If f(x) = (1 + x)¹⁷ (1 + x)¹⁹ (1 + x)²³ \(1 + x)²³ (1 + x)²⁹ (1 + x)³⁴ \(1 + x)⁴¹ (1 + x)⁴³ (1 + x)⁴⁷ = A + Bx

NCERT Exemplar · Q47

f(x) = \begin{vmatrix} (1 + x)^{17} & (1 + x)^{19} & (1 + x)^{23} \\ (1 + x)^{23} & (1 + x)^{29} & (1 + x)^{34} \\ (1 + x)^{41} & (1 + x)^{43} & (1 + x)^{47} \end{vmatrix} = A + Bx + Cx^2 + \ldots

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Concept understanding — Determinant Equality Equation

Determinant Equality Equation

Sometimes a determinant is not just a number to compute — it is set equal to a given value, and that equality becomes an equation you must solve. The unknown sits inside the matrix, so you first evaluate the determinant as an expression in that unknown, then solve the resulting ordinary equation.

Note

Core idea: a determinant containing a variable is just a polynomial in disguise. "Expand the determinant, set it equal to the given value, solve" — that is the whole recipe.

The basic move

Suppose you are told

\begin{vmatrix} x & 2 \\ 3 & x \end{vmatrix} = 10.

Expand the left side: $x \cdot x - 2 \cdot 3 = x^2 - 6$ . Now it is an equation you already know how to handle:

x^2 - 6 = 10 \;\Rightarrow\; x^2 = 16 \;\Rightarrow\; x = \pm 4.

A $2\times 2$ gives a quadratic; a $3\times 3$ typically gives a cubic. The number of solutions matches the degree of the polynomial you get.

Tip

Before expanding a $3\times 3$ , use row/column operations to create zeros. Fewer non-zero entries means a much shorter polynomial to solve — the value of the determinant is unchanged when you add a multiple of one row to another.

The important special case: equals zero

Most board problems set the determinant to $0$ :

\begin{vmatrix} 1 & 1 & 1 \\ 2 & x & 4 \\ 4 & x^2 & 16 \end{vmatrix} = 0.

Expanding gives a polynomial in $x$ ; its roots are the required values. Geometrically, a determinant being zero means the rows (or columns) are linearly dependent — the matrix is singular — so these equations often ask "for what value does the system collapse?"

A classic application: three points on one line

Three points $A(x_1, y_1)$ , $B(x_2, y_2)$ , $C(x_3, y_3)$ are collinear exactly when the area of triangle $ABC$ is zero. Since that area is $\tfrac{1}{2}$ of a determinant, the collinearity condition is a determinant equation: …

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The constant term $A$ of the determinant polynomial is just the determinant evaluated at $x=0$ , which simplifies to a $3\times3$ determinant of powers of $1$ . That determinant is zero because the rows become linearly dependent — specifically, the second row is a scalar multiple of the first. So $A = 0$ .

We are asked for the constant term $A$ in the expansion of $f(x)$ as a polynomial in $x$ . The determinant is a polynomial in $x$ because each entry is a binomial expansion in $x$ . The constant term of any polynomial $P(x)$ is simply $P(0)$ . So instead of expanding the whole determinant, we just plug $x=0$ into every entry.

1. Evaluate at $x=0$ .

When $x=0$ , each $(1+x)^n$ becomes $1^n = 1$ . So the matrix becomes:

\begin{vmatrix} 1^{17} & 1^{19} & 1^{23} \\ 1^{23} & 1^{29} & 1^{34} \\ 1^{41} & 1^{43} & 1^{47} \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix}

2. Recognize the structure.

All nine entries are $1$ . This is a matrix where every row is identical — the first row is $(1,1,1)$ , and so are the second and third rows.

3. Determinant of a matrix with two equal rows is zero. …

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

Showing the 12 most recent of 19 on this concept.

KEAM 2026Set eng-2026-04204 marksMCQ
$\begin{vmatrix} x+2 & 8 & 9 \\ 4 & x+6 & 9 \\ 4 & 8 & x+7 \end{vmatrix} = (x-2)^2(ax+b)$
›Reveal solutionSolution

Subtracting adjacent rows factors out $(x-2)^2$ , leaving a $3\times3$ determinant equal to $x+19$ . Thus the determinant is $(x-2)^2(x+19)$ , giving $a=1,\ b=19$ .

