Determinants for JEE: Properties, Adjoint, Inverse and Systems

Every square matrix has a single number attached to it, its determinant. That number decides whether the matrix has an inverse, whether a system of equations has one solution, none or infinitely many, and what area three points enclose. JEE asks about it in two ways: compute it cleverly, or use its rules (product, adjoint, inverse) without computing anything at all.

This note follows straight on from the Matrices note. It builds the value of a determinant, its properties, the product rule, area and collinearity, the adjoint and its chain of rules, the inverse, and systems of linear equations. Every worked example is a JEE Main 2026 question unless it says otherwise, and every answer was re-derived from scratch.

Value of a determinant

For a $2\times2$ matrix,

$$\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc$$

For a $3\times3$ matrix, the minor $M_{ij}$ of the entry $a_{ij}$ is the $2\times2$ determinant left after deleting row $i$ and column $j$. The cofactor is the minor with a sign:

$$C_{ij}=(-1)^{i+j}M_{ij}$$ Left: a 3 by 3 matrix with row 2 and column 3 shaded and the four remaining entries boxed as the minor of a23. Right: the checkerboard of signs plus, minus, plus for the cofactors

The determinant is the sum of the entries of any one row or column, each times its own cofactor. Along row 1:

$$\Delta=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}$$ $$\Delta=a_{11}M_{11}-a_{12}M_{12}+a_{13}M_{13}$$

Expand along the row or column with the most zeros. One more fact is used in the adjoint section: entries of one row times the cofactors of a different row always add to $0$.

$$a_{11}C_{21}+a_{12}C_{22}+a_{13}C_{23}=0$$

Example 1 (JEE Main 2026, 4 April Shift 2). Let

$$A=\begin{bmatrix}1&2&7\\4&-2&8\\3&8&-7\end{bmatrix}$$

and $\det(A-\alpha I)=0$ for a real number $\alpha$. If the largest possible value of $\alpha$ is $p$, then the circle $(x-p)^2+(y-2p)^2=320$ meets the coordinate axes at

▸ SHOW SOLUTION

Step 1. Expand the determinant

$\det(A-\alpha I)$ is

$$\begin{vmatrix}1-\alpha&2&7\\4&-2-\alpha&8\\3&8&-7-\alpha\end{vmatrix}$$

Expanding along row 1 and collecting powers of $\alpha$:

$$\det(A-\alpha I)=-\alpha^3-8\alpha^2+88\alpha+320$$

Step 2. Find the roots

$\alpha=8$ gives $-512-512+704+320=0$, so $\alpha-8$ is a factor:

$$-(\alpha-8)(\alpha^2+16\alpha+40)=0$$$$\alpha=8,\quad \alpha=-8\pm2\sqrt6$$

Both other roots are negative, so $p=8$.

Step 3. Meet the axes

The circle is $(x-8)^2+(y-16)^2=320$.

On $y=0$:

$$(x-8)^2=64\implies x=0,\ 16$$

On $x=0$:

$$(y-16)^2=256\implies y=0,\ 32$$

The origin is counted in both lists, so the distinct points are $(0,0)$, $(16,0)$ and $(0,32)$.

Circle with centre (8, 16) passing through the origin and meeting the x-axis again at (16, 0) and the y-axis again at (0, 32)

Answer: (3) $3$ points

Properties of determinants

These are what make a determinant quick to evaluate. Each is stated for rows; every one holds for columns as well.

