5\end{bmatrix}$ (Example 1), find $\operatorname{adj}A$ and verify that $A(\operatorname{adj}A)=|A|\,I$.
Row 1 From Example 1: $A_{11}=11$, $A_{12}=-8$, $A_{13}=6$.
Row 2 $A_{21}=-\begin{vmatrix}2&-1\\1&5\end{vmatrix}=-11$, $A_{22}=+\begin{vmatrix}1&-1\\-2&5\end{vmatrix}=3$, $A_{23}=-\begin{vmatrix}1&2\\-2&1\end{vmatrix}=-5$.
Row 3 $A_{31}=+\begin{vmatrix}2&-1\\3&4\end{vmatrix}=11$, $A_{32}=-\begin{vmatrix}1&-1\\0&4\end{vmatrix}=-4$, $A_{33}=+\begin{vmatrix}1&2\\0&3\end{vmatrix}=3$.
Transpose Cofactors of row $i$ go into column $i$: $$\operatorname{adj}A=\begin{bmatrix}11&-11&11\\-8&3&-4\\6&-5&3\end{bmatrix}.$$
Multiply Row 1 of $A$ times the columns of $\operatorname{adj}A$: $11-16-6=-11$; $\ -11+6+5=0$; $\ 11-8-3=0$. Row 2: $0-24+24=0$; $\ 0+9-20=-11$; $\ 0-12+12=0$. Row 3: $-22-8+30=0$; $\ 22+3-25=0$; $\ -22-4+15=-11$. So $$A(\operatorname{adj}A)=\begin{bmatrix}-11&0&0\\0&-11&0\\0&0&-11\end{bmatrix}=-11\,I,$$ and $|A|=-11$ (Example 1) ✓.
Interactive 1 — Cofactor Builder

Press Play (or Next) to visit the nine entries of $A$ one by one. For each entry the lab strikes out its row and column, evaluates the blue minor, attaches the chessboard sign and writes the cofactor into the cofactor matrix. When all nine are in, the cofactor matrix is transposed — row $i$ becomes column $i$ — to give $\operatorname{adj}A$, and the lab checks $A(\operatorname{adj}A)=|A|\,I$.

entry—
minor $M_{ij}$—
cofactor $A_{ij}$—
$A(\operatorname{adj}A)$—

Singular Matrices and the Inverse by the Adjoint

📘 Definition 4.6 — Singular and non-singular matrices
A square matrix $A$ is singular if $|A|=0$, and non-singular if $|A|\ne0$.

To connect singularity with inverses we need one more fact, which we state without proof (for $2\times2$ matrices you can check it by multiplying out — Exercise III).

📐 Theorem 4.5 — Determinant of a product
If $A$ and $B$ are square matrices of the same order, then $|AB|=|A|\,|B|$.

Recall from Section 3.4 that a square matrix $A$ is invertible when there is a matrix $B$ of the same order with $AB=BA=I$; then $B$ is unique and is written $A^{-1}$.

📐 Theorem 4.6 — Invertible if and only if non-singular
A square matrix $A$ is invertible if and only if it is non-singular, i.e. $|A|\ne0$. In that case $$A^{-1}=\frac1{|A|}\operatorname{adj}A.$$

Proof. Suppose $A$ is invertible. Then $AA^{-1}=I$, and Theorem 4.5 gives $|A|\,|A^{-1}|=|I|=1$. A product of two numbers equal to $1$ cannot have a zero factor, so $|A|\ne0$. Conversely, suppose $|A|\ne0$ and put $B=\tfrac1{|A|}\operatorname{adj}A$. By Theorem 4.4, $$AB=\tfrac1{|A|}\,A(\operatorname{adj}A)=\tfrac1{|A|}\,|A|\,I=I,$$ and in the same way $BA=I$. So $A$ is invertible and, since the inverse is unique, $A^{-1}=B$. $\square$

The proof also shows that $|A^{-1}|=\dfrac1{|A|}$ for every non-singular $A$. For a $2\times2$ matrix the formula reads $$\begin{bmatrix}a&b\\c&d\end{bmatrix}^{-1}=\frac1{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$ whenever $ad-bc\ne0$, which explains the denominators you met in Section 3.4.

