Matrices for JEE: Types, Powers, Transpose and Inverse

A matrix is a rectangular arrangement of numbers that is handled as one object. You add matrices, multiply them, take powers of them and invert them, and the rules look like ordinary algebra with two important differences: the order of multiplication matters, and a product can be zero when neither factor is zero.

This note builds the chapter in the order JEE uses it: what a matrix is, the special types, the operations, transpose and the symmetric and skew-symmetric split, powers, the equation every $2\times2$ matrix satisfies, the inverse, and elementary row operations. Every worked example is a JEE Main 2026 or JEE Advanced 2026 question unless it says otherwise, and every answer was re-derived from scratch.

What a matrix is

A matrix with $m$ rows and $n$ columns is said to have order $m\times n$. The entry in row $i$ and column $j$ is written $a_{ij}$, and the whole matrix is written $A=[a_{ij}]_{m\times n}$.

A 3 by 4 matrix with row 2 and column 3 shaded and the entry a23 marked where they cross

$$A=\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\end{bmatrix}_{2\times3}$$

The first index is always the row and the second is always the column. A matrix of order $m\times n$ has $mn$ entries, so if each entry can be chosen from a set of $k$ numbers, the number of such matrices is

$$k^{mn}$$

Two matrices are equal only when they have the same order and every corresponding entry is equal. So one matrix equation of order $m\times n$ is really $mn$ ordinary equations.

Types of matrices

Five 3 by 3 matrices side by side: diagonal, scalar, identity, upper triangular and lower triangular, with the zero entries shown faint

In a square matrix the entries $a_{11},a_{22},\ldots,a_{nn}$ form the principal diagonal, and their sum is the trace:

$$\operatorname{tr}(A)=a_{11}+a_{22}+\cdots+a_{nn}$$

Trace is linear and does not care about the order of a product:

$$\operatorname{tr}(A+B)=\operatorname{tr}A+\operatorname{tr}B$$ $$\operatorname{tr}(kA)=k\operatorname{tr}A$$ $$\operatorname{tr}(AB)=\operatorname{tr}(BA)$$

Addition, scalar multiple and product

Matrices of the same order are added entry by entry, and $kA$ multiplies every entry by $k$. Addition is commutative and associative, $O$ is the additive identity, and $k(A+B)=kA+kB$.

The product $AB$ is defined only when the number of columns of $A$ equals the number of rows of $B$. If $A$ is $m\times n$ and $B$ is $n\times p$, then $AB$ is $m\times p$ and

$$(AB)_{ij}=a_{i1}b_{1j}+a_{i2}b_{2j}+\cdots+a_{in}b_{nj}$$

That is, row $i$ of $A$ against column $j$ of $B$, term by term, then add.

Row 2 of A shaded, column 1 of B shaded, and the single entry they produce in AB shaded at row 2 column 1

Multiplication is associative, $A(BC)=(AB)C$, and distributes over addition, $A(B+C)=AB+AC$. Also $AI=IA=A$ for a square $A$.

The column picture

Multiplying a matrix by the standard column $e_1=\begin{bmatrix}1&0&0\end{bmatrix}^T$ picks out its first column, $e_2$ its second, and $e_3$ its third:

$$Me_1=\text{column 1 of }M$$ $$Me_2=\text{column 2 of }M$$

So if a question tells you what $M$ does to $e_1$, $e_2$, $e_3$, it has handed you $M$ column by column. Similarly $e_i^TM$ is row $i$ of $M$.