Apply $R_1\to R_1-R_2$ and $R_2\to R_2-R_3$ :
$R_1-R_2=(x-2,\; 2-x,\; 0)=(x-2)(1,-1,0),$ $R_2-R_3=(0,\; x-2,\; 2-x)=(x-2)(0,1,-1).$
Factoring $(x-2)$ from each of the first two rows:

$\Delta=(x-2)^2\begin{vmatrix}1 & -1 & 0\\ 0 & 1 & -1\\ 4 & 8 & x+7\end{vmatrix}.$ …
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KEAM 2025Set eng-2025-04274 marksMCQ
$\begin{vmatrix} 1 & 0 & 0 \\ x & x+2 & 0 \\ x^2 & x & x+3 \end{vmatrix} = 0$
›Reveal solutionSolution

The matrix is lower triangular, so its determinant is the product of the diagonal: $(x+2)(x+3)=0\Rightarrow x=-2,-3$ .

The matrix

$\begin{vmatrix}1&0&0\\x&x+2&0\\x^2&x&x+3\end{vmatrix}$ …
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KEAM 2025Set eng-2025-04254 marksMCQ
$A = \begin{bmatrix} a & -1 & -a \\ 0 & 1 & -1 \\ 1 & 0 & 4 \end{bmatrix}$
›Reveal solutionSolution

Cofactor-expand along the first row and solve $5a+1=26$ .

For $A = \begin{bmatrix} a & -1 & -a \\ 0 & 1 & -1 \\ 1 & 0 & 4 \end{bmatrix}$ , expand along the first row:

$|A| = a\begin{vmatrix}1 & -1\\ 0 & 4\end{vmatrix} - (-1)\begin{vmatrix}0 & -1\\ 1 & 4\end{vmatrix} + (-a)\begin{vmatrix}0 & 1\\ 1 & 0\end{vmatrix}.$ …
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KEAM 2025Set eng-2025-04284 marksMCQ
$\begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 2$
›Reveal solutionSolution

The determinant equals $abc-(a+b+c)+2$ . Setting it to $2$ gives $abc=a+b+c$ , whose positive-integer solution is $\{1,2,3\}$ , summing to $6$ .

Expanding the determinant:

$\begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix}=a(bc-1)-1(c-1)+1(1-b)=abc-a-b-c+2.$ …
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KEAM 2024Set eng-2024-06054 marksMCQ
$\begin{vmatrix}x & 2 & -1\\ 1 & x & 5\\ 3 & 2 & x\end{vmatrix}=0$
›Reveal solutionSolution

The determinant equals $x^3-9x+28$ ; its real root is $x=-4$ .

$\begin{vmatrix}x&2&-1\\1&x&5\\3&2&x\end{vmatrix}=x(x^2-10)-2(x-15)+(-1)(2-3x).$
$=x^3-10x-2x+30-2+3x=x^3-9x+28.$ …
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KEAM 2026Set eng-2026-04174 marksMCQ
$\begin{vmatrix} 2 & -2 & 4 \\ -5 & x+2 & -10 \\ -1 & 1 & x+1 \end{vmatrix}=0$
›Reveal solutionSolution

The determinant simplifies to $2x^2-18=0$ , whose roots are $x=\pm3$ .

Expand along the first row:
$\Delta=2\big[(x+2)(x+1)+10\big]+2\big[-5(x+1)-10\big]+4\big[-5+(x+2)\big].$
Compute each bracket:
$(x+2)(x+1)+10=x^2+3x+12,\quad -5(x+1)-10=-5x-15,\quad -5+(x+2)=x-3.$
So …
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KEAM 2022Set eng-2022-P2-B14 marksMCQ
$x$
›Reveal solutionSolution
$x=-2$ or $x=4$ .

Concept and Intuition

Expand the determinant and factor the resulting quadratic in $x$ .

Step-by-Step Solution

Expand along the first row: $x(2\cdot1-(-x)\cdot1)-4(2\cdot1-(-x)\cdot1)+0$ .

