Prove that $\begin{vmatrix} bc - a^2 & ca - b^2 & ab - c^2 \\ ca - b^2 & ab - c^2 & bc - a^2 \\ ab - c^2 & bc - a^2 & ca - b^2 \end{vmatrix}$ is divisible by $a + b + c$ and find the quotient.

Prove that bc - a² ca - b² ab - c² ca - b² ab - c² – 2nd PUC Mathematics

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\begin{vmatrix} bc - a^2 & ca - b^2 & ab - c^2 \\ ca - b^2 & ab - c^2 & bc - a^2 \\ ab - c^2 & bc - a^2 & ca - b^2 \end{vmatrix}

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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 is a circulant in $p=bc-a^2,\,q=ca-b^2,\,r=ab-c^2$ , equal to $(a+b+c)^2(a^2+b^2+c^2-ab-bc-ca)^2$ ; dividing by $a+b+c$ leaves $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)^2$ .

The idea

Each row is a cyclic shift of $p,q,r$ . Such a circulant has the standard value $3pqr-p^3-q^3-r^3$ , which factors as $-(p+q+r)(p^2+q^2+r^2-pq-qr-rp)$ . We then substitute back in $a,b,c$ .

Step 1 — Name the entries

p=bc-a^2,\quad q=ca-b^2,\quad r=ab-c^2,\qquad \Delta=\begin{vmatrix}p&q&r\\q&r&p\\r&p&q\end{vmatrix}.

Step 2 — Circulant value

\Delta=3pqr-p^3-q^3-r^3=-(p+q+r)\big(p^2+q^2+r^2-pq-qr-rp\big).

Step 3 — The sum $p+q+r$

p+q+r=(bc+ca+ab)-(a^2+b^2+c^2)=-\big(a^2+b^2+c^2-ab-bc-ca\big).

Write $S=a^2+b^2+c^2-ab-bc-ca$ , so $p+q+r=-S$ .

Step 4 — The second factor

Compute one difference:

p-q=(bc-a^2)-(ca-b^2)=c(b-a)+(b-a)(b+a)=(b-a)(a+b+c).

Cyclically, $q-r=(c-b)(a+b+c)$ and $r-p=(a-c)(a+b+c)$ . Then …

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

COMEDK 2024Set 2024-A1 markMCQ
$\left|\begin{array}{ccc} \cos (\alpha+\beta) & -\sin (\alpha+\beta) & \cos 2 \beta \\ \sin \alpha & \cos \alpha & \sin \beta \\ -\cos \alpha & \sin \alpha & \cos \beta \end{array}\right|$
›Reveal solutionSolution

Expanding along the first row collapses the determinant to $1+\cos 2\beta$ , which contains no $\alpha$ — so it is independent of $\alpha$ : option (D).

Cofactor expansion along row 1

The three minors are
$M_{11}=\begin{vmatrix}\cos\alpha&\sin\beta\\\sin\alpha&\cos\beta\end{vmatrix}=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\cos(\alpha+\beta),$ $M_{12}=\begin{vmatrix}\sin\alpha&\sin\beta\\-\cos\alpha&\cos\beta\end{vmatrix}=\sin\alpha\cos\beta+\cos\alpha\sin\beta=\sin(\alpha+\beta),$ $M_{13}=\begin{vmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{vmatrix}=\sin^2\alpha+\cos^2\alpha=1.$
With the cofactor sign pattern $(+,-,+)$ and the row-1 entries $\cos(\alpha+\beta),\ -\sin(\alpha+\beta),\ \cos2\beta$ : …
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COMEDK 2025Set 2025-A1 markMCQ
$a_{21}$
›Reveal solutionSolution
The cofactor of $a_{21}$ is found by taking $(-1)^{2+1}$ times the determinant of the submatrix obtained by deleting row 2 and column 1. The result is $-4$ , so the correct option is (C).

The cofactor of an element in a matrix is not just the minor (the determinant of the submatrix left after removing that element’s row and column). It also includes a sign factor $(-1)^{i+j}$ , where $i$ and $j$ are the row and column indices. This sign alternates like a chessboard pattern. For $a_{21}$ (row 2, column 1), the sign is negative because $2+1=3$ is odd. So we compute the minor and then flip its sign.

Identify the element and its position. The element $a_{21}$ is in row 2, column 1. In the given matrix

$\Delta = \begin{vmatrix} 1 & 4 & 4 \\ -3 & 5 & 9 \\ 2 & 1 & 2 \end{vmatrix},$
$a_{21} = -3$ . But the cofactor depends only on position, not on the value of the element itself.

Delete row 2 and column 1. Removing row 2 and column 1 leaves the submatrix:

$\begin{pmatrix} 4 & 4 \\ 1 & 2 \end{pmatrix}.$

Compute the minor $M_{21}$ . The minor is the determinant of that $2\times2$ submatrix:

$M_{21} = \begin{vmatrix} 4 & 4 \\ 1 & 2 \end{vmatrix} = (4)(2) - (4)(1) = 8 - 4 = 4.$

Apply the sign factor. …
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Q.Prove that $\begin{vmatrix} bc - a^2 & ca - b^2 & ab - c^2 \\ ca - b^2 & ab - c^2 & bc - a^2 \\ ab - c^2 & bc - a^2 & ca - b^2 \end{vmatrix}$ is divisible by $a + b + c$ and find the quotient.

Concept understanding — Determinant Evaluation Using Identities

Determinant Evaluation Using Identities

The geometric intuition

The three row (or column) operations

The strategy

Worked example

The idea

Step 1 — Name the entries

Step 2 — Circulant value

Step 3 — The sum $p+q+r$

Step 4 — The second factor

Method: Evaluating a Circulant Determinant to Test Divisibility

Steps

Common Mistakes

The idea

Step 1 — Name the entries

Step 2 — Circulant value

Step 3 — The sum p+q+rp+q+rp+q+r

Step 4 — The second factor

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Step 3 — The sum $p+q+r$

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