Using properties of determinants, show that $\begin{vmatrix} 3a & -a+b & -a+c \\ -b+a & 3b & -b+c \\ -c+a & -c+b & 3c \end{vmatrix} = 3(a + b + c)(ab + bc + ca)$.

Using properties of determinants, show that 3a -a+b -a+c \-b+a 3b -b+c \-c+a -c+b 3c = 3(a + b + c)(ab + bc +

Question

\begin{vmatrix} 3a & -a+b & -a+c \\ -b+a & 3b & -b+c \\ -c+a & -c+b & 3c \end{vmatrix} = 3(a + b + c)(ab + bc + ca)

CBSECBSE Class XII Board 2019Subjective· 4mImportance★★★★★

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Concept understanding — Determinant Evaluation Using Identities

Determinant Evaluation Using Identities

Expanding a $4\times4$ or $5\times5$ determinant term by term is painful and error-prone. The smarter route is to transform the determinant into an easy form using properties (the "identities") that change its value in a known, controlled way — then read the answer off a triangular matrix.

The geometric intuition

A determinant measures the signed "volume" of the box spanned by the rows in $n$ -dimensional space. Sliding one row parallel to another doesn't change that volume; swapping two rows flips its sign; scaling a row scales the volume. The algebraic identities are just these facts translated into rules.

The three row (or column) operations

\det \to -\det

The identical rules hold for columns. There is also row-wise linearity: if a row is a sum $R_i = R_i' + R_i''$ , the determinant splits into the sum of two determinants with all other rows fixed.

Watch out

Row-wise linearity is not $\det(A+B)=\det A+\det B$ — that is false. The splitting works one row at a time.

The strategy

Use operation 3 to create zeros in a row or column (value unchanged).
Factor out common factors with operation 2.
Swap rows if needed to reach upper-triangular form (track the sign change).
The determinant is then the product of the diagonal entries.

Worked example

\det\begin{pmatrix}1&2&3\\4&5&6\\7&8&10\end{pmatrix}.

Apply $R_2\to R_2-4R_1$ and $R_3\to R_3-7R_1$ (no change), then $R_3\to R_3-2R_2$ : …

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The determinant simplifies by applying column operations that exploit symmetry, reducing it to a form where the factor $(a+b+c)$ can be pulled out, and the remaining determinant evaluates to $3(ab+bc+ca)$ .

We are given the determinant:

\Delta = \begin{vmatrix} 3a & -a+b & -a+c \\ -b+a & 3b & -b+c \\ -c+a & -c+b & 3c \end{vmatrix}

The goal is to show $\Delta = 3(a+b+c)(ab+bc+ca)$ using properties of determinants — no expansion by brute force.

Why use identities?

A determinant with symmetric off-diagonal terms often hides a common factor. By adding or subtracting rows and columns, we can create zeros or proportional entries, making the factor $(a+b+c)$ pop out. The remaining $2 \times 2$ determinant then yields the product $ab+bc+ca$ .

Add all columns to the first column.

Let $C_1 \to C_1 + C_2 + C_3$ .

The first column becomes:
- Row 1: $3a + (-a+b) + (-a+c) = 3a - a + b - a + c = a + b + c$
- Row 2: $(-b+a) + 3b + (-b+c) = -b + a + 3b - b + c = a + b + c$
- Row 3: $(-c+a) + (-c+b) + 3c = -c + a - c + b + 3c = a + b + c$
So:

\Delta = \begin{vmatrix} a+b+c & -a+b & -a+c \\ a+b+c & 3b & -b+c \\ a+b+c & -c+b & 3c \end{vmatrix}

Factor $(a+b+c)$ out of the first column. Since every entry in $C_1$ is the same, we get:

\Delta = (a+b+c) \begin{vmatrix} 1 & -a+b & -a+c \\ 1 & 3b & -b+c \\ 1 & -c+b & 3c \end{vmatrix}

Simplify the second and third columns using row operations.

Subtract row 1 from rows 2 and 3 (this does not change the determinant).

Let $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$ :
- New row 2:
  
  $(1-1,\; 3b - (-a+b),\; -b+c - (-a+c)) = (0,\; 3b + a - b,\; -b + c + a - c) = (0,\; a+2b,\; a-b)$
- New row 3:
  
  $(1-1,\; -c+b - (-a+b),\; 3c - (-a+c)) = (0,\; -c+b + a - b,\; 3c + a - c) = (0,\; a-c,\; a+2c)$
Row 1 stays as $(1,\; -a+b,\; -a+c)$ . So:

\Delta = (a+b+c) \begin{vmatrix} 1 & -a+b & -a+c \\ 0 & a+2b & a-b \\ 0 & a-c & a+2c \end{vmatrix}

Expand along the first column. The only non-zero entry in column 1 is the top $1$ , and its cofactor is the $2 \times 2$ determinant of the bottom-right block:

\Delta = (a+b+c) \cdot 1 \cdot \begin{vmatrix} a+2b & a-b \\ a-c & a+2c \end{vmatrix}

Evaluate the $2 \times 2$ determinant. Compute directly: …

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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 59 on this concept.