🧭 Problem-solving strategy — finding $A^{-1}$ by the adjoint
  1. Find $|A|$. If $|A|=0$, stop: $A$ is singular and has no inverse.
  2. Find all the cofactors $A_{ij}=(-1)^{i+j}M_{ij}$ — nine of them for order 3. Write them in a $3\times3$ array as you go.
  3. Transpose the array to get $\operatorname{adj}A$ (row $i$ of cofactors becomes column $i$).
  4. Divide every entry by $|A|$ (or keep $\tfrac1{|A|}$ outside the matrix).
  5. Check one or two entries of $AA^{-1}$ — the diagonal ones should be $1$, the others $0$. A wrong sign in one cofactor shows up at once.
✏️ Example 3 — Singular or not?
(a) Is $\begin{bmatrix}6&4\\3&2\end{bmatrix}$ singular?  (b) For which value of $k$ is $\begin{bmatrix}1&1&1\\1&2&3\\1&3&k\end{bmatrix}$ singular?
(a) $6\cdot2-3\cdot4=0$: singular, so it has no inverse. (Its second row is half the first — the two rows carry the same information.)
(b) Along row 1: $1(2k-9)-1(k-3)+1(3-2)=k-5$. Singular exactly when $k=5$.
Check At $k=5$ the third row $(1,3,5)$ equals $2\times$ row 2 $-$ row 1 $=(2-1,4-1,6-1)$ ✓ — again one row is built from the others.
✏️ Example 4 — Inverse of a 2 × 2 matrix
Find the inverse of $A=\begin{bmatrix}3&5\\2&4\end{bmatrix}$ and check your answer.
Determinant $|A|=12-10=2\ne0$, so $A^{-1}$ exists.
Adjoint Swap the diagonal, negate the rest: $\operatorname{adj}A=\begin{bmatrix}4&-5\\-2&3\end{bmatrix}$.
Inverse $$A^{-1}=\frac12\begin{bmatrix}4&-5\\-2&3\end{bmatrix}=\begin{bmatrix}2&-\tfrac52\\-1&\tfrac32\end{bmatrix}.$$
Check $AA^{-1}$: row 1: $3\cdot2+5\cdot(-1)=1$, $\ 3\cdot(-\tfrac52)+5\cdot\tfrac32=0$; row 2: $2\cdot2+4\cdot(-1)=0$, $\ 2\cdot(-\tfrac52)+4\cdot\tfrac32=1$. So $AA^{-1}=I$ ✓.
✏️ Example 5 — Inverse of a 3 × 3 matrix
Find $A^{-1}$ for $A=\begin{bmatrix}2&1&-1\\1&1&0\\3&0&1\end{bmatrix}$.
Determinant Along row 2 (it has a zero; signs $-,+,-$): $-1\begin{vmatrix}1&-1\\0&1\end{vmatrix}+1\begin{vmatrix}2&-1\\3&1\end{vmatrix}-0=-1+5=4\ne0$.
Cofactors Row 1: $A_{11}=+(1-0)=1$, $A_{12}=-(1-0)=-1$, $A_{13}=+(0-3)=-3$. Row 2: $A_{21}=-(1-0)=-1$, $A_{22}=+(2+3)=5$, $A_{23}=-(0-3)=3$. Row 3: $A_{31}=+(0+1)=1$, $A_{32}=-(0+1)=-1$, $A_{33}=+(2-1)=1$.
Adjoint Transpose the cofactor array: $$\operatorname{adj}A=\begin{bmatrix}1&-1&1\\-1&5&-1\\-3&3&1\end{bmatrix}.$$
Inverse $$A^{-1}=\frac14\begin{bmatrix}1&-1&1\\-1&5&-1\\-3&3&1\end{bmatrix}.$$
Check Row 1 of $A$ times column 1 of $\operatorname{adj}A$: $2(1)+1(-1)+(-1)(-3)=4$; times column 2: $2(-1)+1(5)+(-1)(3)=0$. Row 3 times column 3: $3(1)+0+1(1)=4$. So $A(\operatorname{adj}A)=4I$ and $AA^{-1}=I$ ✓.
Interactive 2 — Singular-or-Not Lab

One entry of $A$ is a parameter $k$ (shown in red). Move $k$ or press Play. The lab shows $\operatorname{adj}A$ and the product $A(\operatorname{adj}A)$, which is always $|A|\,I$, and — when $|A|\ne0$ — the inverse $A^{-1}=\tfrac1{|A|}\operatorname{adj}A$ in exact fractions. The strip at the bottom plots $|A|$ against $k$: where the line crosses $0$, $A$ is singular, $A(\operatorname{adj}A)$ collapses to the zero matrix and the inverse disappears.