Property 5 is the one that handles sums. If only one column of a determinant depends on $n$, then summing the determinant over $n$ is the same as summing that column and keeping the others fixed:

$$\sum_n\begin{vmatrix}f(n)&b_1&c_1\\g(n)&b_2&c_2\\h(n)&b_3&c_3\end{vmatrix}=\begin{vmatrix}\sum f(n)&b_1&c_1\\\sum g(n)&b_2&c_2\\\sum h(n)&b_3&c_3\end{vmatrix}$$

Example 2 (JEE Main 2026, 23 January Shift 1). Among the statements

I: if

$$D_1=\begin{vmatrix}1&\cos\alpha&\cos\beta\\\cos\alpha&1&\cos\gamma\\\cos\beta&\cos\gamma&1\end{vmatrix}$$ $$D_0=\begin{vmatrix}0&\cos\alpha&\cos\beta\\\cos\alpha&0&\cos\gamma\\\cos\beta&\cos\gamma&0\end{vmatrix}$$

and $D_1=D_0$,

then $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=\dfrac32$;

II: if the determinant

$$\begin{vmatrix}x^2+x&x+1&x-2\\2x^2+3x-1&3x&3x-3\\x^2+2x+3&2x-1&2x-1\end{vmatrix}$$

equals $px+q$,

then $p^2=196q^2$,

▸ SHOW SOLUTION

Step 1. Statement I

Write $a=\cos\alpha$, $b=\cos\beta$, $c=\cos\gamma$ and expand both:

$$\text{left}=1-a^2-b^2-c^2+2abc$$$$\text{right}=2abc$$

Equal only when $a^2+b^2+c^2=1$, not $\dfrac32$. False.

Step 2. Statement II, find q

Put $x=0$:

$$q=\begin{vmatrix}0&1&-2\\-1&0&-3\\3&-1&-1\end{vmatrix}$$$$q=-1(1+9)-2(1)=-12$$

Step 3. Find p

Put $x=1$:

$$p+q=\begin{vmatrix}2&2&-1\\4&3&0\\6&1&1\end{vmatrix}$$$$p+q=2(3)-2(4)-1(4-18)=12$$

So $p=24$.

Step 4. Check

$$p^2=576,\qquad 196q^2=196\times144$$

Not equal. False.

Answer: (1) both are false

Example 3 (JEE Main 2026, 5 April Shift 2). For $k\in\mathbb N$, define $f:\mathbb N\to\mathbb Z$ by letting $f(n)$ be

$$\begin{vmatrix}n&-1&-5\\-2n^2&3(2k+1)&2k+1\\-3n^3&3k(2k+1)&3k(k+2)+1\end{vmatrix}$$

If the sum of $f(n)$ for $n=1$ to $k$ is $98$, then $k$ is

▸ SHOW SOLUTION

Step 1. Expand along column 1

Only column 1 contains $n$. Its cofactors do not:

$$C_{11}=3(2k+1)(k^2+5k+1)$$$$C_{21}=-(27k^2+9k-1)$$$$C_{31}=14(2k+1)$$$$f(n)=nC_{11}-2n^2C_{21}-3n^3C_{31}$$

Step 2. Sum over n

Use the three standard sums:

$$\sum n=\frac{k(k+1)}{2}$$$$\sum n^2=\frac{k(k+1)(2k+1)}{6}$$$$\sum n^3=\frac{k^2(k+1)^2}{4}$$

Take $\dfrac{k(k+1)(2k+1)}{6}$ out of every term. What is left is

$$9(k^2+5k+1)+2(27k^2+9k-1)$$$$-63k(k+1)=7$$$$\sum_{n=1}^{k}f(n)=7\cdot\frac{k(k+1)(2k+1)}{6}$$

Step 3. Solve

$$7\sum n^2=98$$$$\sum n^2=14=1+4+9$$

So $k=3$.

