Samacheer Kalvi 10th Maths Exercise 3.20 Solutions

Inconsistent means the system has no solution.

No solution happens only when the three planes have no common point.

So the planes do not intersect.

From $3z = 9$, $z = 3$.

Put $z=3$ in $-7y + 7z = 7$: $-7y + 21 = 7 \Rightarrow y = 2$.

Put $y=2,\ z=3$ in $x + y - 3z = -6$: $x + 2 - 9 = -6 \Rightarrow x = 1$.

$x^2 - 2x - 24 = (x-6)(x+4)$, so $(x-6)$ is a factor.

Since $(x-6)$ is the HCF, it must divide $x^2 - kx - 6$.

So $x = 6$ makes it zero: $36 - 6k - 6 = 0$.

$6k = 30 \Rightarrow k = 5$.

$\dfrac{3y-3}{y} \div \dfrac{7y-7}{3y^2} = \dfrac{3(y-1)}{y} \times \dfrac{3y^2}{7(y-1)}$

Cancel $(y-1)$ and one $y$.

$= \dfrac{9y}{7}$

$\left(y + \dfrac{1}{y}\right)^2 = y^2 + 2 + \dfrac{1}{y^2}$, which is $2$ more than $y^2 + \dfrac{1}{y^2}$.

Options (A), (C), (D) all simplify to $y^2 + \dfrac{1}{y^2}$.

So (B) is the one that is not equal.

$x^2 - 25 = (x+5)(x-5)$ and $x^2 + 6x + 5 = (x+5)(x+1)$.

LCM $= (x+5)(x-5)(x+1)$.

$\dfrac{x(x+1) - 8(x-5)}{(x+5)(x-5)(x+1)} = \dfrac{x^2 + x - 8x + 40}{(x^2-25)(x+1)}$

$= \dfrac{x^2 - 7x + 40}{(x^2-25)(x+1)}$

$\dfrac{256x^8y^4z^{10}}{25x^6y^6z^6} = \dfrac{256\,x^2 z^4}{25\,y^2}$

Taking square root: $\dfrac{16}{5}\left|\dfrac{x z^2}{y}\right|$

For $(x^2 + 8)^2 = x^4 + 16x^2 + 64$.

Given expression has $x^4 + 64$, so the missing middle term is $16x^2$.

$(2x-1)^2 = 9 \Rightarrow 2x - 1 = \pm 3$.

$2x - 1 = 3 \Rightarrow x = 2$.

$2x - 1 = -3 \Rightarrow x = -1$.

Let $4x^4 - 24x^3 + 76x^2 + ax + b = (2x^2 - 6x + c)^2$.

Expanding: $4x^4 - 24x^3 + (36 + 4c)x^2 - 12cx + c^2$.

Compare $x^2$ terms: $36 + 4c = 76 \Rightarrow c = 10$.

So $a = -12(10) = -120$ and $b = 10^2 = 100$.

Let $\alpha,\beta$ be roots of $qx^2 + px + r = 0$: $\alpha+\beta = -\dfrac{p}{q}$, $\alpha\beta = \dfrac{r}{q}$.

For $q^2x^2 + p^2x + r^2 = 0$: $\alpha^2 + \beta^2 = -\dfrac{p^2}{q^2}$, $\alpha^2\beta^2 = \dfrac{r^2}{q^2}$.

Now $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = \dfrac{p^2}{q^2} - \dfrac{2r}{q}$.

So $\dfrac{p^2}{q^2} - \dfrac{2r}{q} = -\dfrac{p^2}{q^2} \Rightarrow \dfrac{2p^2}{q^2} = \dfrac{2r}{q} \Rightarrow p^2 = qr$.

Hence $q, p, r$ are in G.P.

A linear equation in two variables has degree one, and its graph is a straight line.

$x^2 + 4x + 4 = (x+2)^2$.

It has equal roots $x = -2, -2$.

So the curve touches the $X$ axis at exactly one point.

$A$ has $3$ rows and $4$ columns, so order of $A$ is $3 \times 4$.

Transpose interchanges rows and columns, so $A^T$ is $4 \times 3$.

$A_{2\times 3} B_{3\times 4}$ gives order $2 \times 4$.

So $AB$ has $4$ columns.

A matrix with number of rows $\ne$ number of columns is a rectangular matrix.

A column matrix has order $m \times 1$.

Its transpose has order $1 \times m$, which is a row matrix.

$2X = \begin{pmatrix} 5 & 7 \\ 9 & 5 \end{pmatrix} - \begin{pmatrix} 1 & 3 \\ 5 & 7 \end{pmatrix} = \begin{pmatrix} 4 & 4 \\ 4 & -2 \end{pmatrix}$

$X = \dfrac{1}{2}\begin{pmatrix} 4 & 4 \\ 4 & -2 \end{pmatrix} = \begin{pmatrix} 2 & 2 \\ 2 & -1 \end{pmatrix}$

Order of $A$ is $3 \times 2$ and order of $B$ is $3 \times 3$.

$A^2 = A\,A$: $(3\times2)(3\times2)$ — columns of first $\ne$ rows of second, not defined.

$B^2$: $(3\times3)(3\times3)$ — defined.

$AB$: $(3\times2)(3\times3)$ — not defined.

$BA$: $(3\times3)(3\times2)$ — defined.

$AB = \begin{pmatrix} 5 & 4 \\ 7 & 0 \end{pmatrix}$, so $AB + C = \begin{pmatrix} 5 & 5 \\ 5 & 5 \end{pmatrix}$ — (i) is correct.

$BC = \begin{pmatrix} 0 & 1 \\ 2 & -3 \\ -4 & 10 \end{pmatrix}$ — (ii) is correct.

$BA$ is of order $3\times3$ and $C$ is $2\times2$, so $BA + C$ is not defined — (iii) is wrong.

$(AB)C = \begin{pmatrix} 5 & 4 \\ 7 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ -2 & 5 \end{pmatrix} = \begin{pmatrix} -8 & 25 \\ 0 & 7 \end{pmatrix}$ — (iv) is wrong.