JAMIA 2022 — Mathematics PYQ

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Question

JAMIA 2022 — Mathematics PYQ

JAMIA | Mathematics | 2022

If $a b c am p; p am p; q am p; r am p; x am p; y am p; z = 16$ , then the value of $p + x q + y r + z am p; a + x am p; b + y am p; c + z am p; a + p am p; b + q am p; c + r$ is

Choose the correct answer:

Explanation

1. Given Information:

Let $\Delta_1 = \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix} = 16$ .

We need to find the value of $\Delta_2 = \begin{vmatrix} p+x & a+x & a+p \\ q+y & b+y & b+q \\ r+z & c+z & c+r \end{vmatrix}$ .

2. Applying Column Operations:

Perform the operation $C_1 \to C_1 + C_2 + C_3$ :

\Delta_2 = \begin{vmatrix} 2(a+p+x) & a+x & a+p \\ 2(b+q+y) & b+y & b+q \\ 2(c+r+z) & c+z & c+r \end{vmatrix}

Take the constant factor $2$ out of the first column:

\Delta_2 = 2 \begin{vmatrix} a+p+x & a+x & a+p \\ b+q+y & b+y & b+q \\ c+r+z & c+z & c+r \end{vmatrix}

3. Simplifying the Columns:

Now, perform $C_2 \to C_2 - C_1$ :

\Delta_2 = 2 \begin{vmatrix} a+p+x & -p & a+p \\ b+q+y & -q & b+q \\ c+r+z & -r & c+r \end{vmatrix}

Next, perform $C_3 \to C_3 - C_1$ :

\Delta_2 = 2 \begin{vmatrix} a+p+x & -p & -x \\ b+q+y & -q & -y \\ c+r+z & -r & -z \end{vmatrix}

Finally, perform $C_1 \to C_1 + C_2 + C_3$ :

\Delta_2 = 2 \begin{vmatrix} a & -p & -x \\ b & -q & -y \\ c & -r & -z \end{vmatrix}

4. Factoring out the Negatives:

Take $-1$ out of the second column and $-1$ out of the third column:

\Delta_2 = 2 \times (-1) \times (-1) \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix}

\Delta_2 = 2 \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix}

5. Final Calculation:

Substitute the given value $\Delta_1 = 16$ :

\Delta_2 = 2 \times 16

\Delta_2 = 32

Final Answer:

The value of the determinant is $32$ .