$k$—
$|A|$—
$A(\operatorname{adj}A)$—
$A^{-1}$—
✏️ Example 6 — Products: determinants and inverses
Let $P=\begin{bmatrix}4&7\\1&2\end{bmatrix}$ and $R=\begin{bmatrix}3&5\\2&4\end{bmatrix}$. (a) Verify that $|PR|=|P|\,|R|$. (b) Verify that $(PR)^{-1}=R^{-1}P^{-1}$.
(a) $PR=\begin{bmatrix}12+14&20+28\\3+4&5+8\end{bmatrix}=\begin{bmatrix}26&48\\7&13\end{bmatrix}$, so $|PR|=338-336=2$. Also $|P|=8-7=1$ and $|R|=2$: $|P|\,|R|=2$ ✓ (Theorem 4.5).
(b) Left side $$(PR)^{-1}=\frac12\begin{bmatrix}13&-48\\-7&26\end{bmatrix}=\begin{bmatrix}\tfrac{13}2&-24\\-\tfrac72&13\end{bmatrix}.$$
(b) Right side $P^{-1}=\begin{bmatrix}2&-7\\-1&4\end{bmatrix}$ ($|P|=1$) and $R^{-1}=\begin{bmatrix}2&-\tfrac52\\-1&\tfrac32\end{bmatrix}$ (Example 4). Then $$R^{-1}P^{-1}=\begin{bmatrix}4+\tfrac52&-14-10\\-2-\tfrac32&7+6\end{bmatrix}=\begin{bmatrix}\tfrac{13}2&-24\\-\tfrac72&13\end{bmatrix}$$ — the same, as Theorem 3.9 of Section 3.4 promises.
Note The order matters: $P^{-1}R^{-1}=\begin{bmatrix}11&-\tfrac{31}2\\-6&\tfrac{17}2\end{bmatrix}$ is not the inverse of $PR$.
⚠️ Common mistakes when studying alone

1. Forgetting to transpose. The adjoint is the transpose of the cofactor matrix. If you skip this step for a non-symmetric matrix, $A\cdot(\text{your answer})$ will not be $|A|\,I$.

2. Minors instead of cofactors. Four of the nine entries need a sign change. Mark the chessboard before you start.

3. The $2\times2$ shortcut on a $3\times3$ matrix. “Swap the diagonal and negate the rest” works for order 2 only.

4. Dividing by $0$. Always find $|A|$ first. If it is $0$, write “$A$ is singular, so $A^{-1}$ does not exist” — do not go on to compute an adjoint and divide.

5. Reversing a product. $(AB)^{-1}=B^{-1}A^{-1}$, not $A^{-1}B^{-1}$ (Example 6).

🚀 Beyond the board syllabus (JEE Main · JEE Advanced)

In JEE Main and JEE Advanced 2026; not examined in the CBSE board exam. JEE questions often ask for determinants of $kA$, $\operatorname{adj}A$ or combinations, for a square matrix $A$ of order $n$:

(i) $|kA|=k^n|A|$ — each of the $n$ rows is multiplied by $k$, and a row factor comes outside the determinant (property P4 of the 🚀 box in Section 4.1).  (ii) $|\operatorname{adj}A|=|A|^{n-1}$ — take determinants in $A(\operatorname{adj}A)=|A|\,I_n$: by Theorem 4.5 and (i), $|A|\,|\operatorname{adj}A|=|A|^n$, and divide by $|A|$ when $|A|\ne0$ (the formula also holds when $|A|=0$).  (iii) Consequently $|A^{-1}|=|A|^{-1}$ and $\operatorname{adj}(\operatorname{adj}A)=|A|^{n-2}A$ for non-singular $A$.

Worked example. $A$ is $3\times3$ with $|A|=-4$. Find $|2A|$, $|\operatorname{adj}A|$ and $|\operatorname{adj}(2A)|$. Solution. $|2A|=2^3(-4)=-32$; $|\operatorname{adj}A|=(-4)^2=16$; $|\operatorname{adj}(2A)|=|2A|^2=(-32)^2=1024$. Check with $A=\operatorname{diag}(1,1,-4)$: $2A=\operatorname{diag}(2,2,-8)$ has determinant $-32$, and its adjoint is $\operatorname{diag}(-16,-16,4)$, with determinant $1024$ ✓.

