If $x$, $y$, $z$ are all different from zero and $\begin{vmatrix} 1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z \end{vmatrix} = 0$, then value of $x^{-1} + y^{-1} + z^{-1}$ is (A) $xyz$ (B) $x^{-1} y^{-1} z^{-1}$ (C) $-x - y - z$ (D) $-1$

If x, y, z are all different from zero and 1 + x 1 1 \1 1 + y 1 \1 1 1 + z = 0, then value of x⁻¹ + y⁻¹ + z⁻¹

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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 determinant equals $xyz\left(1+\tfrac1x+\tfrac1y+\tfrac1z\right)$ ; since it is $0$ and $xyz\neq0$ , the reciprocal sum must be $-1$ — option (D).

The idea

Evaluate the determinant in closed form. It factors as $xyz$ times $\big(1+\sum 1/x\big)$ , so the condition $\Delta=0$ (with none of $x,y,z$ zero) pins the reciprocal sum immediately.

Step 1 — Simplify with column operations

Apply $C_1\to C_1-C_2$ and $C_2\to C_2-C_3$ :

\Delta=\begin{vmatrix}x&0&1\\-y&y&1\\0&-z&1+z\end{vmatrix}.

Step 2 — Expand

Expanding along the first row,

\Delta=x\big[y(1+z)-1\cdot(-z)\big]+1\cdot\big[(-y)(-z)-y\cdot0\big]=x(y+yz+z)+yz=xyz+xy+yz+zx.

Pulling out $x,y,z$ , …

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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 42 on this concept.

CBSE 2026Set 65/2/11 markMCQ
$\begin{vmatrix} -1 & -2 & 5 \\ -2 & a & -1 \\ 0 & 4 & 2a \end{vmatrix} = -86$
›Reveal solutionSolution
Expand the determinant along the first column, set it equal to $-86$ , and solve the resulting quadratic. The sum of roots is -4.

When a determinant equals a specific value, we compute the determinant algebraically (treating any unknowns as variables), then solve the resulting equation. The determinant of a $3 \times 3$ matrix can be found by cofactor expansion along any row or column; choosing the column or row with the most zeros minimizes arithmetic.

Here the first column has a zero in position $(3,1)$ , so expanding along the first column is efficient.

Solution

Expand along the first column

The determinant is:

$\begin{vmatrix} -1 & -2 & 5 \\ -2 & a & -1 \\ 0 & 4 & 2a \end{vmatrix} = (-1) \cdot \begin{vmatrix} a & -1 \\ 4 & 2a \end{vmatrix} - (-2) \cdot \begin{vmatrix} -2 & 5 \\ 4 & 2a \end{vmatrix} + 0 \cdot \begin{vmatrix} -2 & 5 \\ a & -1 \end{vmatrix}$
The signs alternate: $+, -, +$ down the column, and we multiply each by the element in that position.

Compute the $2 \times 2$ determinants

For the first minor:

$\begin{vmatrix} a & -1 \\ 4 & 2a \end{vmatrix} = a(2a) - (-1)(4) = 2a^2 + 4$
For the second minor:
$\begin{vmatrix} -2 & 5 \\ 4 & 2a \end{vmatrix} = (-2)(2a) - (5)(4) = -4a - 20$

Substitute back

$\text{Det} = (-1)(2a^2 + 4) + 2(-4a - 20)$ …
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CBSE 2026Set A1 markMCQ
$\begin{vmatrix} x & 4 \\ 2 & 2x \end{vmatrix} = 0 \Rightarrow x =$
(a) $\pm 2$
(b) $\pm 1$
(c) $\pm 3$
(d) $0$
›Reveal solutionSolution

Expanding the determinant: $2x^2-8=0$ , so $x=\pm 2$ .

Expand:

$\begin{vmatrix} x & 4 \\ 2 & 2x \end{vmatrix} = x\cdot 2x - 4\cdot 2 = 2x^2 - 8.$

Set equal to $0$ : …

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CBSE 2026Set ANNUAL1 markMCQ
$\begin{vmatrix} x & 3 \\ 8 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 2 \end{vmatrix}$
(a) $\sqrt{24}$
(b) $-\sqrt{24}$
(c) $\pm\sqrt{24}$
(d) None of these
›Reveal solutionSolution

Expand both $2\times2$ determinants and equate; the resulting value of $x$ does not match the listed options.

LHS: $\begin{vmatrix} x & 3 \\ 8 & x \end{vmatrix} = x^2 - 24$

RHS: $\begin{vmatrix} 6 & 2 \\ 18 & 2 \end{vmatrix} = 6(2) - 2(18) = 12-36 = -24$

Setting LHS = RHS:

$x^2 - 24 = -24$

$x^2 = 0$

…

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CBSE 2026Set ANNUAL1 markMCQ
$\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}$
(a) 3
(b) $\sqrt{3}$
(c) $-\sqrt{3}$
(d) $\sqrt{3}, -\sqrt{3}$
›Reveal solutionSolution

Evaluate both determinants and equate them to solve for $x$ .