A 3 by 3 matrix M with its three columns in different shades, and arrows from Me1, Me2, Me3 to the matching columns

Example 1 (JEE Main 2026, 5 April Shift 2). Let $M$ be a $3\times3$ matrix such that $Me_1=\begin{bmatrix}1&2&3\end{bmatrix}^T$, $Me_2=\begin{bmatrix}0&1&2\end{bmatrix}^T$ and $Me_3=\begin{bmatrix}-1&1&1\end{bmatrix}^T$, where $e_1,e_2,e_3$ are the standard unit columns. If $M\begin{bmatrix}x&y&z\end{bmatrix}^T=\begin{bmatrix}1&7&11\end{bmatrix}^T$, then $x+y+z$ equals

▸ SHOW SOLUTION

Step 1. Write M column by column

$$M=\begin{bmatrix}1&0&-1\\2&1&1\\3&2&1\end{bmatrix}$$

Step 2. Three equations

$$x-z=1$$$$2x+y+z=7$$$$3x+2y+z=11$$

Step 3. Solve

Third minus second:

$$x+y=4$$

First gives $z=x-1$. Into the second:

$$3x+y=8$$

So $2x=4$, $x=2$, $y=2$, $z=1$.

$$x+y+z=5$$

Answer: (2) $5$

Transpose

The transpose $A^T$ is obtained by turning the rows of $A$ into columns. If $A$ is $m\times n$, then $A^T$ is $n\times m$ and $(A^T)_{ij}=a_{ji}$.

A 2 by 3 matrix and its 3 by 2 transpose, with the first row of A and the first column of the transpose in the same shade

$$(A^T)^T=A,\quad (kA)^T=kA^T$$ $$(A+B)^T=A^T+B^T$$ $$(AB)^T=B^TA^T$$

The order reverses in the last rule. For three factors, $(ABC)^T=C^TB^TA^T$.

One identity is worth keeping ready. The $(j,j)$ entry of $A^TA$ is column $j$ of $A$ against itself, which is the sum of the squares of that column. Adding over all $j$:

$$\operatorname{tr}(A^TA)=\sum_{i,j}a_{ij}^2$$

So $\operatorname{tr}(A^TA)$ is the sum of the squares of every entry of $A$, and it is $0$ only when $A=O$.

Example 2 (JEE Main 2026, 24 January Shift 1). The number of $3\times2$ matrices $A$ which can be formed using the elements of the set $\{-2,-1,0,1,2\}$ such that the sum of all the diagonal elements of $A^TA$ is $5$, is ______.

▸ SHOW SOLUTION

Step 1. Turn the condition into a sum of squares

$A$ has $6$ entries and

$$\operatorname{tr}(A^TA)=5$$

which is the sum of the squares of all $6$ entries.

Each square is $0$, $1$ or $4$.

Step 2. List the ways to make 5

Five squares equal to $1$ and one equal to $0$, or one $4$, one $1$ and four $0$s.

Step 3. Count each case

Five entries $\pm1$ and one $0$: choose the place of $0$ in $6$ ways, signs in $2^5$ ways.

$$6\times32=192$$

One entry $\pm2$, one entry $\pm1$, rest $0$: places in $6\times5$ ways, signs in $2\times2$ ways.

$$30\times4=120$$

Step 4. Add

$$192+120=312$$

Answer: $312$

Symmetric and skew-symmetric matrices

A square matrix is symmetric if $A^T=A$, that is $a_{ij}=a_{ji}$, and skew-symmetric if $A^T=-A$, that is $a_{ij}=-a_{ji}$.

Putting $i=j$ in the skew condition gives $a_{ii}=-a_{ii}$, so every diagonal entry of a skew-symmetric matrix is $0$.

A symmetric matrix with mirror-image entries across the diagonal, and a skew-symmetric matrix with zeros on the diagonal and opposite entries across it

Every square matrix splits in exactly one way into a symmetric part and a skew-symmetric part:

$$A=\frac{A+A^T}{2}+\frac{A-A^T}{2}$$

Proof. Put $P=\dfrac{A+A^T}{2}$ and $Q=\dfrac{A-A^T}{2}$. Then $P^T=\dfrac{A^T+A}{2}=P$ and $Q^T=\dfrac{A^T-A}{2}=-Q$, and $P+Q=A$. If also $A=P_1+Q_1$ with $P_1$ symmetric and $Q_1$ skew, taking transposes gives $A^T=P_1-Q_1$, so $P_1=\dfrac{A+A^T}{2}=P$ and $Q_1=Q$.

