Question

If a, b, c are real then the value of determinant $\begin{array}{ccc}a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ a c & b c & c^{2}+1\end{array}$ = 1 if

• A. a+b+c=0
• B. a+b+c=1
• C. a+b+c=-1
• D. a=b=c=0

Answer 1

Answer: (D)

a=b=c=0

Answer 2

To find the value of the determinant and the condition for it to be 1, we can use properties of determinants.

The given determinant is: $D = \begin{vmatrix} a^2+1 & ab & ac \\ ab & b^2+1 & bc \\ ac & bc & c^2+1 \end{vmatrix}$

Step 1: Transform the determinant

Multiply $C_1$ by $a$, $C_2$ by $b$, $C_3$ by $c$. To keep the determinant value unchanged, we must divide the entire determinant by $abc$.

$D = \frac{1}{abc} \begin{vmatrix} a(a^2+1) & b(ab) & c(ac) \\ a(ab) & b(b^2+1) & c(bc) \\ a(ac) & b(bc) & c(c^2+1) \end{vmatrix}$

$D = \frac{1}{abc} \begin{vmatrix} a^3+a & ab^2 & ac^2 \\ a^2b & b^3+b & bc^2 \\ a^2c & b^2c & c^3+c \end{vmatrix}$

Now, take $a$ common from $R_1$, $b$ common from $R_2$, and $c$ common from $R_3$.

$D = \frac{abc}{abc} \begin{vmatrix} a^2+1 & b^2 & c^2 \\ a^2 & b^2+1 & c^2 \\ a^2 & b^2 & c^2+1 \end{vmatrix}$

$D = \begin{vmatrix} a^2+1 & b^2 & c^2 \\ a^2 & b^2+1 & c^2 \\ a^2 & b^2 & c^2+1 \end{vmatrix}$

Step 2: Simplify the transformed determinant

Let $x = a^2$, $y = b^2$, $z = c^2$. The determinant becomes:

$D = \begin{vmatrix} x+1 & y & z \\ x & y+1 & z \\ x & y & z+1 \end{vmatrix}$

Apply the column operation $C_1 \to C_1 + C_2 + C_3$:

$D = \begin{vmatrix} x+y+z+1 & y & z \\ x+y+z+1 & y+1 & z \\ x+y+z+1 & y & z+1 \end{vmatrix}$

Take $(x+y+z+1)$ common from $C_1$:

$D = (x+y+z+1) \begin{vmatrix} 1 & y & z \\ 1 & y+1 & z \\ 1 & y & z+1 \end{vmatrix}$

Apply row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:

$D = (x+y+z+1) \begin{vmatrix} 1 & y & z \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$

Expand the determinant along $C_1$:

$D = (x+y+z+1) \cdot 1 \cdot (1 \cdot 1 - 0 \cdot 0)$

$D = (x+y+z+1) \cdot 1$

$D = x+y+z+1$

Step 3: Substitute back $x, y, z$ and find the condition

Substitute $x=a^2, y=b^2, z=c^2$ back into the expression for $D$:

$D = a^2+b^2+c^2+1$

The problem states that the value of the determinant is 1.

So, $a^2+b^2+c^2+1 = 1$

$a^2+b^2+c^2 = 0$

Since $a, b, c$ are real numbers, their squares ($a^2, b^2, c^2$) are non-negative. The sum of non-negative terms can be zero if and only if each term is zero.

Therefore, $a^2=0$, $b^2=0$, and $c^2=0$.

This implies $a=0$, $b=0$, and $c=0$.

The condition for the determinant to be 1 is $a=b=c=0$.