Answer: (1) $3$

Example 4 (JEE Main 2026, 24 January Shift 2). Let $P=[p_{ij}]$ and $Q=[q_{ij}]$ be square matrices of order $3$ with $q_{ij}=2^{\,i+j-1}p_{ij}$ and $\det Q=2^{10}$. Then $\det(\operatorname{adj}(\operatorname{adj}P))$ is

▸ SHOW SOLUTION

Step 1. Take factors out of rows and columns

$$q_{ij}=2^{\,i-1}\cdot2^{\,j}\cdot p_{ij}$$

Row $i$ carries $2^{i-1}$, so the rows give $2^{0+1+2}=2^3$. Column $j$ carries $2^j$, so the columns give $2^{1+2+3}=2^6$.

$$\det Q=2^9\det P$$

Step 2. Find det P

$$2^9\det P=2^{10}\implies\det P=2$$

Step 3. Use the adjoint rule

For order $3$, $\lvert\operatorname{adj}(\operatorname{adj}P)\rvert=\lvert P\rvert^4$ (see the adjoint section):

$$2^4=16$$

Answer: (2) $16$

Example 5 (IITianForum). If $0\le[x]<3$, $-2\le[y]<1$ and $1\le[z]<4$, where $[\,\cdot\,]$ is the greatest integer function, find the maximum value of

$$\Delta=\begin{vmatrix}[x]+1&[y]&[z]\\ [x]&[y]+1&[z]\\ [x]&[y]&[z]+1\end{vmatrix}$$

▸ SHOW SOLUTION

Step 1. Short names

Write $p=[x]$, $q=[y]$, $r=[z]$.

Step 2. Add all columns into the first

$C_1\to C_1+C_2+C_3$ makes every entry of column 1 equal to $1+p+q+r$. Take it out:

$$\Delta=(1+p+q+r)\begin{vmatrix}1&q&r\\1&q+1&r\\1&q&r+1\end{vmatrix}$$

Step 3. Clear column 1

$R_2\to R_2-R_1$ and $R_3\to R_3-R_1$:

$$\begin{vmatrix}1&q&r\\0&1&0\\0&0&1\end{vmatrix}=1$$$$\Delta=1+p+q+r$$

Step 4. Largest values

$p\le2$, $q\le0$, $r\le3$:

$$\Delta_{\max}=1+2+0+3=6$$

Answer: $6$

Product rule

For square matrices of the same order,

$$\lvert AB\rvert=\lvert A\rvert\,\lvert B\rvert$$

and so

$$\lvert A^n\rvert=\lvert A\rvert^n,\qquad \lvert A^{-1}\rvert=\frac{1}{\lvert A\rvert}$$

A matrix with $\lvert A\rvert=0$ is singular; one with $\lvert A\rvert\ne0$ is non-singular. Only non-singular matrices have inverses.

Example 6 (JEE Main 2026, 22 January Shift 1). If

$$A=\begin{bmatrix}2&3\\3&5\end{bmatrix}$$

then the determinant of $A^{2025}-3A^{2024}+A^{2023}$ is

▸ SHOW SOLUTION

Step 1. Factor out the lowest power

$$A^{2025}-3A^{2024}+A^{2023}$$$$=A^{2023}(A^2-3A+I)$$

Step 2. Use the 2×2 identity

$\operatorname{tr}A=7$ and $\lvert A\rvert=10-9=1$:

$$A^2-7A+I=O\implies A^2+I=7A$$$$A^2-3A+I=4A$$

Step 3. Product rule

$$\lvert A^{2023}\cdot4A\rvert=\lvert A\rvert^{2023}\cdot4^2\lvert A\rvert$$$$=1\cdot16\cdot1=16$$

Answer: (4) $16$

Example 7 (IITianForum). Let

$$A_1=\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix}\ne0$$ $$A_2=\begin{vmatrix}a+3b&b+2c&c+5a\\b+3c&c+2a&a+5b\\c+3a&a+2b&b+5c\end{vmatrix}$$

Find $\dfrac{A_2}{A_1}$.

▸ SHOW SOLUTION

Step 1. Columns of A₂ from columns of A₁

Call the columns of $A_1$ $C_1$, $C_2$, $C_3$. The columns of $A_2$ are

$$C_1+3C_2,\quad C_2+2C_3,\quad C_3+5C_1$$

Step 2. Write it as a product

Column $j$ of a product $PM$ is $\sum_k(\text{column }k\text{ of }P)\,m_{kj}$, so

$$A_2=A_1\cdot\begin{vmatrix}1&0&5\\3&1&0\\0&2&1\end{vmatrix}$$

Step 3. The small determinant

$$1(1-0)-0+5(6-0)=31$$

Answer: $31$

Do not expand $A_1$ or $A_2$. Spotting that one is the other with columns mixed linearly turns the question into a $3\times3$ of small numbers.