Facts that come up often:

• $A+A^T$ is symmetric and $A-A^T$ is skew-symmetric, for any square $A$.
• $AA^T$ and $A^TA$ are symmetric for any $A$, square or not.
• If $A$ and $B$ are symmetric, $AB$ is symmetric exactly when $AB=BA$.
• For any $B$, $B^TAB$ is symmetric when $A$ is symmetric and skew-symmetric when $A$ is skew.

Example 3 (IITianForum). How many symmetric matrices of order $3$ have every entry from $\{-1,0,1\}$ and trace $0$?

▸ SHOW SOLUTION

Step 1. Count the free entries

A symmetric $3\times3$ matrix is fixed by its $3$ diagonal entries and the $3$ entries above the diagonal. The entries below are copies.

Step 2. Above the diagonal

Any values:

$$3^3=27$$

Step 3. The diagonal

Three entries from $\{-1,0,1\}$ adding to $0$: all three $0$ (one way), or one each of $1$, $0$, $-1$ in some order ($3!=6$ ways).

$$1+6=7$$

Step 4. Multiply

$$7\times27=189$$

Answer: $189$

Example 4 (IITianForum). Let $B$ be a square matrix with $\lvert B\rvert=3$, and let $A$ satisfy

$$3A+B=B^T-2A^T$$

Then $\lvert A+B\rvert$ is

▸ SHOW SOLUTION

Step 1. Collect A on one side

$$3A+2A^T=B^T-B\qquad(1)$$

Step 2. Transpose

$$3A^T+2A=B-B^T\qquad(2)$$

Step 3. Add (1) and (2)

$$5(A+A^T)=O\implies A^T=-A$$

So $A$ is skew-symmetric.

Step 4. Back into (1)

$$3A-2A=B^T-B\implies A=B^T-B$$$$A+B=B^T$$

Step 5. Determinant

$$\lvert A+B\rvert=\lvert B^T\rvert=\lvert B\rvert=3$$

Answer: (3) $3$

Transposing the given equation and adding is the standard first move whenever $A$ and $A^T$ appear together.

Orthogonal matrices

A square matrix is orthogonal if

$$AA^T=I$$

Then $A^T$ is the inverse of $A$, so $A^TA=I$ as well. Read entry by entry, $AA^T=I$ says each row has sum of squares $1$ and any two different rows have dot product $0$. $A^TA=I$ says the same for the columns.

Example 5 (JEE Advanced 2026, Paper 1). For real numbers $\alpha,\beta,\gamma,\delta,\mu$, consider

$$M=\begin{bmatrix}\alpha&\tfrac{1}{\sqrt2}&-\tfrac{1}{\sqrt2}\\\tfrac{1}{\sqrt3}&\beta&\tfrac{1}{\sqrt3}\\\gamma&\delta&\mu\end{bmatrix}$$

with $MM^T=I$. Let $\vec u$, $\vec v$, $\vec w$ be the vectors whose components are the first, second and third columns of $M$. Match List-I with List-II.

▸ SHOW SOLUTION

Step 1. Rows of M

Row 1 has sum of squares $1$:

$$\alpha^2+\frac12+\frac12=1\implies\alpha=0$$

Row 2 has sum of squares $1$ and is perpendicular to row 1:

$$\frac13+\beta^2+\frac13=1,\qquad \frac{\beta}{\sqrt2}-\frac{1}{\sqrt6}=0$$$$\beta=\frac{1}{\sqrt3}$$

Step 2. Columns of M, for (P)

$M^TM=I$ too, so each column has sum of squares $1$:

$$0+\frac13+\gamma^2=1\implies\gamma^2=\frac23$$$$\frac12+\frac13+\delta^2=1\implies\delta^2=\frac16$$$$\gamma^2+\delta^2=\frac56$$

Step 3. (Q)

$x\vec u+y\vec v+z\vec w$ is $M$ times the column $(x,y,z)$. So

$$M\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\1\\0\end{bmatrix}\implies\begin{bmatrix}x\\y\\z\end{bmatrix}=M^T\begin{bmatrix}0\\1\\0\end{bmatrix}$$

That is the second row of $M$ read as a column, so $x=\dfrac{1}{\sqrt3}$.

