Maths MCQ

$\begin{aligned} & f(A)=2 A^2-3 A \\ & =2\left[\begin{array}{cc}-2 & 1 \\ 0 & 3\end{array}\right]\left[\begin{array}{cc}-2 & 1 \\ 0 & 3\end{array}\right]-3\left[\begin{array}{cc}-2 & 1 \\ 0 & 3\end{array}\right] \\ & =2\left[\begin{array}{ll}4 & 1 \\ 0 & 9\end{array}\right]-\left[\begin{array}{cc}-6 & 3 \\ 0 & 9\end{array}\right] \\ & =\left[\begin{array}{cc}8+6 & 2-3 \\ 0-0 & 18-9\end{array}\right]=\left[\begin{array}{cc}14 & -1 \\ 0 & 9\end{array}\right]\end{aligned}$

Hint:

$(A B)^{\top}=B^{\top} A^{\top}$

Hint:

$\begin{aligned} & \left|A^3\right|=125 \\ & |A|^3=53 \ldots \ldots\left[\because\left|A^n\right|=|A|^n, n \in N\right]\end{aligned}$

∴ |A| = 5

$\left|\begin{array}{ll}\alpha & 2 \\ 2 & \alpha\end{array}\right|=5$

$\begin{aligned} & a^2-4=5 \\ & \alpha^2=9\end{aligned}$

∴ α = ± 3

$A-\lambda I=\left[\begin{array}{ll}1 & 3 \\ 2 & 2\end{array}\right]-\lambda\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1-\lambda & 3 \\ 2 & 2-\lambda\end{array}\right]$

Since $A-\lambda I$ is a singular matrix,

$\left|\begin{array}{cc}1-\lambda & 3 \\ 2 & 2-\lambda\end{array}\right|=0$

$\begin{aligned} & (1-\lambda)(2-\lambda)-6=0 \\ & 2-3 \lambda+\lambda^2-6=0 \\ & \lambda^2-3 \lambda-4=0\end{aligned}$

$\left|\begin{array}{lll}x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x\end{array}\right|=0$

Applying $\mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_2+\mathrm{R}_3$, we get

$\left|\begin{array}{ccc}x+9 & x+9 & x+9 \\ 2 & x & 2 \\ 7 & 6 & x\end{array}\right|=0$

Taking $(x+9)$ common from $\mathrm{R}_1$, we get

$(x+9)\left|\begin{array}{lll}1 & 1 & 1 \\ 2 & x & 2 \\ 7 & 6 & x\end{array}\right|$ =0

$\begin{aligned} & (x+9)\left[1\left(x^2-12\right)-1(2 x-14)+1(12-7 x)\right]=0 \\ & (x+9)\left(x^2-9 x+14\right)=0 \\ & (x+9)(x-7)(x-2)=0\end{aligned}$

The other two roots are $x=2$ and $x=7$.

$\left|\begin{array}{ccc}3 & -1 & 4 \\ 1 & 2 & -3 \\ 6 & 5 & \lambda\end{array}\right|=0$

Applying $R_3 \rightarrow R_3-\left(R_1+3 R_2\right)$, we get

$\left|\begin{array}{ccc}3 & -1 & 4 \\ 1 & 2 & -3 \\ 0 & 0 & \lambda+5\end{array}\right|=0$

$\begin{array}{ll}\therefore & 3[2(\lambda+5)-0]-(-1)(\lambda+5-0)+4(0-0)=0 \\ \therefore & 7 \lambda+35=0 \\ \therefore & \lambda=-\frac{35}{7}=-5\end{array}$

Hint:

Let $D=\left|\begin{array}{lll}b+c & \mathrm{c}+\mathrm{a} & \mathrm{a}+\mathrm{b} \\ \mathrm{q}+\mathrm{r} & \mathrm{r}+\mathrm{p} & \mathrm{p}+\mathrm{q} \\ y+z & z+x & x+y\end{array}\right|$