Area of a triangle and collinearity

The area of the triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ is

$$\text{Area}=\frac12\left\lvert\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}\right\rvert$$

The determinant itself can be negative; the area is its absolute value. The three points are collinear exactly when the determinant is $0$.

Two panels: a shaded triangle with vertices A, B and C, and three points A, B, C on one straight line where the triangle has collapsed

Example 8 (IITianForum). The points $A(1,1)$, $B(3,4)$ and $C(k,7)$ form a triangle of area $6$. Find $k$. For which $k$ are the three points collinear?

▸ SHOW SOLUTION

Step 1. The determinant

$$\begin{vmatrix}1&1&1\\3&4&1\\k&7&1\end{vmatrix}$$$$=(4-7)-(3-k)+(21-4k)$$$$=15-3k$$

Step 2. Area 6

$$\frac12\lvert15-3k\rvert=6\implies\lvert15-3k\rvert=12$$$$k=1\ \text{or}\ k=9$$

Step 3. Collinear

$$15-3k=0\implies k=5$$

Check: $(1,1)$, $(3,4)$, $(5,7)$ all have slope $\dfrac32$ between them.

Answer: $k=1$ or $9$; collinear at $k=5$.

Both values of $k$ are needed: the area condition fixes $\lvert\Delta\rvert$, not its sign, and the two answers sit symmetrically on either side of the collinear value $5$.

Example 9 (IITianForum). A triangle has vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ and sides $a$, $b$, $c$. If

$$\Delta=\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}$$

and $\Delta^2=\mu\,(a+b+c)(b+c-a)(c+a-b)(a+b-c)$, find $\mu$.

▸ SHOW SOLUTION

Step 1. Area from the determinant

Area $S=\dfrac12\lvert\Delta\rvert$, so

$$\Delta^2=4S^2$$

Step 2. Area from Heron's formula

With $s=\dfrac{a+b+c}{2}$, $S^2=s(s-a)(s-b)(s-c)$, which is

$$16S^2=(a+b+c)(b+c-a)$$$$\qquad\times(c+a-b)(a+b-c)$$

Step 3. Compare

$$\Delta^2=4S^2=\frac14\cdot16S^2$$

Answer: $\mu=\dfrac14$

Adjoint of a matrix

The adjoint of a square matrix is the transpose of its matrix of cofactors:

$$\operatorname{adj}A=[C_{ij}]^T$$ Three 3 by 3 grids: A, then the cofactor matrix with each Cij in place, then adj A with the cofactors reflected across the diagonal

For a $2\times2$ matrix this is quick: swap the diagonal entries and change the sign of the other two.

$$\operatorname{adj}\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$

The key identity comes from the two expansion facts in the first section. Row $i$ of $A$ times column $j$ of $\operatorname{adj}A$ is row $i$ against the cofactors of row $j$, which is $\lvert A\rvert$ when $i=j$ and $0$ otherwise:

$$A(\operatorname{adj}A)=(\operatorname{adj}A)A=\lvert A\rvert I$$

Take determinants of both sides, for order $n$:

$$\lvert A\rvert\,\lvert\operatorname{adj}A\rvert=\lvert A\rvert^n$$

Everything in the table below follows from these two lines.

Example 10 (IITianForum). A square matrix $A$ of order $n$ has $\lvert A\rvert=4$ and $\lvert\operatorname{adj}A\rvert=2^{14}$. The number of entries in $A$ is

▸ SHOW SOLUTION

Step 1. Use the adjoint rule

$$\lvert\operatorname{adj}A\rvert=\lvert A\rvert^{n-1}=4^{n-1}=2^{2n-2}$$

Step 2. Solve

$$2n-2=14\implies n=8$$

An $8\times8$ matrix has $64$ entries.