CBSE 2026Set 65/1/11 markMCQ
$\Delta_1 = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{vmatrix}$
›Reveal solutionSolution

The first determinant is diagonal; the second requires one row interchange to reach diagonal form. Each interchange flips the sign; evaluating both determinants gives $\Delta_2 = -2\Delta_1$ . The answer is (B).

Why determinants change under row operations

A determinant measures the signed volume of the parallelepiped spanned by the row vectors. When you swap two rows, you reflect the figure across a hyperplane—the volume stays the same in magnitude but the orientation reverses, flipping the sign.

The diagonal determinant is the easiest to compute: the product of the diagonal entries. The second determinant looks scrambled, but a single row swap will bring it into a form we recognize.

Step-by-step evaluation

1. Compute $\Delta_1$ directly.

The matrix is diagonal:
$\Delta_1 = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{vmatrix} = 1 \cdot 2 \cdot 3 = 6.$
2. Recognize the structure of $\Delta_2$ .
$\Delta_2 = \begin{vmatrix} 0 & 2 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 6 \end{vmatrix}.$
The first two rows are out of order compared to a diagonal form. Swap rows 1 and 2 to bring the $1$ into the top-left position.

3. Apply the row-interchange property.

Swapping rows 1 and 2:
$\Delta_2 = -\begin{vmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 6 \end{vmatrix}.$
The negative sign comes from the single interchange.

4. Evaluate the new diagonal determinant.
$\begin{vmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 6 \end{vmatrix} = 1 \cdot 2 \cdot 6 = 12.$
So $\Delta_2 = -12$ . …
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CBSE 2026Set A1 markMCQ
$\begin{vmatrix} 23 & 12 & 11 \\ 36 & 10 & 26 \\ 63 & 26 & 37 \end{vmatrix} =$
(a) $1$
(b) $-1$
(c) $0$
(d) $2$
›Reveal solutionSolution

The determinant equals $0$ because one column is the sum of the other two.

Inspect the columns of

$\begin{vmatrix} 23 & 12 & 11 \\ 36 & 10 & 26 \\ 63 & 26 & 37 \end{vmatrix}.$

Check: $12+11 = 23$ , $10+26 = 36$ , $26+37 = 63$ .

…

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CBSE 2026Set A1 markMCQ
$\begin{vmatrix} \cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ \end{vmatrix} =$
(a) $1$
(b) $0$
(c) $-1$
(d) $\frac{1}{2}$
›Reveal solutionSolution

The determinant equals $\cos 90^\circ = 0$ .

Expand:

$\begin{vmatrix} \cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ \end{vmatrix} = \cos 15^\circ\cos 75^\circ - \sin 15^\circ\sin 75^\circ.$

…

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CBSE 2026Set A1 markMCQ
$\begin{vmatrix} a+ib & c+id \\ -c+id & a-ib \end{vmatrix} =$
(a) $a^2 + b^2 + c^2 + d^2$
(b) $a^2 - b^2 - c^2 - d^2$
(c) $a^2 - b^2 + c^2 + d^2$
(d) $a^2 + b^2 + c^2 - d^2$
›Reveal solutionSolution

Expand the $2\times2$ determinant and simplify the complex products.

$\begin{vmatrix} a+ib & c+id \\ -c+id & a-ib \end{vmatrix} = (a+ib)(a-ib) - (c+id)(-c+id).$

First term: $(a+ib)(a-ib) = a^2 - (ib)^2 = a^2 + b^2$ . …

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

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

$\Delta = (x^2-x+1)(x+1) - (x-1)(x+1)$

$(x^2-x+1)(x+1) = x^3+1$ (the middle terms cancel).

$(x-1)(x+1) = x^2-1$ .

…

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CBSE 2026Set ANNUAL1 mark
$\Delta = \begin{vmatrix}1 & 2 & 4\\ -1 & 3 & 0\\ 4 & 1 & 0\end{vmatrix}$
›Reveal solutionSolution

Expand the $3\times3$ determinant along the first row.

$\Delta=\begin{vmatrix}1&2&4\\-1&3&0\\4&1&0\end{vmatrix}$

Expanding along row 1: …

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CBSE 2026Set ANNUAL1 mark
$\Delta = \begin{vmatrix}0 & \sin\alpha & -\cos\alpha\\ -\sin\alpha & 0 & \sin\beta\\ \cos\alpha & -\sin\beta & 0\end{vmatrix}$
›Reveal solutionSolution

The matrix is skew-symmetric (each $a_{ij}=-a_{ji}$ ) and every odd-order skew-symmetric matrix has determinant $0$ .