Applying $\mathrm{C}_1 \rightarrow \mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3$, we get

$D=\left|\begin{array}{lll}2(a+b+c) & c+a & a+b \\ 2(p+q+r) & r+p & p+q \\ 2(x+y+z) & z+x & x+y\end{array}\right|$

$\mathrm{D}=2\left|\begin{array}{lll}\mathrm{a}+\mathrm{b}+\mathrm{c} & \mathrm{c}+\mathrm{a} & \mathrm{a}+\mathrm{b} \\ \mathrm{p}+\mathrm{q}+\mathrm{r} & \mathrm{r}+\mathrm{p} & \mathrm{p}+\mathrm{q} \\ x+y+z & \mathrm{z}+x & x+y\end{array}\right|$

Applying $\mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_1$ and $\mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_1$, we get

$D=2\left|\begin{array}{lll}a+b+c & -b & -c \\ p+q+r & -q & -r \\ x+y+z & -y & -z\end{array}\right|$

Applying $C_1 \rightarrow C_1+C_2+C_3$, we get

$\mathrm{D}=2\left|\begin{array}{lll}a & -\mathrm{b} & -\mathrm{c} \\ \mathrm{p} & -\mathrm{q} & -\mathrm{r} \\ x & -y & -z\end{array}\right|$

Taking $(-1)$ common from $C_2$ and $C_3$, we get

$\mathrm{D}=2(-1)(-1)\left|\begin{array}{lll}\mathrm{a} & \mathrm{b} & \mathrm{c} \\ \mathrm{p} & \mathrm{q} & \mathrm{r} \\ x & y & \mathrm{z}\end{array}\right|=2\left|\begin{array}{lll}\mathrm{a} & \mathrm{b} & \mathrm{c} \\ \mathrm{p} & \mathrm{q} & \mathrm{r} \\ x & y & \mathrm{z}\end{array}\right|$

The given system of equations will have a non-zero solution, if

$\left|\begin{array}{ccc}a^3 & (a+1)^3 & (a+2)^3 \\ a & a+1 & a+2 \\ 1 & 1 & 1\end{array}\right|=0$

Applying $\mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_1, \mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_2$, we get

$\left|\begin{array}{ccc}a^3 & 3 a^2+3 a+1 & 3 a^2+9 a+7 \\ a & 1 & 1 \\ 1 & 0 & 0\end{array}\right|=0$

$\begin{aligned} a^3(0-0)-\left(3 a^2+3 a\right. & +1)(0-1) \\ & +\left(3 a^2+9 a+7\right)(0-1)=0\end{aligned}$

$-6 a-6=0, \quad \therefore \quad a=-1$

Hint:

$\left|\begin{array}{lll}x^k & x^{k+2} & x^{k+3} \\ y^k & y^{k+2} & y^{k+3} \\ z^k & z^{k+2} & z^{k+3}\end{array}\right|=x^k y^k z^k\left|\begin{array}{ccc}1 & x^2 & x^3 \\ 1 & y^2 & y^3 \\ 1 & z^2 & z^3\end{array}\right|$

$(x y z)^k(x-y)(y-z)(z-x)(x y+y z+z x)$

$=(x y z)^{k+1}(x-y)(y-z)(z-x)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$

$k+1=0, \quad \therefore \quad k=-1$

Hint:

Applying $R_3 \rightarrow R_3-\left(R_1+R_2\right)$, we get

$\left|\begin{array}{ccc}a & b & a+b \\ b & c & b+c \\ 0 & 0 & -(a+2 b+c)\end{array}\right|=0$

$\begin{aligned} & \therefore a[-c(a+2 b+c)-0]-b[-b(a+2 b+c)-0]+(a+b)(0-0)=0 \\ & \therefore\left(-a c+b^2\right)(a+2 b+c)=0 \\ & \therefore-a c+b^2=0 \text { or } a+2 b+c=0 \\ & \therefore b^2=a c \\ & \therefore a, b, c \text { are in G.P. }\end{aligned}$