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.

KCET 2024Set A-11 markMCQ
$f(x) = \frac{\begin{vmatrix} \cos x & x & 1 \\ 2\sin x & x & 2x \\ \sin x & x & x \end{vmatrix}}{x^2}$
›Reveal solutionSolution

Column 2 is $(x,x,x)^T$ , so pull $x$ out; the determinant reduces to $x(\sin x - x\cos x)$ , and dividing by $x^2$ leaves $\frac{\sin x}{x}-\cos x\to 0$ .

Step 1 — Exploit the structure of the determinant.
$D=\begin{vmatrix}\cos x & x & 1\\ 2\sin x & x & 2x\\ \sin x & x & x\end{vmatrix}$
Every entry of the second column is $x$ . A determinant is linear in each column, so $x$ can be taken out of that column:
$D=x\begin{vmatrix}\cos x & 1 & 1\\ 2\sin x & 1 & 2x\\ \sin x & 1 & x\end{vmatrix}$
Step 2 — Expand the reduced determinant along the first column.
$\begin{vmatrix}\cos x & 1 & 1\\ 2\sin x & 1 & 2x\\ \sin x & 1 & x\end{vmatrix} =\cos x\,(x-2x)-1\,(2x\sin x-2x\sin x)+1\,(2\sin x-\sin x)$ $=-x\cos x-0+\sin x=\sin x-x\cos x$
Hence
$D=x\,(\sin x-x\cos x)=x\sin x-x^{2}\cos x$
Step 3 — Divide by $x^{2}$ and take the limit.

$\frac{D}{x^{2}}=\frac{x\sin x-x^{2}\cos x}{x^{2}}=\frac{\sin x}{x}-\cos x$ …
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COMEDK 2024Set 2024-E1 markMCQ
$A=\left[\begin{array}{ccc}0 & x & 16 \\ x & 5 & 7 \\ 0 & 9 & x\end{array}\right]$
›Reveal solutionSolution
A singular matrix has determinant zero. Setting the determinant of the given 3×3 matrix to zero yields a quadratic in $x$ , whose solutions are $x = -12$ and $x = 12$ . Among the options, only $-12$ appears, so the answer is (A).

Concept & Intuition

A singular matrix is one that does not have an inverse — its determinant is exactly zero. For a 3×3 matrix, the determinant is a polynomial in its entries. Here, the matrix contains the variable $x$ in three places, so setting $\det(A) = 0$ gives an equation we can solve for $x$ . The trick is to compute the determinant carefully, especially noticing the zeros in the first column — they will simplify the expansion.

Step-by-step solution

Write down the matrix

$A = \begin{pmatrix} 0 & x & 16 \\ x & 5 & 7 \\ 0 & 9 & x \end{pmatrix}$

Expand the determinant along the first column (because it has two zeros, making the calculation quick). The first column entries are: $a_{11}=0$ , $a_{21}=x$ , $a_{31}=0$ . Only the term from $a_{21}$ survives. The sign factor for row 2, column 1 is $(-1)^{2+1} = -1$ . So:

$\det(A) = 0 \cdot C_{11} + x \cdot (-1) \cdot M_{21} + 0 \cdot C_{31}$
where $M_{21}$ is the minor (determinant of the submatrix after removing row 2 and column 1).

Find the minor $M_{21}$ Remove row 2 and column 1:

$\begin{pmatrix} 0 & x & 16 \\ x & 5 & 7 \\ 0 & 9 & x \end{pmatrix} \longrightarrow \begin{pmatrix} x & 16 \\ 9 & x \end{pmatrix}$
The determinant of this 2×2 matrix is:
$M_{21} = (x)(x) - (16)(9) = x^2 - 144$

Assemble the full determinant

$\det(A) = -x \cdot (x^2 - 144) = -x(x^2 - 144)$

Set the determinant to zero (singular condition)
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COMEDK 2023Set 2023-E1 markMCQ
$\left[\begin{array}{ccc}2+x & 3 & 4 \\ 1 & -1 & 2 \\ x & 1 & -5\end{array}\right]$
›Reveal solutionSolution

Set equal to zero: 25 + 13x = 0 => x = -25/13.