Answer: (3) $64$

Example 11 (JEE Main 2026, 23 January Shift 1). Let $A$ be a $3\times3$ matrix with $\lvert A\rvert=6$. If

$$\left\lvert\operatorname{adj}\big(3\operatorname{adj}(A^2\operatorname{adj}(2A))\big)\right\rvert=2^m3^n$$

for $m,n\in\mathbb N$, then $m+n$ is ______.

▸ SHOW SOLUTION

Step 1. Innermost

$$\lvert2A\rvert=8\cdot6=48$$$$\lvert\operatorname{adj}(2A)\rvert=48^2=2^8\cdot3^2$$

Step 2. Next layer

$$\lvert A^2\operatorname{adj}(2A)\rvert=36\cdot2^8\cdot3^2=2^{10}\cdot3^4$$

Call this matrix $N$.

Step 3. Adjoint, then the scalar 3

$$\lvert\operatorname{adj}N\rvert=\lvert N\rvert^2=2^{20}\cdot3^8$$$$\lvert3\operatorname{adj}N\rvert=27\cdot2^{20}\cdot3^8=2^{20}\cdot3^{11}$$

Step 4. Outer adjoint

$$\left(2^{20}\cdot3^{11}\right)^2=2^{40}\cdot3^{22}$$$$m+n=40+22=62$$

Answer: $62$

Example 12 (JEE Main 2026, 22 January Shift 1). Let $A$ be a $3\times3$ matrix with $A+A^T=O$. If

$$A\begin{bmatrix}1\\-1\\0\end{bmatrix}=\begin{bmatrix}3\\3\\2\end{bmatrix},\quad A^2\begin{bmatrix}1\\-1\\0\end{bmatrix}=\begin{bmatrix}-3\\19\\-24\end{bmatrix}$$

and $\det\big(\operatorname{adj}(2\operatorname{adj}(A+I))\big)=2^\alpha3^\beta11^\gamma$ with $\alpha,\beta,\gamma$ non-negative integers, then $\alpha+\beta+\gamma$ is ______.

▸ SHOW SOLUTION

Step 1. Write the skew-symmetric A

$$A=\begin{bmatrix}0&a&b\\-a&0&c\\-b&-c&0\end{bmatrix}$$

Step 2. First condition

$$A\begin{bmatrix}1\\-1\\0\end{bmatrix}=\begin{bmatrix}-a\\-a\\-b+c\end{bmatrix}=\begin{bmatrix}3\\3\\2\end{bmatrix}$$

So $a=-3$ and $c-b=2$.

Step 3. Second condition

$A^2v=A(Av)$, so apply $A$ to $(3,3,2)$:

$$A\begin{bmatrix}3\\3\\2\end{bmatrix}=\begin{bmatrix}3a+2b\\-3a+2c\\-3b-3c\end{bmatrix}=\begin{bmatrix}-3\\19\\-24\end{bmatrix}$$

$3a+2b=-3$ gives $b=3$, and then $c=5$. (The other two rows check.)

Step 4. Determinant of A + I

$$\lvert A+I\rvert=\begin{vmatrix}1&-3&3\\3&1&5\\-3&-5&1\end{vmatrix}$$

For a skew-symmetric $3\times3$ matrix this is always $1+a^2+b^2+c^2$:

$$\lvert A+I\rvert=1+9+9+25=44$$

Step 5. Peel the layers

$$\lvert\operatorname{adj}(A+I)\rvert=44^2$$$$\lvert2\operatorname{adj}(A+I)\rvert=8\cdot44^2$$$$\lvert\operatorname{adj}(\ldots)\rvert=\left(8\cdot44^2\right)^2=2^6\cdot44^4$$