Check: $a_{12}=\sin\alpha=-a_{21}$ , $a_{13}=-\cos\alpha=-a_{31}$ , $a_{23}=\sin\beta=-a_{32}$ , and all diagonal entries are $0$ — so the matrix is skew-symmetric.

…

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CBSE 2026Set ANNUAL1 markMCQ
$\begin{vmatrix} \cos 30^\circ & \sin 30^\circ \\ \sin 30^\circ & \cos 30^\circ \end{vmatrix} =$
(a) $\dfrac{1}{2}$
(b) $\dfrac{\sqrt{3}}{2}$
(c) 0
(d) None of these
›Reveal solutionSolution

This determinant has the form $\cos^2\theta - \sin^2\theta = \cos 2\theta$ .

$\begin{vmatrix} \cos 30^\circ & \sin 30^\circ \\ \sin 30^\circ & \cos 30^\circ \end{vmatrix} = \cos 30^\circ \cdot \cos 30^\circ - \sin 30^\circ \cdot \sin 30^\circ = \cos^2 30^\circ - \sin^2 30^\circ$

…

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CBSE 2026Set ANNUAL1 mark
$Q.Evaluate the determinant \Delta = \begin{vmatrix}1 & 2 & 4\ -1 & 3 & 0\ 4 & 1 & 0\end{vmatrix}.$
›Reveal solutionSolution

Expand the $3\times3$ determinant along the first row (or any row/column) using cofactors.

Working: Expanding along Row 1:

$\Delta = \begin{vmatrix}1 & 2 & 4\\ -1 & 3 & 0\\ 4 & 1 & 0\end{vmatrix}$ …

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CBSE 2025Set 65/4/11 markMCQ
$\det(M) = m$
›Reveal solutionSolution
The key idea is that $MN = mI$ implies $N = m M^{-1}$ , so $\det(N) = m^3 \det(M^{-1}) = m^3 \cdot \frac{1}{m} = m^2$ . The correct option is (D).

The problem gives us two square matrices $M$ and $N$ of order 3, with $\det(M) = m$ and $MN = mI$ , where $I$ is the $3 \times 3$ identity matrix. We need $\det(N)$ .

The central concept here is the relationship between matrix multiplication and determinants. When two matrices multiply to give a scalar times the identity, that scalar is intimately connected to the determinant of the first matrix. The equation $MN = mI$ is not just a product — it tells us that $N$ is essentially a scaled inverse of $M$ .

Why? Because if $MN = mI$ , then multiplying both sides on the left by $M^{-1}$ (assuming $M$ is invertible) gives $N = m M^{-1}$ . But we must first check: is $M$ invertible? Yes — since $\det(M) = m \neq 0$ (the problem doesn't state $m \neq 0$ explicitly, but if $m = 0$ , then $MN = 0$ , which would make $N$ singular and the answer ambiguous; in standard exam contexts, $m$ is taken as a non-zero scalar, often a real number, and the options suggest $m \neq 0$ ). So $M^{-1}$ exists.

Now, the determinant of a scalar multiple of a matrix: for an $n \times n$ matrix $A$ , $\det(kA) = k^n \det(A)$ . Here $n = 3$ , so $\det(m M^{-1}) = m^3 \det(M^{-1})$ .

And we know $\det(M^{-1}) = \frac{1}{\det(M)} = \frac{1}{m}$ .

Putting it together:

From $MN = mI$ , take determinant on both sides: $\det(MN) = \det(mI)$ .

$\det(MN) = \det(M) \cdot \det(N) = m \cdot \det(N)$ . …
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CBSE 2025Set E1 markMCQ
$\begin{vmatrix} 21 & 11 & 10 \\ 25 & 15 & 10 \\ 64 & 27 & 37 \end{vmatrix} =$
(a) $1190$
(b) $841$
(c) $0$
(d) $1$
›Reveal solutionSolution

A column that is the sum of the other two makes the determinant zero.

Examine the columns of

$\begin{vmatrix} 21 & 11 & 10 \\ 25 & 15 & 10 \\ 64 & 27 & 37 \end{vmatrix}.$

Check $C_2+C_3$ against $C_1$ :

$11+10=21,\quad 15+10=25,\quad 27+37=64.$

…

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CBSE 2025Set E1 markMCQ
$\begin{vmatrix} 10 & 4 \\ 13 & 5 \end{vmatrix} =$
(a) $102$
(b) $2$
(c) $-2$
(d) $-102$
›Reveal solutionSolution

Cross-multiply: $ad-bc=-2$ .

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

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