$=x(2+x)-4(2+x)=(2+x)(x-4)$ .

Set $=0$ : $(2+x)(x-4)=0$ .

Roots: $x=-2,\ x=4$ .

Common Mistakes …
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KEAM 2021Set eng-2021-P2-B14 marksMCQ
$\begin{vmatrix}1 & 2 & 1\\ 0 & x & -3\\ 2 & -1 & x\end{vmatrix}=0$
›Reveal solutionSolution
The determinant vanishes for $x=5$ and $x=-3$ .

Concept and Intuition

Setting a $3\times 3$ determinant to zero produces a polynomial equation in $x$ ; here it turns out quadratic.

Step-by-Step Solution

Expand along row 1: $1(x\cdot x-(-3)(-1))-2(0\cdot x-(-3)\cdot 2)+1(0\cdot(-1)-x\cdot 2)$ .

$=(x^2-3)-2(6)+(-2x)=x^2-2x-15$ .

Solve $x^2-2x-15=0\Rightarrow(x-5)(x+3)=0$ . …
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KEAM 2024Set eng-2024-06064 marksMCQ
$x$
›Reveal solutionSolution

Expand the determinant along row 1 and set equal to 6. …

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KEAM 2024Set eng-2024-06054 marksMCQ
$A=\begin{pmatrix}x & 2\\ 2 & x\end{pmatrix}$
›Reveal solutionSolution

$\det(A^2)=(\det A)^2$ ; solving $(x^2-4)^2=25$ gives $x=\pm3$ .

$\det A=x\cdot x-2\cdot2=x^2-4$ .

$\det(A^2)=(\det A)^2=(x^2-4)^2=25$ , so $x^2-4=\pm5$ . …
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KEAM 2024Set eng-2024-06094 marksMCQ
$\begin{vmatrix} x & 1 & 2 \\ 4 & 1 & x \\ 1 & -1 & 3 \end{vmatrix}=-10$
›Reveal solutionSolution

The determinant equals $x^2+4x-22$ ; setting it to $-10$ gives $x=2$ or $x=-6$ .

Expand along the first row:
$\begin{vmatrix}x&1&2\\4&1&x\\1&-1&3\end{vmatrix}=x(1\cdot3-x\cdot(-1))-1(4\cdot3-x\cdot1)+2(4\cdot(-1)-1\cdot1).$
$=x(3+x)-(12-x)+2(-5)=x^2+3x-12+x-10=x^2+4x-22.$ …
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KEAM 2026Set eng-2026-04184 marksMCQ
$A=\begin{pmatrix}9 & 2 & k\\ 1 & -1 & -3\\ k-1 & 1 & 3\end{pmatrix}$
›Reveal solutionSolution

$\det A=k^2-6k=k(k-6)=0\Rightarrow k=0,6$ .

A homogeneous system $AX=0$ has a nontrivial solution iff $\det A=0$ .

Expanding along the first row of $A=\begin{pmatrix}9&2&k\\1&-1&-3\\k-1&1&3\end{pmatrix}$ :

$\det A=9\big[(-1)(3)-(-3)(1)\big]-2\big[(1)(3)-(-3)(k-1)\big]+k\big[(1)(1)-(-1)(k-1)\big].$ …
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←Previous questionQ46 — 0 x - y x - z y - x 0 y - z z - x z - y 0 = ____ .→Next questionQ48 — (A³)⁻¹ = (A⁻¹)³, where A is a square matrix and |A| ≠ 0.

Q.If $f(x) = \begin{vmatrix} (1 + x)^{17} & (1 + x)^{19} & (1 + x)^{23} \\ (1 + x)^{23} & (1 + x)^{29} & (1 + x)^{34} \\ (1 + x)^{41} & (1 + x)^{43} & (1 + x)^{47} \end{vmatrix} = A + Bx + Cx^2 + \ldots$ , then $A =$ ________ .

Concept understanding — Determinant Equality Equation

Determinant Equality Equation

The basic move

The important special case: equals zero

A classic application: three points on one line

Method: Extracting a Specific Coefficient from a Determinant-Valued Polynomial

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

Concept and Intuition

Step-by-Step Solution

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