Exercises

I
Minors and cofactors.
  • 1 Find all minors and cofactors of $\begin{bmatrix}2&-5\\6&1\end{bmatrix}$.
  • 2 Find all minors and cofactors of $\begin{bmatrix}-1&4\\3&0\end{bmatrix}$.
  • 3 Find all nine cofactors of $A=\begin{bmatrix}1&0&2\\3&-1&4\\0&5&-2\end{bmatrix}$, and use row 1 to find $|A|$.
  • 4 For $B=\begin{bmatrix}4&-1&2\\0&3&1\\5&2&-3\end{bmatrix}$, find the cofactors of the entries of row 2 and hence $|B|$.
  • 5 For the matrix $B$ of part 4, verify Theorem 4.3(b): $b_{11}B_{21}+b_{12}B_{22}+b_{13}B_{23}=0$.
  • 6 In $\begin{bmatrix}x&2&1\\3&-1&4\\x&5&2\end{bmatrix}$ the minor $M_{23}$ equals $12$. Find $x$ and the cofactor $A_{23}$.
1 $M_{11}=1$, $M_{12}=6$, $M_{21}=-5$, $M_{22}=2$; $A_{11}=1$, $A_{12}=-6$, $A_{21}=5$, $A_{22}=2$.
2 $M_{11}=0$, $M_{12}=3$, $M_{21}=4$, $M_{22}=-1$; $A_{11}=0$, $A_{12}=-3$, $A_{21}=-4$, $A_{22}=-1$.
3 Row 1: $A_{11}=(2-20)=-18$, $A_{12}=-(-6-0)=6$, $A_{13}=(15-0)=15$. Row 2: $A_{21}=-(0-10)=10$, $A_{22}=(-2-0)=-2$, $A_{23}=-(5-0)=-5$. Row 3: $A_{31}=(0+2)=2$, $A_{32}=-(4-6)=2$, $A_{33}=(-1-0)=-1$. Then $|A|=1(-18)+0(6)+2(15)=12$.
4 $B_{21}=-\begin{vmatrix}-1&2\\2&-3\end{vmatrix}=-(3-4)=1$, $B_{22}=+\begin{vmatrix}4&2\\5&-3\end{vmatrix}=-22$, $B_{23}=-\begin{vmatrix}4&-1\\5&2\end{vmatrix}=-13$. $|B|=0(1)+3(-22)+1(-13)=-79$.
5 $4(1)+(-1)(-22)+2(-13)=4+22-26=0$ ✓.
6 Delete row 2 and column 3: $M_{23}=\begin{vmatrix}x&2\\x&5\end{vmatrix}=5x-2x=3x=12$, so $x=4$, and $A_{23}=(-1)^5M_{23}=-12$.
II
Adjoints and the identity $A(\operatorname{adj}A)=|A|\,I$.
  • 1 Find $\operatorname{adj}\begin{bmatrix}5&-3\\2&7\end{bmatrix}$.
  • 2 Find $\operatorname{adj}\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}$.
  • 3 Find $\operatorname{adj}A$ for the matrix $A$ of Exercise I.3 and verify that $A(\operatorname{adj}A)=12I$.
  • 4 Find $\operatorname{adj}C$ for $C=\begin{bmatrix}2&3&1\\1&2&1\\1&1&1\end{bmatrix}$ and verify Theorem 4.4.
  • 5 For the singular matrix $S=\begin{bmatrix}1&1&1\\1&2&3\\1&3&5\end{bmatrix}$ (Example 3 with $k=5$), find $\operatorname{adj}S$ and show that $S(\operatorname{adj}S)=O$.
  • 6 For $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$, verify Theorem 4.4 by multiplying out, and show that $\operatorname{adj}(\operatorname{adj}A)=A$.