Step 4. (R)

$\vec u\cdot(\vec v\times\vec w)$ is the determinant with $\vec u,\vec v,\vec w$ as rows, which is $\det(M^T)=\det M$. From $MM^T=I$, $(\det M)^2=1$.

$$\lvert\vec u\cdot(\vec v\times\vec w)\rvert=1$$

Step 5. (S)

$$\vec u\times(\vec v\times\vec w)=(\vec u\cdot\vec w)\vec v-(\vec u\cdot\vec v)\vec w$$

Different columns are perpendicular, so both dot products are $0$ and the value is $0$.

Answer: (1) P→5, Q→4, R→2, S→1

Powers of a square matrix

$A^2=AA$, $A^3=A^2A$, and so on, with $A^0=I$. Powers of the same matrix always commute, $A^mA^n=A^nA^m=A^{m+n}$. Three named types appear constantly:

The method for a high power: write A as I + N

If $I$ and $N$ commute, which they always do, the binomial theorem works for $(I+N)^n$ exactly as for numbers. When $N^2=O$ every term from $N^2$ on vanishes:

$$(I+N)^n=I+nN\qquad\text{when }N^2=O$$

When $N^2\ne O$ but $N^3=O$:

$$(I+N)^n=I+nN+\binom n2N^2$$

So for a high power, subtract $I$ and square what is left. If it dies quickly, the power is a short polynomial in $n$.

The 2 by 2 matrix N with rows (2, minus 4) and (1, minus 2), its square shown as the zero matrix, and A to the n written as I plus n N

Example 6 (JEE Main 2026, 28 January Shift 2). Let

$$A=\begin{bmatrix}3&-4\\1&-1\end{bmatrix}$$

and $B$ be a matrix such that $A^{100}=100B+I$. Then the sum of all the elements of $B^{100}$ is ______.

▸ SHOW SOLUTION

Step 1. Subtract I

$$N=A-I=\begin{bmatrix}2&-4\\1&-2\end{bmatrix}$$$$N^2=\begin{bmatrix}4-4&-8+8\\2-2&-4+4\end{bmatrix}=O$$

Step 2. The power

$$A^{100}=(I+N)^{100}=I+100N$$

Comparing with $100B+I$ gives $B=N$.

Step 3. Power of B

$N^2=O$, so $B^{100}=N^{100}=O$ and the sum of its elements is $0$.

Answer: $0$

Example 7 (JEE Main 2026, 21 January Shift 2). For

$$A=\begin{bmatrix}3&-4\\1&-1\end{bmatrix},\quad B=\begin{bmatrix}-29&49\\-13&18\end{bmatrix}$$

if $(A^{15}+B)X=O$ where $X$ is the column with entries $x$, $y$, then which is true?

▸ SHOW SOLUTION

Step 1. Same A as Example 6

$A=I+N$ with $N^2=O$, so

$$A^{15}=I+15N=\begin{bmatrix}31&-60\\15&-29\end{bmatrix}$$

Step 2. Add B

$$A^{15}+B=\begin{bmatrix}2&-11\\2&-11\end{bmatrix}$$

Step 3. Solve

Both rows give $2x-11y=0$, so $(x,y)$ is a multiple of $(11,2)$. Only option (3) fits.

