Question

If $\Delta = \begin{vmatrix} a & b & c \\ c & a & b \\ b & c & a \end{vmatrix}$, then the value of $\begin{vmatrix} a^2-bc & b^2-ca & c^2-ab \\ c^2-ab & a^2-bc & b^2-ca \\ b^2-ca & c^2-ab & a^2-bc \end{vmatrix}$ is-

• A. $\Delta^2$
• B. $2\Delta^2$
• C. $\Delta^3$
• D. none of

Answer 1

Answer: (A)

$\Delta^2$

Answer 2

To solve this problem, we need to evaluate the given determinant in terms of $\Delta$.

Let the first matrix be $M$: $M = \begin{pmatrix} a & b & c \\ c & a & b \\ b & c & a \end{pmatrix}$ The determinant $\Delta = |M| = \begin{vmatrix} a & b & c \\ c & a & b \\ b & c & a \end{vmatrix}$.

Now, let's look at the second determinant, let's call it $D'$: $D' = \begin{vmatrix} a^2-bc & b^2-ca & c^2-ab \\ c^2-ab & a^2-bc & b^2-ca \\ b^2-ca & c^2-ab & a^2-bc \end{vmatrix}$

We need to find the cofactors of the matrix $M$. The cofactor $A_{ij}$ of an element $m_{ij}$ is given by $(-1)^{i+j}$ times the determinant of the submatrix obtained by removing the $i$-th row and $j$-th column.

Let's calculate the cofactors of $M$: $A_{11} = (-1)^{1+1} \begin{vmatrix} a & b \\ c & a \end{vmatrix} = a^2 - bc$ $A_{12} = (-1)^{1+2} \begin{vmatrix} c & b \\ b & a \end{vmatrix} = -(ca - b^2) = b^2 - ca$ $A_{13} = (-1)^{1+3} \begin{vmatrix} c & a \\ b & c \end{vmatrix} = c^2 - ab$

$A_{21} = (-1)^{2+1} \begin{vmatrix} b & c \\ c & a \end{vmatrix} = -(ab - c^2) = c^2 - ab$ $A_{22} = (-1)^{2+2} \begin{vmatrix} a & c \\ b & a \end{vmatrix} = a^2 - bc$ $A_{23} = (-1)^{2+3} \begin{vmatrix} a & b \\ b & c \end{vmatrix} = -(ac - b^2) = b^2 - ca$

$A_{31} = (-1)^{3+1} \begin{vmatrix} b & c \\ a & b \end{vmatrix} = b^2 - ac = b^2 - ca$ $A_{32} = (-1)^{3+2} \begin{vmatrix} a & c \\ c & b \end{vmatrix} = -(ab - c^2) = c^2 - ab$ $A_{33} = (-1)^{3+3} \begin{vmatrix} a & b \\ c & a \end{vmatrix} = a^2 - bc$

Now, let's form the cofactor matrix $C$ of $M$: $C = \begin{pmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{pmatrix} = \begin{pmatrix} a^2-bc & b^2-ca & c^2-ab \\ c^2-ab & a^2-bc & b^2-ca \\ b^2-ca & c^2-ab & a^2-bc \end{pmatrix}$

Notice that the determinant $D'$ is exactly the determinant of the cofactor matrix $C$, i.e., $D' = |C|$.

We know a property of determinants and adjoint matrices: For any square matrix $M$ of order $n$, the determinant of its adjoint matrix, $|\text{adj}(M)|$, is equal to $|M|^{n-1}$. Also, the adjoint matrix is the transpose of the cofactor matrix, i.e., $\text{adj}(M) = C^T$. Therefore, $|\text{adj}(M)| = |C^T|$.

Since the determinant of a matrix is equal to the determinant of its transpose, $|C^T| = |C|$. So, we have $D' = |C| = |C^T| = |\text{adj}(M)|$.

For the given matrix $M$, its order is $n=3$. Thus, $|\text{adj}(M)| = |M|^{3-1} = |M|^2$. Since $|M| = \Delta$, we have $|\text{adj}(M)| = \Delta^2$.

Combining these results, we get: $D' = \Delta^2$.

The final answer is $\Delta^2$.