1 Swap the diagonal, negate the rest: $\begin{bmatrix}7&3\\-2&5\end{bmatrix}$.
2 $\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$.
3 Transposing the cofactors of I.3: $$\operatorname{adj}A=\begin{bmatrix}-18&10&2\\6&-2&2\\15&-5&-1\end{bmatrix}.$$ Row 1 of $A$ times the columns: $-18+0+30=12$, $10+0-10=0$, $2+0-2=0$; row 2: $-54-6+60=0$, $30+2-20=12$, $6-2-4=0$; row 3: $0+30-30=0$, $0-10+10=0$, $0+10+2=12$. So $A(\operatorname{adj}A)=12I$ ✓.
4 $|C|=2(2-1)-3(1-1)+1(1-2)=1$. Cofactors by rows: $(1,0,-1)$, $(-2,1,1)$, $(1,-1,1)$, so $$\operatorname{adj}C=\begin{bmatrix}1&-2&1\\0&1&-1\\-1&1&1\end{bmatrix},$$ and $C(\operatorname{adj}C)=I=|C|\,I$ ✓ (e.g. row 1 times column 2: $-4+3+1=0$).
5 Cofactors by rows: $(1,-2,1)$, $(-2,4,-2)$, $(1,-2,1)$; the cofactor matrix is symmetric, so $$\operatorname{adj}S=\begin{bmatrix}1&-2&1\\-2&4&-2\\1&-2&1\end{bmatrix}.$$ Each row of $S$ times $(1,-2,1)$: $1-2+1=0$, $1-4+3=0$, $1-6+5=0$, and the other columns are multiples of this one. So $S(\operatorname{adj}S)=O=0\cdot I$ ✓.
6 $\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}=\begin{bmatrix}ad-bc&0\\0&ad-bc\end{bmatrix}$, and similarly in the other order. Applying the shortcut twice swaps the diagonal back and negates the off-diagonal entries twice: $\operatorname{adj}(\operatorname{adj}A)=A$.
III
Inverses. Where the inverse exists, find it; otherwise say why not.
  • 1 $\begin{bmatrix}2&5\\1&3\end{bmatrix}$
  • 2 $\begin{bmatrix}4&-3\\2&1\end{bmatrix}$
  • 3 $\begin{bmatrix}-2&6\\1&-3\end{bmatrix}$
  • 4 $\begin{bmatrix}1&0&0\\2&1&0\\3&2&1\end{bmatrix}$
  • 5 $\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}$
  • 6 $\begin{bmatrix}2&-1&3\\1&1&1\\1&-1&1\end{bmatrix}$
  • 7 For which values of $k$ is $\begin{bmatrix}k-1&2\\3&k\end{bmatrix}$ singular?
  • 8 For which $x$ does $\begin{bmatrix}1&2&3\\4&5&6\\7&8&x\end{bmatrix}$ have no inverse?
  • 9 For the matrix of part 2, verify that $|A^{-1}|=\dfrac1{|A|}$.
  • 10 With $A=\begin{bmatrix}2&5\\1&3\end{bmatrix}$ and $B=\begin{bmatrix}1&1\\1&2\end{bmatrix}$, verify $(AB)^{-1}=B^{-1}A^{-1}$.
  • 11 Prove Theorem 4.5 for $2\times2$ matrices: for $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ and $B=\begin{bmatrix}p&q\\r&s\end{bmatrix}$, show $|AB|=|A|\,|B|$.
Check: inverse by the adjoint, in exact fractions
—