Answer: (3) $x=11,\ y=2$

Example 8 (JEE Main 2026, 6 April Shift 2). Let

$$A=\begin{bmatrix}1&0&0\\3&1&0\\9&3&1\end{bmatrix}$$

and $B=[b_{ij}]=A^{99}-I$. Then $\dfrac{b_{31}-b_{21}}{b_{32}}$ is

▸ SHOW SOLUTION

Step 1. Subtract I and find the powers of N

$$N=\begin{bmatrix}0&0&0\\3&0&0\\9&3&0\end{bmatrix},\qquad N^2=\begin{bmatrix}0&0&0\\0&0&0\\9&0&0\end{bmatrix}$$

and $N^3=O$.

Step 2. The power

$$A^{99}=I+99N+\binom{99}{2}N^2$$$$B=99N+4851N^2$$

Step 3. Read the three entries

$$b_{21}=99\times3=297$$$$b_{32}=99\times3=297$$$$b_{31}=99\times9+4851\times9=44550$$

Step 4. The ratio

$$\frac{44550-297}{297}=\frac{44253}{297}=149$$

Answer: (3) $149$

Example 9 (JEE Advanced 2026, Paper 1). Let

$$M=\begin{bmatrix}2&-1\\1&0\end{bmatrix}$$

and let $p,q,r,s,a,b,c,d$ be integers with

$$M^{26}=\begin{bmatrix}p&q\\r&s\end{bmatrix},\quad \sum_{k=1}^{26}M^k=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$

Which of the following is (are) TRUE?

(1) There is an invertible real $2\times2$ matrix $N$ with $MN=NJ$, where

$$J=\begin{bmatrix}1&1\\0&1\end{bmatrix}$$

(2) $a=378$

(3) For any integers $m,n$ there are unique integers $x,y$ with $px+qy=m$ and $rx+sy=n$

(4) For each $t>0$ the system $(a+t)x+by=1$, $cx+(d+t)y=-1$ has a unique solution

▸ SHOW SOLUTION

Step 1. Write M as I + N

$$M-I=\begin{bmatrix}1&-1\\1&-1\end{bmatrix},\qquad (M-I)^2=O$$$$M^k=I+k(M-I)=\begin{bmatrix}1+k&-k\\k&1-k\end{bmatrix}$$

Step 2. M to the 26 and the sum

$$M^{26}=\begin{bmatrix}27&-26\\26&-25\end{bmatrix}$$$$\sum_{k=1}^{26}M^k=26I+351(M-I)$$$$\sum_{k=1}^{26}M^k=\begin{bmatrix}377&-351\\351&-325\end{bmatrix}$$

So $a=377$ and (2) is false.

Step 3. Option (3)

$$ps-qr=27(-25)+26\cdot26=1$$

By Cramer's rule the unique solution is $x=sm-qn$, $y=pn-rm$, both integers. True.

Step 4. Option (4)

$$(377+t)(t-325)+351^2$$$$=t^2+52t+676$$$$t^2+52t+676=(t+26)^2>0$$

The coefficient determinant is never $0$ for $t>0$, so the solution is unique. True.

Step 5. Option (1)

Put

$$N=\begin{bmatrix}p_1&q_1\\r_1&s_1\end{bmatrix}$$

and compare entries of $MN$ and $NJ$:

$$MN=\begin{bmatrix}2p_1-r_1&2q_1-s_1\\p_1&q_1\end{bmatrix}$$$$NJ=\begin{bmatrix}p_1&p_1+q_1\\r_1&r_1+s_1\end{bmatrix}$$

This needs $r_1=p_1$ and $q_1=p_1+s_1$. Take $p_1=1$, $s_1=0$:

$$N=\begin{bmatrix}1&1\\1&0\end{bmatrix},\qquad \det N=-1\ne0$$

So $N$ is invertible. True.