Concept: a matrix is singular iff its determinant is zero.

det = | 2+x 3 4 ; 1 -1 2 ; x 1 -5 |

Expand along the first row:

(2+x) * [(-1)(-5) - (2)(1)] - 3 * [(1)(-5) - (2)(x)] + 4 * [(1)(1) - (-1)(x)]

= (2+x)(5 - 2) - 3(-5 - 2x) + 4(1 + x)

= 3(2 + x) + 15 + 6x + 4 + 4x …
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KCET 2018Set A-11 markMCQ
$A$
›Reveal solutionSolution

Use $|kA| = k^{n}|A|$ for an $n\times n$ matrix; here $n = 3$ , so the factor is $5^{3} = 125$ .

Step 1 — What scalar multiplication does.

$5A$ means every one of the nine entries is multiplied by $5$ . In particular, each of the 3 rows is scaled by $5$ .

Step 2 — The determinant property.

A determinant is a multilinear function of its rows: if one row is multiplied by $k$ , the determinant is multiplied by $k$ (this is the standard property $|R_i \to kR_i| = k|A|$ ). Applying it once per row for a $3\times3$ matrix:
$|5A| = 5 \cdot 5 \cdot 5 \cdot |A| = 5^{3}|A|$
Step 3 — General rule and evaluation.
$\boxed{|kA| = k^{n}|A| \text{ for } A \text{ of order } n\times n}$
$|5A| = 5^{3}|A| = 125\,|A|$ …
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COMEDK 2023Set 2023-M1 markMCQ
$A=\left[\begin{array}{cc}k+1 & 2 \\ 4 & k-1\end{array}\right]$
›Reveal solutionSolution

A singular matrix has zero determinant: $(k+1)(k-1)-8=0\Rightarrow k^2=9\Rightarrow k=\pm3$ .

$A=\begin{bmatrix}k+1&2\\4&k-1\end{bmatrix}$ singular means $\det A=0$ :

$(k+1)(k-1)-(2)(4)=0$ …
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COMEDK 2026Set 2026-M1 markMCQ
$x=4$
›Reveal solutionSolution
The determinant equation simplifies to a quadratic whose roots are the two values of $x$ that satisfy it; given one root is $4$ , the other root is found via Vieta’s formulas to be $-1$ .

We are given that $x = 4$ satisfies
$\begin{vmatrix} x & 3 \\ 1 & x-2 \end{vmatrix} = 5.$
The determinant of a $2\times2$ matrix $\begin{pmatrix}a & b \\ c & d\end{pmatrix}$ is $ad - bc$ . So the equation becomes
$x(x-2) - (3)(1) = 5.$

Simplify the determinant equation

$x(x-2) - 3 = 5 \quad\Longrightarrow\quad x^2 - 2x - 3 = 5.$
Bring all terms to one side:
$x^2 - 2x - 8 = 0.$

Recognize the quadratic

The equation $x^2 - 2x - 8 = 0$ is a quadratic in $x$ . Its two roots are the values of $x$ that make the original determinant equal to $5$ . We are told one root is $4$ .

Use Vieta’s formulas

For a quadratic $x^2 + bx + c = 0$ , the sum of the roots is $-b$ and the product is $c$ . Here $b = -2$ and $c = -8$ .

Let the roots be $r_1 = 4$ and $r_2$ . Then

$r_1 + r_2 = 2 \quad\Longrightarrow\quad 4 + r_2 = 2 \quad\Longrightarrow\quad r_2 = -2.$ …
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COMEDK 2022Set 20221 markMCQ
$A = \left[ {\begin{matrix} {2 - k} & 2 \\ 1 & {3 - k} \\ \end{matrix} } \right]$
›Reveal solutionSolution

So the value of 5k - k^2 is 4 (this holds for either root k = 1 or k = 4).

Concept: A matrix is singular iff its determinant is zero.

A = [[2 - k, 2], [1, 3 - k]]

|A| = (2 - k)(3 - k) - (2)(1) = 6 - 2k - 3k + k^2 - 2 = k^2 - 5k + 4 …
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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

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