$44^4=2^8\cdot11^4$, so the value is $2^{14}\cdot11^4$.

$$\alpha+\beta+\gamma=14+0+4=18$$

Answer: $18$

Example 13 (JEE Main 2026, 4 April Shift 1). Let

$$A=\begin{bmatrix}1&1&2\\-2&0&1\\1&3&5\end{bmatrix}$$

Then the sum of all elements of $\operatorname{adj}\big(\operatorname{adj}\big(2(\operatorname{adj}A)^{-1}\big)\big)$ is

▸ SHOW SOLUTION

Step 1. Determinant of A

$$\lvert A\rvert=1(0-3)-1(-10-1)+2(-6)$$$$\lvert A\rvert=-3+11-12=-4$$

Step 2. Inverse of the adjoint

From $(\operatorname{adj}A)A=\lvert A\rvert I$:

$$(\operatorname{adj}A)^{-1}=\frac{A}{\lvert A\rvert}=-\frac{A}{4}$$

So $X=2(\operatorname{adj}A)^{-1}=-\dfrac{A}{2}$.

Step 3. Double adjoint

For order $3$, $\operatorname{adj}(\operatorname{adj}X)=\lvert X\rvert X$.

$$\lvert X\rvert=\left(-\frac12\right)^3(-4)=\frac12$$$$\operatorname{adj}(\operatorname{adj}X)=\frac12\cdot\left(-\frac A2\right)=-\frac A4$$

Step 4. Sum of elements

Elements of $A$ add to $12$, so the answer is $-\dfrac{12}{4}=-3$.

Answer: (4) $-3$

Example 14 (JEE Main 2026, 6 April Shift 1). Let

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

satisfy

$$A^2+\alpha\operatorname{adj}(\operatorname{adj}A)$$ $$+\beta\operatorname{adj}A\cdot\operatorname{adj}(\operatorname{adj}A)=C$$

where

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

for some real $\alpha,\beta$. Then $(\alpha-\beta)^2$ is ______.

▸ SHOW SOLUTION

Step 1. Determinant

Expand along row 3:

$$\lvert A\rvert=1\cdot\begin{vmatrix}-1&1\\1&0\end{vmatrix}=-1$$

Step 2. Simplify both adjoint terms

$$\operatorname{adj}(\operatorname{adj}A)=\lvert A\rvert A=-A$$$$\operatorname{adj}A\cdot(-A)=-\lvert A\rvert I=I$$

So the equation is $A^2-\alpha A+\beta I=C$.

Step 3. Compute A squared

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

Step 4. Compare two entries

Entry $(1,2)$: $-1-\alpha=-2$, so $\alpha=1$.

Entry $(3,3)$: $1-\alpha+\beta=-1$, so $\beta=-1$.

$$(\alpha-\beta)^2=4$$

Answer: $4$

Inverse of a matrix

Dividing $A(\operatorname{adj}A)=\lvert A\rvert I$ by $\lvert A\rvert$ gives the inverse whenever $\lvert A\rvert\ne0$:

$$A^{-1}=\frac{1}{\lvert A\rvert}\operatorname{adj}A$$

For a $2\times2$ matrix:

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}^{-1}=\frac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$

A matrix equation is solved by multiplying by the inverse on the correct side:

$$PA=B\implies P=BA^{-1}$$ $$AQ=B\implies Q=A^{-1}B$$

These are different matrices in general. Their traces, though, are equal, because $\operatorname{tr}(XY)=\operatorname{tr}(YX)$.

Example 15 (JEE Main 2026, 2 April Shift 2). Consider

$$A=\begin{bmatrix}2&-2\\4&-2\end{bmatrix},\quad B=\begin{bmatrix}3&9\\1&3\end{bmatrix}$$

If $PA=B$ and $AQ=B$, then the absolute value of the sum of the diagonal elements of $2(P+Q)$ is ______.