LHS: $\begin{vmatrix}2 & 4\\ 5 & 1\end{vmatrix}=2(1)-4(5)=2-20=-18$

RHS: $\begin{vmatrix}2x & 4\\ 6 & x\end{vmatrix}=2x(x)-4(6)=2x^2-24$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If |x 0; 1 x| = |16 0; 8 4| (2\times2 determinants) then value of x is:$
(a) 3
(b) 2
(c) 4
(d) 8
›Reveal solutionSolution

Expand both $2\times2$ determinants and equate them to get $x^2 = 64$ .

For a $2\times2$ determinant $\begin{vmatrix}a & b\\ c& d\end{vmatrix} = ad-bc$ .

Left side: $\begin{vmatrix}x & 0\\ 1 & x\end{vmatrix} = x\cdot x - 0\cdot 1 = x^2$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q.If the determinant \begin{vmatrix}2x & 4\ 2 & 1\end{vmatrix} = 0, then the value of x will be:$
(a) 2
(b) 4
(c) 6
(d) 8
›Reveal solutionSolution

Expand the $2\times2$ determinant and solve the resulting linear equation for $x$ .

Working:

$\begin{vmatrix}2x & 4\\ 2 & 1\end{vmatrix} = (2x)(1) - (4)(2) = 2x - 8$

…

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CBSE 2026Set ANNUAL1 markMCQ
$\begin{vmatrix} 3 & x \\ x & 1 \end{vmatrix} = \begin{vmatrix} 3 & 2 \\ 4 & 1 \end{vmatrix}$
(a) $\pm 2\sqrt{2}$
(b) $\pm\sqrt{2}$
(c) 2
(d) -2
›Reveal solutionSolution

Expand both $2\times2$ determinants and equate, then solve the resulting quadratic in $x$ .

Left-hand side:

$\begin{vmatrix} 3 & x \\ x & 1 \end{vmatrix} = 3(1) - x(x) = 3 - x^2$

Right-hand side: …

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CBSE 2025Set E1 markMCQ
$\begin{vmatrix} x & 15 \\ 4 & 4 \end{vmatrix} = 0 \ \Rightarrow x =$
(a) $15$
(b) $-15$
(c) $12$
(d) $60$
›Reveal solutionSolution

Expand the determinant, set it to zero and solve for $x$ ; $x=15$ .

$\begin{vmatrix} x & 15 \\ 4 & 4 \end{vmatrix}=(x)(4)-(15)(4)=4x-60.$

…

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CBSE 2025Set A1 markMCQ
$\begin{vmatrix}1 & x \\ x & 1\end{vmatrix} = \begin{vmatrix}0 & 2 \\ 1 & 2\end{vmatrix}$
(a) $0$
(b) $\pm 1$
(c) $\pm\sqrt{3}$
(d) $\pm\sqrt{2}$
›Reveal solutionSolution

Expand both $2\times2$ determinants and equate.

Left side: $\begin{vmatrix}1 & x \\ x & 1\end{vmatrix} = 1(1) - x(x) = 1 - x^2$

Right side: $\begin{vmatrix}0 & 2 \\ 1 & 2\end{vmatrix} = 0(2) - 2(1) = -2$

…

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CBSE 2025Set ANNUAL1 mark
$\begin{vmatrix}2 & 1\\ 1 & 2\end{vmatrix}-k\begin{vmatrix}1 & 0\\ 0 & 1\end{vmatrix}=0$
›Reveal solutionSolution

Evaluate both determinants and solve the resulting linear equation for $k$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$x$
(a) $1$
(b) $0$
(c) $-1$
(d) $2$
›Reveal solutionSolution

A singular matrix has determinant zero; set |A| = 0 and solve for x.

$A = \begin{bmatrix} x & 2 \\ 2 & 4 \end{bmatrix}$

$|A| = x(4) - 2(2) = 4x - 4$

…

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CBSE 2025Set ANNUAL1 markMCQ
$\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$
(a) $2$
(b) $4$
(c) $0$
(d) $1$
›Reveal solutionSolution

Evaluate both 2×2 determinants and equate them.

$\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = 2(5)-3(4) = 10-12 = -2$

…

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←Previous questionQ34 — If A and B are invertible matrices, then which of the following is not…→Next questionQ36 — The value of the determinant x x + y x + 2y x + 2y x x + y x + y x + 2…

Q.If $x$ , $y$ , $z$ are all different from zero and $\begin{vmatrix} 1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z \end{vmatrix} = 0$ , then value of $x^{-1} + y^{-1} + z^{-1}$ is
(A) $xyz$
(B) $x^{-1} y^{-1} z^{-1}$
(C) $-x - y - z$
(D) $-1$

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

The idea

Step 1 — Simplify with column operations

Step 2 — Expand

Method: Evaluating a Determinant Whose Rows Differ by a Common Additive Shift

Steps

Common Mistakes

Solution

The idea

Step 1 — Simplify with column operations

Step 2 — Expand

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