For order 2 only the first two entries of rows 1 and 2 are used. Entries may be fractions such as -3/4.

1 $|A|=6-5=1$: $A^{-1}=\begin{bmatrix}3&-5\\-1&2\end{bmatrix}$.
2 $|A|=4+6=10$: $A^{-1}=\tfrac1{10}\begin{bmatrix}1&3\\-2&4\end{bmatrix}=\begin{bmatrix}\tfrac1{10}&\tfrac3{10}\\-\tfrac15&\tfrac25\end{bmatrix}$.
3 $|A|=6-6=0$: singular, no inverse (row 1 is $-2\times$ row 2).
4 $|A|=1$ (triangular). Cofactors by rows: $(1,-2,1)$, $(0,1,-2)$, $(0,0,1)$; transposing, $$A^{-1}=\begin{bmatrix}1&0&0\\-2&1&0\\1&-2&1\end{bmatrix}.$$
5 $|A|=1(6-4)+1(4-0)+0=6$. Cofactors by rows: $(2,-4,2)$, $(2,2,-1)$, $(-4,-4,5)$; $$A^{-1}=\frac16\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}.$$
6 $|A|=2(1+1)+1(1-1)+3(-1-1)=-2$. Cofactors by rows: $(2,0,-2)$, $(-2,-1,1)$, $(-4,1,3)$; $$A^{-1}=-\frac12\begin{bmatrix}2&-2&-4\\0&-1&1\\-2&1&3\end{bmatrix}=\begin{bmatrix}-1&1&2\\0&\tfrac12&-\tfrac12\\1&-\tfrac12&-\tfrac32\end{bmatrix}.$$
7 $k(k-1)-6=k^2-k-6=(k-3)(k+2)=0$: $k=3$ or $k=-2$.
8 $1(5x-48)-2(4x-42)+3(32-35)=-3x+27=0$: $x=9$.
9 $|A^{-1}|=\tfrac1{10}\cdot\tfrac25-\tfrac3{10}\cdot\left(-\tfrac15\right)=\tfrac2{50}+\tfrac3{50}=\tfrac1{10}=\tfrac1{|A|}$ ✓.
10 $AB=\begin{bmatrix}7&12\\4&7\end{bmatrix}$, $|AB|=1$, $(AB)^{-1}=\begin{bmatrix}7&-12\\-4&7\end{bmatrix}$. $B^{-1}=\begin{bmatrix}2&-1\\-1&1\end{bmatrix}$, $A^{-1}=\begin{bmatrix}3&-5\\-1&2\end{bmatrix}$, and $B^{-1}A^{-1}=\begin{bmatrix}6+1&-10-2\\-3-1&5+2\end{bmatrix}=\begin{bmatrix}7&-12\\-4&7\end{bmatrix}$ ✓.
11 $AB=\begin{bmatrix}ap+br&aq+bs\\cp+dr&cq+ds\end{bmatrix}$. Then $|AB|=(ap+br)(cq+ds)-(aq+bs)(cp+dr)$. The terms $apcq$ and $brds$ cancel against $aqcp$ and $bsdr$, leaving $adps+bcqr-adqr-bcps=(ad-bc)(ps-qr)=|A|\,|B|$.
IV
Reasoning and applications.
  • 1 $A$ is a $3\times3$ matrix with $|A|=6$. Find $|A^{-1}|$ and $|A^2|$.
  • 2 Show that if $A$ is non-singular and $AB=AC$, then $B=C$. Then show by an example with $A=\begin{bmatrix}1&1\\1&1\end{bmatrix}$ that this can fail when $A$ is singular.
  • 3 Matrix code. Letters are numbered A $=1$, …, Z $=26$, and a pair of letters $(p,q)$ is sent as $K\begin{bmatrix}p\\q\end{bmatrix}$ with key $K=\begin{bmatrix}2&1\\3&2\end{bmatrix}$. Encode “HI”, find $K^{-1}$, and decode the received pair $(31,49)$.
  • 4 Show that $R=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$ (a rotation of the plane) is non-singular for every $\theta$, and that $R^{-1}=R'$.
  • 5 Three-letter code. A group of three letters is sent as $K\begin{bmatrix}p\\q\\r\end{bmatrix}$ with $K=\begin{bmatrix}1&0&0\\2&1&0\\3&2&1\end{bmatrix}$. Use Exercise III.4 to decode the received triple $(3,7,31)$.
1 $|A^{-1}|=\tfrac1{|A|}=\tfrac16$; by Theorem 4.5, $|A^2|=|A|\,|A|=36$.
2 Multiply $AB=AC$ on the left by $A^{-1}$ (it exists by Theorem 4.6): $IB=IC$, so $B=C$. Example: $B=\begin{bmatrix}1&0\\0&0\end{bmatrix}$, $C=\begin{bmatrix}0&0\\1&0\end{bmatrix}$ give $AB=AC=\begin{bmatrix}1&0\\1&0\end{bmatrix}$ although $B\ne C$; here $|A|=0$.
3 H $=8$, I $=9$: $K\begin{bmatrix}8\\9\end{bmatrix}=\begin{bmatrix}25\\42\end{bmatrix}$. $|K|=1$, so $K^{-1}=\begin{bmatrix}2&-1\\-3&2\end{bmatrix}$, and $K^{-1}\begin{bmatrix}31\\49\end{bmatrix}=\begin{bmatrix}62-49\\-93+98\end{bmatrix}=\begin{bmatrix}13\\5\end{bmatrix}$: “ME”.
4 $|R|=\cos^2\theta+\sin^2\theta=1\ne0$. $R^{-1}=\tfrac11\operatorname{adj}R=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}$, which is $R'$. (Undoing a rotation by $\theta$ is a rotation by $-\theta$.)
5 $K^{-1}\begin{bmatrix}3\\7\\31\end{bmatrix}=\begin{bmatrix}3\\-6+7\\3-14+31\end{bmatrix}=\begin{bmatrix}3\\1\\20\end{bmatrix}$: C, A, T — “CAT”. Check: $K(3,1,20)=(3,\,7,\,9+2+20)=(3,7,31)$ ✓.
🎯 Exam corner Board · 1 markBoard · 3 marksJEE MainJEE Advanced