Answer: (1), (3), (4)

The equation every 2×2 matrix satisfies

For

$$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$

multiply out $A^2$ and check directly that

$$A^2-(a+d)A+(ad-bc)I=O$$

That is,

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

This one line does three jobs. It lowers any power: $A^2$ is a combination of $A$ and $I$, so $A^3$, $A^4$, and every higher power is too. It gives the inverse when $\det A\ne0$:

$$A^{-1}=\frac{(\operatorname{tr}A)I-A}{\det A}$$

And it decodes a given equation. If $A$ is not a multiple of $I$ and $A^2-pA+qI=O$, subtracting the identity above gives $(\operatorname{tr}A-p)A=(\det A-q)I$, which forces $\operatorname{tr}A=p$ and $\det A=q$.

Example 10 (IITianForum). A square matrix $P$ satisfies $P^2=I+P$. If $P^n=8I+13P$, then $n$ is

▸ SHOW SOLUTION

Step 1. One multiplication rule

Every power is $aI+bP$. Multiplying by $P$:

$$P(aI+bP)=aP+bP^2$$$$=bI+(a+b)P$$

Step 2. Climb

Power	$I$	$P$
$P^2$	$1$	$1$
$P^3$	$1$	$2$
$P^4$	$2$	$3$
$P^5$	$3$	$5$
$P^6$	$5$	$8$
$P^7$	$8$	$13$

swipe for more →

Answer: (3) $7$

The coefficients are consecutive Fibonacci numbers, because each new row is (previous $P$-coefficient, sum of the previous two).

Example 11 (JEE Main 2026, 4 April Shift 1). Let $S$ be the set of all $2\times2$ matrices

$$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$

with $a,b,c,d\in\{0,1,2,3,4\}$ and $A^2-4A+3I=O$. The number of matrices in $S$ whose diagonal sum is $4$ is

▸ SHOW SOLUTION

Step 1. Use the 2×2 identity with trace 4

$$A^2-4A+(\det A)I=O$$

Subtract the given equation:

$$(\det A-3)I=O\implies ad-bc=3$$

Conversely, trace $4$ and determinant $3$ give the equation back. So count $a+d=4$ with $ad-bc=3$.

Step 2. Go through the diagonal

$(a,d)$	$ad$	needed $bc$	pairs $(b,c)$
$(0,4)$, $(4,0)$	$0$	$-3$	$0$
$(1,3)$, $(3,1)$	$3$	$0$	$9$ each
$(2,2)$	$4$	$1$	$1$

swipe for more →

For $bc=0$ with $b,c\in\{0,\ldots,4\}$: $5+5-1=9$ pairs. For $bc=1$: only $(1,1)$.

Step 3. Add

$$9+9+1=19$$

Answer: (4) $19$

Example 12 (JEE Advanced 2026, Paper 2). Let

$$S=\begin{bmatrix}0&-1\\1&0\end{bmatrix},\quad T=\begin{bmatrix}1&1\\0&1\end{bmatrix}$$ $$ST=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$

Let $H=\{x+iy:y>0\}$. Which of the following is (are) TRUE?

(1) $\dfrac{b+ia}{d+ic}=i$

(2) If $\omega=\dfrac{-1+i\sqrt3}{2}$, then $\dfrac{a\omega+b}{c\omega+d}=\omega$

(3) If $m>2$ is an integer with $(ST)^2=(ST)^m$, then $m$ is a multiple of $8$

(4) If $z\in H$, then $\dfrac{az+b}{cz+d}\in H$

▸ SHOW SOLUTION

Step 1. The product

$$ST=\begin{bmatrix}0&-1\\1&1\end{bmatrix}$$

so $a=0$, $b=-1$, $c=1$, $d=1$.

Step 2. Option (1)

$$\frac{-1}{1+i}=\frac{-1+i}{2}\ne i$$

False.

Step 3. Option (2)

$1+\omega=-\omega^2$, so

$$\frac{-1}{\omega+1}=\frac{1}{\omega^2}=\omega$$

using $\omega^3=1$. True.

