If a + x = b + y = c + z + 1, where a, b, c, x, y, z are non-zero distinct real numbers, then | \beginmatrix x & a + y & x + a y & b + y & y + b z & c + y & z + c \endmatrix | is equal to :

Matrices and Determinants PYQ — JEE Main 2020 · 5 Sep · Shift 2 · Q23 with Solution

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Correct answer: B

Let the common value be $a+x=b+y=c+z+1=k.$ Then

\quad y=k-b, \quad z=k-c-1.$$ 2. The determinant is $$\Delta=\begin{vmatrix} x & a+y & x+a\\ y & b+y & y+b\\ z & c+y & z+c \end{vmatrix}.$$ Now simplify the entries: - Since $x+a=k$, we get third entry of row 1 as $k$. - Since $y+b=k$, we get third entry of row 2 as $k$. - Since $z+c=k-1$, we get third entry of row 3 as $k-1$. Also, $$a+y=a+(k-b)=k+(a-b),$$ $$b+y=b+(k-b)=k,$$ $$c+y=c+(k-b)=k+(c-b).$$ So $$\Delta=\begin{vmatrix} k-a & k+a-b & k\\ k-b & k & k\\ k-c-1 & k+c-b & k-1 \end{vmatrix}.$$ 3. Perform the column operation $$C_2\to C_2-C_3.$$ Then $$\Delta=\begin{vmatrix} k-a & a-b & k\\ k-b & 0 & k\\ k-c-1 & c-b+1 & k-1 \end{vmatrix}.$$ 4. Now perform $$R_1\to R_1-R_2, \qquad R_3\to R_3-R_2.$$ This gives $$\Delta=\begin{vmatrix} b-a & a-b & 0\\ k-b & 0 & k\\ b-c-1 & c-b+1 & -1 \end{vmatrix}.$$ Notice that $$a-b=-(b-a), \qquad c-b+1=-(b-c-1).$$ So row 1 becomes $$[b-a, -(b-a), 0],$$ and row 3 becomes $$[b-c-1, -(b-c-1), -1].$$ 5. Expand along the first row: $$\Delta=(b-a)\begin{vmatrix}0 & k\\ -(b-c-1) & -1\end{vmatrix}-(a-b)\begin{vmatrix}k-b & k\\ b-c-1 & -1\end{vmatrix}.$$ But this is a bit longer. A cleaner way is to use row 1 directly: Since row 1 is $[(b-a),-(b-a),0]$, factor $(b-a)$: $$\Delta=(b-a)\begin{vmatrix}1 & -1 & 0\\ k-b & 0 & k\\ b-c-1 & -(b-c-1) & -1\end{vmatrix}.$$ Now do $$C_1\to C_1+C_2.$$ Then $$\Delta=(b-a)\begin{vmatrix}0 & -1 & 0\\ k-b & 0 & k\\ 0 & -(b-c-1) & -1\end{vmatrix}.$$ Expanding along the first row, $$\Delta=(b-a)\left[(-1)^{1+2}(-1)\begin{vmatrix}k-b & k\\ 0 & -1\end{vmatrix}\right].$$ Since $(-1)^{3}=-1$, this becomes $$\Delta=(b-a)\begin{vmatrix}k-b & k\\ 0 & -1\end{vmatrix}.$$ Therefore, $$\Delta=(b-a)ig((k-b)(-1)-0\big)=-(b-a)(k-b).$$ But $k-b=y$, so $$\Delta=-(b-a)y=y(a-b).$$ 6. Therefore, $$\boxed{y(a-b)}.$$ So the correct option is **B**.

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View written solutionFree

Correct answer: B

Let the common value be $a+x=b+y=c+z+1=k.$ Then

\quad y=k-b, \quad z=k-c-1.$$ 2. The determinant is $$\Delta=\begin{vmatrix} x & a+y & x+a\\ y & b+y & y+b\\ z & c+y & z+c \end{vmatrix}.$$ Now simplify the entries: - Since $x+a=k$, we get third entry of row 1 as $k$. - Since $y+b=k$, we get third entry of row 2 as $k$. - Since $z+c=k-1$, we get third entry of row 3 as $k-1$. Also, $$a+y=a+(k-b)=k+(a-b),$$ $$b+y=b+(k-b)=k,$$ $$c+y=c+(k-b)=k+(c-b).$$ So $$\Delta=\begin{vmatrix} k-a & k+a-b & k\\ k-b & k & k\\ k-c-1 & k+c-b & k-1 \end{vmatrix}.$$ 3. Perform the column operation $$C_2\to C_2-C_3.$$ Then $$\Delta=\begin{vmatrix} k-a & a-b & k\\ k-b & 0 & k\\ k-c-1 & c-b+1 & k-1 \end{vmatrix}.$$ 4. Now perform $$R_1\to R_1-R_2, \qquad R_3\to R_3-R_2.$$ This gives $$\Delta=\begin{vmatrix} b-a & a-b & 0\\ k-b & 0 & k\\ b-c-1 & c-b+1 & -1 \end{vmatrix}.$$ Notice that $$a-b=-(b-a), \qquad c-b+1=-(b-c-1).$$ So row 1 becomes $$[b-a, -(b-a), 0],$$ and row 3 becomes $$[b-c-1, -(b-c-1), -1].$$ 5. Expand along the first row: $$\Delta=(b-a)\begin{vmatrix}0 & k\\ -(b-c-1) & -1\end{vmatrix}-(a-b)\begin{vmatrix}k-b & k\\ b-c-1 & -1\end{vmatrix}.$$ But this is a bit longer. A cleaner way is to use row 1 directly: Since row 1 is $[(b-a),-(b-a),0]$, factor $(b-a)$: $$\Delta=(b-a)\begin{vmatrix}1 & -1 & 0\\ k-b & 0 & k\\ b-c-1 & -(b-c-1) & -1\end{vmatrix}.$$ Now do $$C_1\to C_1+C_2.$$ Then $$\Delta=(b-a)\begin{vmatrix}0 & -1 & 0\\ k-b & 0 & k\\ 0 & -(b-c-1) & -1\end{vmatrix}.$$ Expanding along the first row, $$\Delta=(b-a)\left[(-1)^{1+2}(-1)\begin{vmatrix}k-b & k\\ 0 & -1\end{vmatrix}\right].$$ Since $(-1)^{3}=-1$, this becomes $$\Delta=(b-a)\begin{vmatrix}k-b & k\\ 0 & -1\end{vmatrix}.$$ Therefore, $$\Delta=(b-a)ig((k-b)(-1)-0\big)=-(b-a)(k-b).$$ But $k-b=y$, so $$\Delta=-(b-a)y=y(a-b).$$ 6. Therefore, $$\boxed{y(a-b)}.$$ So the correct option is **B**.

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