If | \beginmatrix a - b - c & 2a & 2a 2b & b - c - a & 2b 2c & 2c & c - a - b \endmatrix | = (a + b + c) (x + a + b + c)2, x e 0, then x is equal to :

Matrices and Determinants PYQ — JEE Main 2019 · 11 Jan · Shift 2 · Q32 with Solution

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Correct answer: A

Let

a-b-c & 2a & 2a\\ 2b & b-c-a & 2b\\ 2c & 2c & c-a-b \end{vmatrix}.$$ We are told it can be written as $$D=(a+b+c)(x+a+b+c)^2, \qquad x\ne 0.$$ We need to find $x$. --- ## 1. Rewrite the matrix neatly Let $$s=a+b+c.$$ Then $$a-b-c=2a-s,\qquad b-c-a=2b-s,\qquad c-a-b=2c-s.$$ So the determinant becomes $$D=\begin{vmatrix} 2a-s & 2a & 2a\\ 2b & 2b-s & 2b\\ 2c & 2c & 2c-s \end{vmatrix}.$$ This matrix can be written as

\begin{pmatrix}2a\2b\2c\end{pmatrix} \begin{pmatrix}1&1&1\end{pmatrix}-sI.

Indeed, the matrix $$\begin{pmatrix}2a\\2b\\2c\end{pmatrix}\begin{pmatrix}1&1&1\end{pmatrix}$$ has rows $(2a,2a,2a)$, $(2b,2b,2b)$, $(2c,2c,2c)$, and subtracting $sI$ gives exactly the required matrix. --- ## 2. Observe eigenvalues of the rank-one part Consider $$M=\begin{pmatrix}2a\\2b\\2c\end{pmatrix}\begin{pmatrix}1&1&1\end{pmatrix}.$$ This is a rank-one matrix of the form $uv^T$. For such a matrix: - one eigenvalue is $v^Tu$, - the other two eigenvalues are $0,0$. Here, $$u=\begin{pmatrix}2a\\2b\\2c\end{pmatrix}, \qquad v=\begin{pmatrix}1\\1\\1\end{pmatrix}.$$ So $$v^Tu=2a+2b+2c=2s.$$ Hence eigenvalues of $M$ are $$2s,\ 0,\ 0.$$ Now our determinant is for the matrix $$M-sI.$$ Therefore its eigenvalues are $$2s-s=s,\qquad 0-s=-s,\qquad 0-s=-s.$$ Thus, $$D=s(-s)(-s)=s^3.$$ So, $$D=(a+b+c)^3.$$ --- ## 3. Compare with the given form Given $$D=(a+b+c)(x+a+b+c)^2.$$ Since $s=a+b+c$, this is $$s(x+s)^2.$$ But we found $$D=s^3.$$ So $$s(x+s)^2=s^3.$$ If $s\neq 0$, divide by $s$: $$ (x+s)^2=s^2. $$ Hence $$x+s=\pm s.$$ So, $$x=0 \quad \text{or} \quad x=-2s.$$ Given $x\ne 0$, we must have $$x=-2s=-2(a+b+c).$$ --- ## 4. Check options - **A:** $-2(a+b+c)$ ✅ - **B:** $2(a+b+c)$ ❌ - **C:** $abc$ ❌ - **D:** $-(a+b+c)$ ❌ Therefore, the correct answer is $$\boxed{-2(a+b+c)}.$$ --- ## 5. Comparison with stored answer Stored correct answer: **A** Our derived answer is also **A**.

View written solutionFree

Correct answer: A

Let

a-b-c & 2a & 2a\\ 2b & b-c-a & 2b\\ 2c & 2c & c-a-b \end{vmatrix}.$$ We are told it can be written as $$D=(a+b+c)(x+a+b+c)^2, \qquad x\ne 0.$$ We need to find $x$. --- ## 1. Rewrite the matrix neatly Let $$s=a+b+c.$$ Then $$a-b-c=2a-s,\qquad b-c-a=2b-s,\qquad c-a-b=2c-s.$$ So the determinant becomes $$D=\begin{vmatrix} 2a-s & 2a & 2a\\ 2b & 2b-s & 2b\\ 2c & 2c & 2c-s \end{vmatrix}.$$ This matrix can be written as

\begin{pmatrix}2a\2b\2c\end{pmatrix} \begin{pmatrix}1&1&1\end{pmatrix}-sI.

Indeed, the matrix $$\begin{pmatrix}2a\\2b\\2c\end{pmatrix}\begin{pmatrix}1&1&1\end{pmatrix}$$ has rows $(2a,2a,2a)$, $(2b,2b,2b)$, $(2c,2c,2c)$, and subtracting $sI$ gives exactly the required matrix. --- ## 2. Observe eigenvalues of the rank-one part Consider $$M=\begin{pmatrix}2a\\2b\\2c\end{pmatrix}\begin{pmatrix}1&1&1\end{pmatrix}.$$ This is a rank-one matrix of the form $uv^T$. For such a matrix: - one eigenvalue is $v^Tu$, - the other two eigenvalues are $0,0$. Here, $$u=\begin{pmatrix}2a\\2b\\2c\end{pmatrix}, \qquad v=\begin{pmatrix}1\\1\\1\end{pmatrix}.$$ So $$v^Tu=2a+2b+2c=2s.$$ Hence eigenvalues of $M$ are $$2s,\ 0,\ 0.$$ Now our determinant is for the matrix $$M-sI.$$ Therefore its eigenvalues are $$2s-s=s,\qquad 0-s=-s,\qquad 0-s=-s.$$ Thus, $$D=s(-s)(-s)=s^3.$$ So, $$D=(a+b+c)^3.$$ --- ## 3. Compare with the given form Given $$D=(a+b+c)(x+a+b+c)^2.$$ Since $s=a+b+c$, this is $$s(x+s)^2.$$ But we found $$D=s^3.$$ So $$s(x+s)^2=s^3.$$ If $s\neq 0$, divide by $s$: $$ (x+s)^2=s^2. $$ Hence $$x+s=\pm s.$$ So, $$x=0 \quad \text{or} \quad x=-2s.$$ Given $x\ne 0$, we must have $$x=-2s=-2(a+b+c).$$ --- ## 4. Check options - **A:** $-2(a+b+c)$ ✅ - **B:** $2(a+b+c)$ ❌ - **C:** $abc$ ❌ - **D:** $-(a+b+c)$ ❌ Therefore, the correct answer is $$\boxed{-2(a+b+c)}.$$ --- ## 5. Comparison with stored answer Stored correct answer: **A** Our derived answer is also **A**.