For some a, b, let f(x)=|\beginarrayccca+\frac\sin xx & 1 & ~b a & 1+\frac\sin xx & ~b a & 1 & ~b+\frac\sin xx\endarray|, x eq 0, \lim \limits_x arrow 0 f(x)=\lambda+\mu a+ u b. Then (\lambda+\mu+v)^2 is equal to :

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Correct answer: B

Rewrite the determinant neatly

Let $s=\frac{\sin x}{x}.$ Then

a+s & 1 & b\\ a & 1+s & b\\ a & 1 & b+s \end{vmatrix}, \qquad x\ne 0.$$ As $x\to 0$, we know $$\frac{\sin x}{x}\to 1,$$ so we can evaluate the limit by continuity of the determinant. 2. **Compute the determinant in terms of $s$** Observe the matrix as

\begin{pmatrix} a+s & 1 & b\ a & 1+s & b\ a & 1 & b+s \end{pmatrix}

\begin{pmatrix} a & 1 & b\ a & 1 & b\ a & 1 & b \end{pmatrix} + \begin{pmatrix} s & 0 & 0\ 0 & s & 0\ 0 & 0 & s \end{pmatrix}.

A quick determinant evaluation is easiest by row operations. Apply: - $R_1 \to R_1-R_3$ - $R_2 \to R_2-R_3$ Then

f(x)=\begin{vmatrix} s & 0 & -s\ 0 & s & -s\ a & 1 & b+s \end{vmatrix}.

Factor $s$ from the first two rows:

f(x)=s^2\begin{vmatrix} 1 & 0 & -1\ 0 & 1 & -1\ a & 1 & b+s \end{vmatrix}.

Now expand this determinant:

\begin{vmatrix} 1 & 0 & -1\ 0 & 1 & -1\ a & 1 & b+s \end{vmatrix} =1\begin{vmatrix}1 & -1\ 1 & b+s\end{vmatrix}+(-1)\begin{vmatrix}0 & 1\ a & 1\end{vmatrix}.

Compute each minor:

\begin{vmatrix}1 & -1\ 1 & b+s\end{vmatrix}=1(b+s)-(-1)(1)=b+s+1,

\begin{vmatrix}0 & 1\ a & 1\end{vmatrix}=0\cdot 1-1\cdot a=-a.

So

\begin{vmatrix} 1 & 0 & -1\ 0 & 1 & -1\ a & 1 & b+s \end{vmatrix}=(b+s+1)+(-1)(-a)=a+b+s+1.

Hence $$f(x)=s^2(a+b+s+1).$$ 3. **Take the limit as $x\to 0$** Since $s\to 1$,

\lim_{x\to 0} f(x)=1^2(a+b+1+1)=a+b+2.

Given

\lim_{x\to 0} f(x)=\lambda+\mu a+\nu b,

we compare coefficients: $$\lambda=2,\qquad \mu=1,\qquad \nu=1.$$ 4. **Find the required value**

(\lambda+\mu+\nu)^2=(2+1+1)^2=4^2=16.

5. **Compare with stored answer** Derived answer: **16** Stored correct answer: **B = 16** So the stored answer is correct.

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View written solutionFree

Correct answer: B

Rewrite the determinant neatly

Let $s=\frac{\sin x}{x}.$ Then

a+s & 1 & b\\ a & 1+s & b\\ a & 1 & b+s \end{vmatrix}, \qquad x\ne 0.$$ As $x\to 0$, we know $$\frac{\sin x}{x}\to 1,$$ so we can evaluate the limit by continuity of the determinant. 2. **Compute the determinant in terms of $s$** Observe the matrix as

\begin{pmatrix} a+s & 1 & b\ a & 1+s & b\ a & 1 & b+s \end{pmatrix}

\begin{pmatrix} a & 1 & b\ a & 1 & b\ a & 1 & b \end{pmatrix} + \begin{pmatrix} s & 0 & 0\ 0 & s & 0\ 0 & 0 & s \end{pmatrix}.

A quick determinant evaluation is easiest by row operations. Apply: - $R_1 \to R_1-R_3$ - $R_2 \to R_2-R_3$ Then

f(x)=\begin{vmatrix} s & 0 & -s\ 0 & s & -s\ a & 1 & b+s \end{vmatrix}.

Factor $s$ from the first two rows:

f(x)=s^2\begin{vmatrix} 1 & 0 & -1\ 0 & 1 & -1\ a & 1 & b+s \end{vmatrix}.

Now expand this determinant:

\begin{vmatrix} 1 & 0 & -1\ 0 & 1 & -1\ a & 1 & b+s \end{vmatrix} =1\begin{vmatrix}1 & -1\ 1 & b+s\end{vmatrix}+(-1)\begin{vmatrix}0 & 1\ a & 1\end{vmatrix}.

Compute each minor:

\begin{vmatrix}1 & -1\ 1 & b+s\end{vmatrix}=1(b+s)-(-1)(1)=b+s+1,

\begin{vmatrix}0 & 1\ a & 1\end{vmatrix}=0\cdot 1-1\cdot a=-a.

So

\begin{vmatrix} 1 & 0 & -1\ 0 & 1 & -1\ a & 1 & b+s \end{vmatrix}=(b+s+1)+(-1)(-a)=a+b+s+1.

Hence $$f(x)=s^2(a+b+s+1).$$ 3. **Take the limit as $x\to 0$** Since $s\to 1$,

\lim_{x\to 0} f(x)=1^2(a+b+1+1)=a+b+2.

Given

\lim_{x\to 0} f(x)=\lambda+\mu a+\nu b,

we compare coefficients: $$\lambda=2,\qquad \mu=1,\qquad \nu=1.$$ 4. **Find the required value**

(\lambda+\mu+\nu)^2=(2+1+1)^2=4^2=16.

5. **Compare with stored answer** Derived answer: **16** Stored correct answer: **B = 16** So the stored answer is correct.

Matrices and Determinants question

Matrices and Determinants question

\begin{pmatrix} a+s & 1 & b\ a & 1+s & b\ a & 1 & b+s \end{pmatrix}

More from Matrices and Determinants

Matrices and Determinants question

Matrices and Determinants question

\begin{pmatrix} a+s & 1 & b\ a & 1+s & b\ a & 1 & b+s \end{pmatrix}

More from Matrices and Determinants