Step 4. Option (3)

Let $P=ST$. Trace $1$, determinant $1$:

$$P^2-P+I=O$$

Multiply by $P+I$:

$$P^3+I=O\implies P^3=-I,\quad P^6=I$$

$P\ne\pm I$ and $P^2\ne I$, so the powers of $P$ repeat with period exactly $6$. $P^m=P^2$ needs $m-2$ to be a multiple of $6$, so $m=8,14,20,\ldots$ Since $14$ is not a multiple of $8$, false.

Six points on a circle labelled P, P squared, up to P to the sixth equal to I, with arrows showing the cycle closing after six steps

Step 5. Option (4)

With $z=x+iy$,

$$\frac{az+b}{cz+d}=\frac{-1}{z+1}=\frac{-(x+1)+iy}{(x+1)^2+y^2}$$

Its imaginary part has the sign of $y$, which is positive. True.

Answer: (2), (4)

Inverse of a matrix

A square matrix $A$ is invertible if there is a matrix $B$ with

$$AB=BA=I$$

Then $B$ is unique and is written $A^{-1}$. Not every square matrix has one: a matrix with determinant $0$ never does. How to compute $A^{-1}$ through the adjoint belongs to the Determinants note; here are the rules that need no computation.

$$(A^{-1})^{-1}=A,\qquad (kA)^{-1}=\frac1kA^{-1}$$ $$(AB)^{-1}=B^{-1}A^{-1}$$ $$(A^T)^{-1}=(A^{-1})^T$$

If $AB=I$ for square matrices of the same order, then $BA=I$ automatically, so one side is enough.

Example 13 (JEE Main 2026, 28 January Shift 1). Let $A$, $B$, $C$ be $2\times2$ real matrices with $B=(I+A)^{-1}$ and $A+C=I$. If

$$BC=\begin{bmatrix}1&-5\\-1&2\end{bmatrix},\quad CB\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}12\\-6\end{bmatrix}$$

then $x_1+x_2$ is

▸ SHOW SOLUTION

Step 1. Show CB = BC

$C=I-A$. The products $(I+A)(I-A)$ and $(I-A)(I+A)$ both equal $I-A^2$. Multiply that equality by $(I+A)^{-1}$ on the left and on the right:

$$(I-A)(I+A)^{-1}=(I+A)^{-1}(I-A)$$

That is $CB=BC$.

Step 2. Solve

$$x_1-5x_2=12$$$$-x_1+2x_2=-6$$

Adding, $-3x_2=6$, so $x_2=-2$ and $x_1=2$.

$$x_1+x_2=0$$

Answer: (2) $0$

Example 14 (JEE Main 2026, 23 January Shift 2). Let

$$A=\begin{bmatrix}0&2&-3\\-2&0&1\\3&-1&0\end{bmatrix}$$

and $B$ be a matrix such that $B(I-A)=I+A$. Then the sum of the diagonal elements of $B^TB$ is ______.

▸ SHOW SOLUTION

Step 1. A is skew-symmetric

$A^T=-A$. Also $I-A$ and $I+A$ are invertible (each has determinant $1+4+9+1=15$), so

$$B=(I+A)(I-A)^{-1}$$

Step 2. Transpose B

$$B^T=\big((I-A)^{-1}\big)^T(I+A)^T$$$$B^T=(I+A)^{-1}(I-A)$$

using $(I-A)^T=I+A$ and $(I+A)^T=I-A$.

Step 3. Multiply

Write $P=I+A$ and $Q=I-A$, so $B=PQ^{-1}$ and $B^T=P^{-1}Q$:

$$B^TB=P^{-1}QPQ^{-1}$$

$P$ and $Q$ commute, so $QP=PQ$ and everything cancels:

$$B^TB=P^{-1}PQQ^{-1}$$$$B^TB=I$$

Step 4. Trace

$$\operatorname{tr}(B^TB)=\operatorname{tr}(I_3)=3$$

Answer: $3$

So $B$ is orthogonal. A skew-symmetric $A$ always gives an orthogonal $(I+A)(I-A)^{-1}$, and the entries of $A$ never mattered.

