Evaluate $\begin{vmatrix}0&1\\-1&0\end{vmatrix}$.

Evaluate 01; -10 .

Exercise: Determinant of 2×2 and 3×3 ... · Q10

\begin{vmatrix}0&1\\-1&0\end{vmatrix}

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Concept understanding — Determinant of a Matrix

A determinant is a single number computed from a square array of numbers. For order 2, $\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc$ . For order 3, expand along any row or column using cofactors: expanding along row 1,

D=a_{11}\begin{vmatrix}a_{22}&a_{23}\\a_{32}&a_{33}\end{vmatrix}-a_{12}\begin{vmatrix}a_{21}&a_{23}\\a_{31}&a_{33}\end{vmatrix}+a_{13}\begin{vmatrix}a_{21}&a_{22}\\a_{31}&a_{32}\end{vmatrix}

Every determinant can be expanded along any of its 3 rows or 3 columns and gives the same value — a fact that follows from Property 1 (transpose invariance) of §4.2 and underlies why we're free to pick whichever row/column has the most zeros. The determinant of a matrix (as opposed to a bare grid of numbers) is defined only when the matrix is square — replace its square brackets with vertical bars to get $|A|=\det(A)$ . If $|A|=0$ , $A$ is called singular; otherwise non-singular. This single number packs in a huge amount of information: it tells you whether a linear system has a unique solution (Cramer's Rule, §4.3.1), whether three points are collinear (§4.3.3), and whether a matrix can be "undone" (has an inverse) — the last of these is developed further in the Class-12 continuation of this topic.

"Determinant of a matrix formula for order 2 and 3" and "determinants class 12 important questions" are among the most searched topics tied to the NCERT/CBSE Class 12 Mathematics Determinants chapter, since this concept underlies Cramer's-rule-style systems, collinearity checks and matrix invertibility questions in both board exams and JEE Main. Choosing a row or column with the most zeros before expanding, as noted here, is a genuine time-saving exam technique rather than just a textbook remark.

$\begin{vmatrix}0&1\\-1&0\end{vmatrix}=0(0)-1(-1)=0+1=1$ . [!ANSWER] $\begin{vmatrix}0&1\\-1&0\end{vmatrix}=1$ .

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

CBSE 2026Set ANNUAL1 markMCQ
$A$
(a) $\det A = -\det A'$
(b) $\det A = 2\det A'$
(c) $\det A = \det A'$
(d) $\det A = 3\det A'$
›Reveal solutionSolution

A standard property of determinants: $\det A=\det A'$ for any square matrix $A$ .

One of the fundamental properties of determinants is that the determinant of a matrix and its transpose are always equal, i.e.

$\det A=\det A'$

This is because expanding $\det A$ along its rows is identical in value to expanding $\det A'$ along its columns (which were the rows of $A$ ) — cofactor expansion gives the same sum of terms either way. This holds for every square matrix $A$ , regardless of order.

✓Final answer
$\det A=\det A'$
CBSE 2025Set A1 mark
$A$
›Reveal solutionSolution

Use the identity $\det(A^{-1}) = 1/\det(A)$ and check it against the claim $\det(A^{-1})=\det(A)$ .

For an invertible matrix $A$ , $A A^{-1} = I$ , so taking determinants:

$\det(A)\det(A^{-1}) = \det(I) = 1 \implies \det(A^{-1}) = \dfrac{1}{\det A}$

For this to equal $\det A$ , we'd need $\det A = \dfrac{1}{\det A}$ , i.e. $(\det A)^2 = 1$ , i.e. $\det A = \pm1$ . This is not true for every invertible $2\times2$ matrix (e.g. $A = \begin{bmatrix}2&0\\0&1\end{bmatrix}$ has $\det A = 2$ but $\det(A^{-1}) = 1/2\ne 2$ ). So the statement is false in general.

✓Final answer
$False.$
CBSE 2025Set A1 mark
$A$
›Reveal solutionSolution

"Singular" is defined as $|A|=0$ .

A square matrix $A$ is called singular if and only if $|A| = 0$ (equivalently, it has no inverse). This is the standard definition, so the statement is true.

✓Final answer
$True.$
CBSE 2023Set A1 markMCQ
$A$
(a) $5|A|$
(b) $25|A|$
(c) $125|A|$
(d) $15|A|$
›Reveal solutionSolution

Scaling every entry of a $2\times2$ matrix by 5 scales its determinant by $5^2=25$ .

For a square matrix $A$ of order $n$ and scalar $k$ , the rule is $|kA|=k^n|A|$ (each of the $n$ rows contributes one factor of $k$ ). Here $A$ is $2\times2$ so $n=2$ and $k=5$ :

$|5A|=5^2|A|=25|A|.$

✓Final answer
$25|A|$
CBSE 2022Set ANNUAL1 markMCQ
$Q.Let A be a square matrix of order 3 \times 3, then |kA| is equal to -$
(a) k|A|
(b) k^2|A|
(c) 3k|A|
(d) k^3|A|
›Reveal solutionSolution

For an $n\times n$ matrix, $|kA| = k^n|A|$ .

Here $n=3$ , so $|kA| = k^3|A|$ .

✓Final answer
$k^3|A|$
CBSE 2022Set HE2191 markMCQ
$A$
(a) $K|A|$
(b) $K^3|A|$
(c) $K^2|A|$
(d) $3K|A|$
›Reveal solutionSolution

Multiplying every entry of an $n\times n$ matrix by $K$ scales the determinant by $K^n$ .

If $A$ is a square matrix of order $n$ , then $|KA| = K^n |A|$ , because $K$ can be pulled out of each of the $n$ rows, contributing a factor $K$ each time. For $A$ of order $3\times 3$ , $n=3$ , so

$|KA| = K^3|A|$

✓Final answer
$K^3|A|$
CBSE 2020Set HE8231 markMCQ
$A$
(a) $\det(A)$
(b) $\dfrac{1}{\det(A)}$
(c) $1$
(d) $0$
›Reveal solutionSolution

$\det(A^{-1})=\dfrac{1}{\det(A)}$ .

For any invertible matrix $A$ , $AA^{-1}=I$ . Taking determinants of both sides and using the property $\det(PQ)=\det(P)\det(Q)$ :

$\det(A)\det(A^{-1})=\det(I)=1.$

Since $A$ is invertible, $\det(A)\ne0$ , so we can divide:

$\det(A^{-1})=\dfrac{1}{\det(A)}.$

This holds for a square invertible matrix of any order, including order $2$ .

✓Final answer
$\dfrac{1}{\det(A)}$

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Q.Evaluate ∣01−10∣\begin{vmatrix}0&1\\-1&0\end{vmatrix}​0−1​10​​.

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Q.Evaluate $\begin{vmatrix}0&1\\-1&0\end{vmatrix}$ .

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