▸ SHOW SOLUTION

Step 1. Inverse of A

$$\lvert A\rvert=-4+8=4$$$$A^{-1}=\frac14\begin{bmatrix}-2&2\\-4&2\end{bmatrix}$$

Step 2. P and Q have the same trace

$P=BA^{-1}$ and $Q=A^{-1}B$, so

$$\operatorname{tr}P=\operatorname{tr}(BA^{-1})=\operatorname{tr}(A^{-1}B)=\operatorname{tr}Q$$

Step 3. Compute one of them

$$A^{-1}B=\frac14\begin{bmatrix}-4&-12\\-10&-30\end{bmatrix}$$$$\operatorname{tr}Q=\frac{-4-30}{4}=-\frac{17}{2}$$

Step 4. Trace of 2(P + Q)

$$2(\operatorname{tr}P+\operatorname{tr}Q)=2\left(-17\right)=-34$$

Answer: $34$

Example 16 (JEE Main 2026, 22 January Shift 2). If $X=\begin{bmatrix}x&y&z\end{bmatrix}^T$ is a solution of $AX=B$, where

$$\operatorname{adj}A=\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix},\quad B=\begin{bmatrix}4\\0\\2\end{bmatrix}$$

then $\lvert x+y+z\rvert$ is

▸ SHOW SOLUTION

Step 1. Get |A| from the adjoint

$$\lvert\operatorname{adj}A\rvert=4(10)-2(-20)+2(10)$$$$\lvert\operatorname{adj}A\rvert=100$$

For order $3$, $\lvert\operatorname{adj}A\rvert=\lvert A\rvert^2$, so $\lvert A\rvert=\pm10$.

Step 2. Solve

$$X=A^{-1}B=\frac{1}{\lvert A\rvert}(\operatorname{adj}A)B$$$$(\operatorname{adj}A)B=\begin{bmatrix}20\\-10\\10\end{bmatrix}$$

Step 3. Add

$$x+y+z=\frac{20-10+10}{\pm10}=\pm2$$

Answer: (4) $2$

The sign of $\lvert A\rvert$ is never fixed by the adjoint, which is exactly why the question asks for the absolute value.

Systems of linear equations

Three equations in $x,y,z$ with coefficient matrix $A$ and right side $B$ are written $AX=B$. Let $\Delta=\lvert A\rvert$, and let $\Delta_x$, $\Delta_y$, $\Delta_z$ be $\Delta$ with the $x$, $y$ or $z$ column replaced by $B$.

Cramer's rule. If $\Delta\ne0$, the system has exactly one solution:

$$x=\frac{\Delta_x}{\Delta},\quad y=\frac{\Delta_y}{\Delta},\quad z=\frac{\Delta_z}{\Delta}$$ Three panels in two variables: lines crossing at one point, parallel lines that never meet, and two lines lying on top of each other

The picture shows the same three outcomes for two lines in the plane. In three variables each equation is a plane, and the outcomes are the same: one point, nothing, or a whole line or plane of solutions.

The homogeneous system $AX=O$ always has $x=y=z=0$. It has a non-trivial solution, and then infinitely many, exactly when $\Delta=0$.

A flowchart: compute Delta; if non-zero, unique solution; if zero, homogeneous systems have infinitely many, otherwise write row 3 as a combination of rows 1 and 2 and compare right sides to decide between infinitely many and none

Example 17 (JEE Main 2026, 21 January Shift 2). If the system $3x+y+4z=3$, $2x+\alpha y-z=-3$, $x+2y+z=4$ has no solution, then $\alpha$ is

▸ SHOW SOLUTION

Step 1. Set Δ = 0

$$\Delta=3(\alpha+2)-3+4(4-\alpha)$$$$\Delta=19-\alpha$$

No solution needs $\Delta=0$, so $\alpha=19$.

