Let A=[[|gathered 1 3 -1 2 1 0 1 -1 gathered]]| be a singular matrix.

Question

Let $A=[[|\begin{gathered} 1 \quad  3 \quad  -1 \ 2 \quad  1 \quad  \alpha \ 0 \quad  1 \quad  -1 \end{gathered}]]|$ be a singular matrix. Let $f(x)={\int }_{x}^{0}({t}^{2}+2t+3)dt,x\in [1,\alpha ]$. If M and m are respectively the maximum and the minimum values of $f$ in $[1,\alpha ]$, then $3(M-m)$ is equal to :

Accepted Answer

Since $A$ is a singular matrix, its determinant must be zero.

We have

$A=[[|\begin{gathered} 1 \quad  3 \quad  -1 \ 2 \quad  1 \quad  \alpha \ 0 \quad  1 \quad  -1 \end{gathered}]]|$

So,

$|A|=0$

Now let us calculate the determinant by expanding along the first row:

$|A|=1|\begin{array}{cc} 1 & \alpha \ 1 & -1 \end{array}|-3|\begin{array}{cc} 2 & \alpha \ 0 & -1 \end{array}|+(-1)|\begin{array}{cc} 2 & 1 \ 0 & 1 \end{array}|$

$=1[(1)(-1)-(\alpha )(1)]-3[(2)(-1)-(\alpha )(0)]-1[(2)(1)-(1)(0)]$

$=(-1-\alpha )-3(-2)-2$

$=-1-\alpha +6-2$

$=3-\alpha $

Since $|A|=0$,

$3-\alpha =0\Rightarrow \alpha =3$

Now,

$f(x)={\int }_{x}^{0}({t}^{2}+2t+3)\, dt$

Integrating,

$f(x)={[\frac{{t}^{3}}{3}+{t}^{2}+3t]}_{x}^{0}=\frac{{x}^{3}}{3}+{x}^{2}+3x$

We need maximum and minimum values of $f(x)$ in $[1,\alpha ]=[1,3]$.

Now check whether $f(x)$ is increasing or decreasing.

${f}^{'}(x)={x}^{2}+2x+3$

${f}^{'}(x)={x}^{2}+2x+3=(x+1{)}^{2}+2>0\, x$

So $f(x)$ is strictly increasing on $[1,3]$.

Hence,

minimum value at $x=1$

maximum value at $x=3$

Now,

$m=f(1)=\frac{1}{3}+1+3=\frac{13}{3}$

and

$M=f(3)=\frac{27}{3}+9+9=9+9+9=27$

Therefore,

$M-m=27-\frac{13}{3}=\frac{81-13}{3}=\frac{68}{3}$

So,

$3(M-m)=3\cdot \frac{68}{3}=68$

Hence the correct answer is:

$68$

So, the correct option is .

Solutions · 1