Example 15 (IITianForum). Let $P$ and $Q$ be invertible $2\times2$ real matrices with

$$P^{-1}+Q^{-1}=(P+Q)^{-1}$$

If $\lvert P\rvert=7$, then $\lvert Q\rvert$ is ______.

▸ SHOW SOLUTION

Step 1. Multiply out

$$(P+Q)(P^{-1}+Q^{-1})=I$$$$I+PQ^{-1}+QP^{-1}+I=I$$

Step 2. One matrix X

Put $X=PQ^{-1}$, so $QP^{-1}=X^{-1}$:

$$X+X^{-1}+I=O$$

Multiply by $X$:

$$X^2+X+I=O$$

Step 3. Multiply by X − I

$$(X-I)(X^2+X+I)=X^3-I=O$$$$X^3=I$$

Step 4. Determinants

$$\lvert X\rvert^3=1\implies\lvert X\rvert=1$$

since $\lvert X\rvert$ is real. And $\lvert X\rvert=\dfrac{\lvert P\rvert}{\lvert Q\rvert}$, so $\lvert Q\rvert=7$.

Answer: $7$

The order of the matrices was never used: for any order, the condition forces $\lvert P\rvert=\lvert Q\rvert$.

Elementary row operations

Three operations on the rows of a matrix are called elementary:

1. $R_i\leftrightarrow R_j$, swap two rows.
2. $R_i\to kR_i$ with $k\ne0$, multiply a row by a non-zero number.
3. $R_i\to R_i+kR_j$, add a multiple of another row.

The 3 by 3 identity matrix with three arrows to the three results: two rows swapped, one row multiplied by k, and k times row 3 added to row 1

Each one can be undone by an operation of the same kind, and their effect on a determinant is fixed: a swap changes the sign, scaling by $k$ multiplies it by $k$, and adding a multiple of another row changes nothing. Starting from $I$, whose determinant is $1$, the determinant can therefore never become $0$.

Conversely, every matrix with non-zero determinant can be reduced to $I$ by row operations, and running those operations backwards builds it from $I$. So:

$$\text{reachable from } I \iff \det\ne0$$

The same row operations, applied to $[A\mid I]$ until the left half becomes $I$, turn the right half into $A^{-1}$. That is the elementary-operations method for the inverse.

Example 16 (JEE Advanced 2026, Paper 1). Which one of the following matrices can be obtained by performing elementary row transformations on the $3\times3$ identity matrix?

$$(1)\ \begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}\quad(2)\ \begin{bmatrix}1&1&1\\2&3&4\\1&2&1\end{bmatrix}$$ $$(3)\ \begin{bmatrix}1&1&1\\2&3&4\\2&5&8\end{bmatrix}\quad(4)\ \begin{bmatrix}1&1&1\\-1&1&2\\0&2&3\end{bmatrix}$$

▸ SHOW SOLUTION

Step 1. Use the determinant test

Only a matrix with non-zero determinant can come from $I$.

Step 2. Check each option

(1): all rows equal, determinant $0$.

(3): $R_3=3R_2-4R_1$, determinant $0$.

(4): $R_3=R_1+R_2$, determinant $0$.

(2):

$$1(3-8)-1(2-4)+1(4-3)$$$$=-5+2+1=-2\ne0$$

Answer: (2)

Which result to use

A decision chart: from what the question gives, arrows lead to the column picture, trace of A transpose A, writing A as I plus N, the 2 by 2 identity, commuting polynomials in A, or the determinant test

Quick recall

Practise these methods on real exam questions in the doMath Matrices chapter, and continue with the Determinants note, which takes over from the inverse and builds the adjoint, Cramer's rule and systems of equations.