How this is examined: Board papers ask for a cofactor or a $2\times2$ adjoint (1 mark), use $A(\operatorname{adj}A)=|A|\,I$ in MCQs and Assertion–Reason, and set a $3\times3$ inverse for 3 marks — often as the first half of a 5-mark question that then solves a system (Section 4.3). JEE asks for determinants of $kA$ and $\operatorname{adj}A$ and for relations such as $A^{-1}=\alpha A+\beta I$.

E1. Board · 1 mark If $A=\begin{bmatrix}2&-3\\4&1\end{bmatrix}$, then $\operatorname{adj}A$ is  (A) $\begin{bmatrix}1&3\\-4&2\end{bmatrix}$  (B) $\begin{bmatrix}1&-4\\3&2\end{bmatrix}$  (C) $\begin{bmatrix}2&-4\\3&1\end{bmatrix}$  (D) $\begin{bmatrix}-1&-3\\4&-2\end{bmatrix}$

Idea Swap the diagonal entries $2,1$ and negate $-3,4$: $\operatorname{adj}A=\begin{bmatrix}1&3\\-4&2\end{bmatrix}$.
Answer (A). Trap (B) is the cofactor matrix $\begin{bmatrix}1&-4\\3&2\end{bmatrix}$ — it has not been transposed. (C) is the transpose $A'$; (D) is $-\operatorname{adj}A$.

E2. Board · 1 mark Assertion–Reason. Assertion (A): the matrix $\begin{bmatrix}3&-6\\-1&2\end{bmatrix}$ has no inverse. Reason (R): a square matrix is invertible if and only if it is non-singular.  (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

Idea R is Theorem 4.6. The determinant is $3\cdot2-(-1)(-6)=6-6=0$, so the matrix is singular and, by R, not invertible.
Answer (A).

E3. Board · 1 mark If $A$ is a square matrix of order $3$ and $A(\operatorname{adj}A)=\begin{bmatrix}-4&0&0\\0&-4&0\\0&0&-4\end{bmatrix}$, then $|A|$ equals  (A) $-4$  (B) $-64$  (C) $4$  (D) $16$

Idea By Theorem 4.4, $A(\operatorname{adj}A)=|A|\,I$. The given matrix is $-4I$, so $|A|=-4$.
Answer (A). Trap (B): $-64$ is the determinant of the product $-4I$, not $|A|$.

E4. Board · 3 marks Find the inverse of $A=\begin{bmatrix}3&0&-1\\2&3&0\\0&4&1\end{bmatrix}$ and verify that $A^{-1}A=I$.

Determinant Along row 1: $3(3-0)-0+(-1)(8-0)=9-8=1\ne0$. (½ mark)
Cofactors Row 1: $A_{11}=3$, $A_{12}=-(2-0)=-2$, $A_{13}=8$. Row 2: $A_{21}=-(0+4)=-4$, $A_{22}=3$, $A_{23}=-(12-0)=-12$. Row 3: $A_{31}=(0+3)=3$, $A_{32}=-(0+2)=-2$, $A_{33}=9$. (1½ marks)
Inverse $$A^{-1}=\frac11\operatorname{adj}A=\begin{bmatrix}3&-4&3\\-2&3&-2\\8&-12&9\end{bmatrix}.$$ (½ mark)
Verify Row 1 of $A^{-1}$ times the columns of $A$: $9-8+0=1$, $0-12+12=0$, $-3+0+3=0$; row 2: $-6+6=0$, $9-8=1$, $2-2=0$; row 3: $24-24=0$, $-36+36=0$, $-8+9=1$. So $A^{-1}A=I$ ✓. (½ mark)

E5. JEE Main $A$ is a square matrix of order $3$ with $|A|=3$. Then $|\operatorname{adj}(2A)|$ equals  (A) $576$  (B) $144$  (C) $72$  (D) $1152$

Method (🚀 box) $|2A|=2^3\cdot3=24$, and $|\operatorname{adj}B|=|B|^{n-1}=|B|^2$ for order $3$, so $|\operatorname{adj}(2A)|=24^2=576$.
Answer (A). Trap (B): $144=12^2$ uses $|2A|=2^2|A|$, the rule for order 2; (C) $=24\cdot3$ mixes $|2A|$ with $|A|$; (D) doubles the correct answer.