Step 2. Confirm it is "no solution"

$$\Delta_x=\begin{vmatrix}3&1&4\\-3&19&-1\\4&2&1\end{vmatrix}$$$$\Delta_x=63-1-328=-266\ne0$$

Answer: (1) $19$

Example 18 (JEE Main 2026, 8 April Shift 2). If the system $x+y+z=6$, $x+2y+5z=10$, $2x+3y+\lambda z=\mu$ has infinitely many solutions, then $\lambda+\mu$ equals

▸ SHOW SOLUTION

Step 1. Set Δ = 0

$$\Delta=(2\lambda-15)-(\lambda-10)+(3-4)$$$$\Delta=\lambda-6$$

So $\lambda=6$.

Step 2. Third row as a combination

The coefficients $(2,3,6)$ are $(1,1,1)+(1,2,5)$, so the third equation must be the sum of the first two:

$$\mu=6+10=16$$$$\lambda+\mu=22$$

Answer: (3) $22$

Example 19 (JEE Main 2026, 23 January Shift 2). The system $x+y+z=6$, $2x+5y+az=36$, $x+2y+3z=b$ has

▸ SHOW SOLUTION

Step 1. Δ at a = 8

$$\Delta=(15-2a)-(6-a)+(4-5)$$$$\Delta=8-a$$

At $a=8$, $\Delta=0$, so (1) and (4) are out.

Step 2. Third row as a combination

Find $p,q$ with $(1,2,3)=p(1,1,1)+q(2,5,8)$:

$$p+2q=1,\quad p+5q=2$$$$q=\frac13,\quad p=\frac13$$

Check the third coordinate: $\dfrac13+\dfrac83=3$.

Step 3. Right sides

Infinitely many solutions need

$$b=\frac13\cdot6+\frac13\cdot36=14$$

At $b=16$ there is no solution.

Answer: (2)

Example 20 (JEE Main 2026, 6 April Shift 2). The sum of all values of $\theta\in[0,2\pi]$ for which the system $x\cos3\theta-8y-12z=0$, $x\cos2\theta+3y+3z=0$, $x+y+3z=0$ has a non-trivial solution is

▸ SHOW SOLUTION

Step 1. Homogeneous, so Δ = 0

$$\Delta=6\cos3\theta+24\cos2\theta-24$$$$\quad-12\cos2\theta+36$$$$\Delta=6\cos3\theta+12\cos2\theta+12$$

Step 2. Write in cos θ

Use $\cos3\theta=4c^3-3c$ and $\cos2\theta+1=2c^2$, with $c=\cos\theta$:

$$\Delta=6(4c^3+4c^2-3c)$$$$\Delta=6c(2c-1)(2c+3)$$

Step 3. Solve

$c=0$ gives $\dfrac\pi2$, $\dfrac{3\pi}{2}$. $c=\dfrac12$ gives $\dfrac\pi3$, $\dfrac{5\pi}{3}$. $c=-\dfrac32$ is impossible.

$$\frac\pi2+\frac{3\pi}2+\frac\pi3+\frac{5\pi}3=4\pi$$

Answer: (4) $4\pi$

Example 21 (JEE Main 2026, 5 April Shift 1). The system $x+2y+tz=0$, $6x+y+5tz=0$, $3x+t^2y+f(t)z=0$, where $f:\mathbb R\to\mathbb R$ is differentiable, has infinitely many solutions for every real $t$. Then $f$

▸ SHOW SOLUTION

Step 1. Homogeneous, so Δ = 0 for every t

$$\Delta=(f-5t^3)-2(6f-15t)$$$$\quad+t(6t^2-3)$$$$\Delta=-11f+t^3+27t$$

Step 2. Solve for f

$$f(t)=\frac{t^3+27t}{11}$$

Step 3. Derivative

$$f'(t)=\frac{3(t^2+9)}{11}>0$$

for every $t$.

Answer: (2) strictly increasing

Which result to use

Quick recall

Practise these methods on real exam questions in the doMath Determinants chapter, and go back to the Matrices note for powers, transpose and the $2\times2$ identity used in Example 6.