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 19 on this concept.

KEAM 2026Set eng-2026-04174 marksMCQ
$\begin{pmatrix} \alpha & 2 & 3 \\ 2 & 3 & -\alpha \\ 3 & 5 & \alpha+1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} 2 \\ 3 \\ 5 \end{pmatrix}$
›Reveal solutionSolution
[!TLDR]

The system lacks a unique solution when $\det A = 0$ ; solving $8\alpha^2-7\alpha-1=0$ over the integers gives $\alpha = 1$ .

Concept

For a square system $A\mathbf{x}=\mathbf{b}$ , Cramer's rule / the invertibility criterion tells us a unique solution exists iff $\det A \neq 0$ . So "no unique solution" requires $\det A = 0$ . This determinant condition is standard in the NCERT/CBSE matrices and determinants chapter.

Solution

Expand the determinant along the first row:
$\det A = \alpha\begin{vmatrix}3 & -\alpha\\ 5 & \alpha+1\end{vmatrix} - 2\begin{vmatrix}2 & -\alpha\\ 3 & \alpha+1\end{vmatrix} + 3\begin{vmatrix}2 & 3\\ 3 & 5\end{vmatrix}.$
Evaluating each minor:
$\begin{vmatrix}3 & -\alpha\\ 5 & \alpha+1\end{vmatrix} = 3(\alpha+1)+5\alpha = 8\alpha+3,$
$\begin{vmatrix}2 & -\alpha\\ 3 & \alpha+1\end{vmatrix} = 2(\alpha+1)+3\alpha = 5\alpha+2,$ …
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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 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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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-2025-04234 marksMCQ
$\begin{bmatrix} 8-k & 2 \\ -2 & 4-k \end{bmatrix}$
›Reveal solutionSolution
Setting the determinant to zero gives (k-6)^2 = 0, so k = 6.

Concept and Intuition

A matrix is singular exactly when its determinant vanishes. For a 2x2 matrix this is a quadratic in k.

Step-by-Step Solution

det = (8-k)(4-k) - (2)(-2) = (8-k)(4-k) + 4.

Expand: 32 - 12k + k^2 + 4 = k^2 - 12k + 36. …
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KEAM 2025Set eng-2025-04234 marksMCQ
$x + y + z = 1$
›Reveal solutionSolution
The coefficient determinant vanishes when 4 + 2m = 0, i.e. m = -2.

Concept and Intuition

A linear system fails to have a unique solution precisely when the determinant of its coefficient matrix is zero, giving a condition on the parameter m.

Step-by-Step Solution

Determinant of [[1,1,1],[2,3,-m],[3,5,3]].

= 1(33 - (-m)5) - 1(23 - (-m)3) + 1(25 - 33). …
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KEAM 2025Set eng-2025-04254 marksMCQ
$A = \begin{bmatrix} 1 & -1 \\ 4\lambda & 8 \end{bmatrix}$
›Reveal solutionSolution

Set the $2\times 2$ determinant to zero.

For $A = \begin{bmatrix}1 & -1 \\ 4\lambda & 8\end{bmatrix}$ to be singular, $\det A = 0$ : …
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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-2025-04274 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-2025-04284 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 2025Set eng-2025-04284 marksMCQ
$x - 2y + z = 5$
›Reveal solutionSolution

The system has a unique solution iff its coefficient determinant is nonzero. That determinant is $3\lambda-3$ , so $\lambda\ne1$ .

The coefficient matrix is
$\begin{bmatrix}1&-2&1\\2&-1&2\\1&2&\lambda\end{bmatrix}.$
Its determinant: …
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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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←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

Concept

Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

The idea

Step 1 — Simplify with column operations

Step 2 — Expand

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