E6. JEE Advanced (Numerical answer.) For $A=\begin{bmatrix}2&-1\\1&3\end{bmatrix}$ there are real numbers $\alpha,\beta$ with $A^{-1}=\alpha A+\beta I$. Find $7(\alpha+\beta)$.

Method $|A|=6+1=7$ and $\operatorname{adj}A=\begin{bmatrix}3&1\\-1&2\end{bmatrix}$. Notice $\operatorname{adj}A=5I-A$ (for any $2\times2$ matrix, $\operatorname{adj}A=(a_{11}+a_{22})I-A$). So $A^{-1}=\tfrac17(5I-A)$: $\alpha=-\tfrac17$, $\beta=\tfrac57$.
Answer $7(\alpha+\beta)=-1+5=\mathbf 4$. Check: $A\cdot\tfrac17(5I-A)=\tfrac17(5A-A^2)$, and $A^2=\begin{bmatrix}3&-5\\5&8\end{bmatrix}$, so $5A-A^2=\begin{bmatrix}7&0\\0&7\end{bmatrix}$ ✓.
🔬 Research–Teaching NexusMini project · about 60 minutes

Research question. What happens to $A^{-1}$ when $A$ is nearly singular — and why do engineers who use GPS, medical scanners or weather models care?

Why it matters. Theorem 4.6 says $A^{-1}$ exists whenever $|A|\ne0$. But the formula divides by $|A|$, so a tiny determinant means huge entries in $A^{-1}$, and tiny errors in the data are blown up into large errors in the answer. A GPS receiver that sees satellites bunched in one part of the sky has a nearly singular geometry matrix and gives a poor position fix; engineers call this “dilution of precision”.

What to do.

  1. In a spreadsheet, set up $A_\varepsilon=\begin{bmatrix}1&1\\1&1+\varepsilon\end{bmatrix}$ for $\varepsilon=1,\ 0.1,\ 0.01,\ 0.001$. Compute $|A_\varepsilon|$ with MDETERM and $A_\varepsilon^{-1}$ with MINVERSE; check one inverse by hand with the adjoint formula.
  2. Solve $A_\varepsilon X=\begin{bmatrix}2\\2+\varepsilon\end{bmatrix}$ (exact answer $x=y=1$) as $X=A_\varepsilon^{-1}B$. Then change the right side by $0.01$ in one entry, as a measurement error would, and record how far $X$ moves.
  3. Plot the size of the change in $X$ against $\tfrac1{|A_\varepsilon|}$. What pattern do you see?
  4. Repeat with a $3\times3$ matrix of your own whose third row is almost the sum of the first two.

Tools: Spreadsheet (MDETERM, MINVERSE, MMULT); calculator for the hand check.

Hand in: Your table and graph, one hand-worked inverse, and a paragraph explaining to a younger student why “$|A|\ne0$” is not the whole story in real measurements.

Going further. Find out how a GPS receiver chooses which satellites to use, and why it prefers satellites spread widely across the sky.

Instructor note: $|A_\varepsilon|=\varepsilon$ and $A_\varepsilon^{-1}=\tfrac1\varepsilon\begin{bmatrix}1+\varepsilon&-1\\-1&1\end{bmatrix}$. Changing the second entry of $B$ by $\delta$ changes $X$ by $\tfrac\delta\varepsilon(-1,1)$, so the change grows exactly like $\tfrac1{|A_\varepsilon|}$ — a clean linear graph. At $\varepsilon=0.001$ a $0.01$ error moves $x$ and $y$ by $10$.

🧪 Self-check quiz

0/5 correct

Q1. The cofactor $A_{12}$ of $\begin{bmatrix}5&3\\2&7\end{bmatrix}$ is

Q2. $\operatorname{adj}\begin{bmatrix}a&b\\c&d\end{bmatrix}$ equals

Q3. A square matrix $A$ is singular when

Q4. $A$ is a $2\times2$ matrix with $|A|=-2$. Then $A(\operatorname{adj}A)$ equals

Q5. The inverse of $\begin{bmatrix}2&1\\5&3\end{bmatrix}$ is

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