Consider the system of equations:
$x-2y+3z=-1$
$-x+y-2z=k$
$x-3y+4z=1$
Statement - 1 : The system of equations has no solution for $k\ne3$.
and
Statement - 2 : The determinant $\left| {\matrix{ 1 & 3 & { - 1} \cr { - 1} & { - 2} & k \cr 1 & 4 & 1 \cr } } \right| \ne 0$, for $k \ne 3$.
The sum of the elements of $\mathrm{U}^{-1}$ is:
-1
0
1
3
$ \begin{aligned} &\mathrm{U}^{-1}=\frac{1}{3}\left[\begin{array}{ccc} -1 & -2 & 0 \\ -7 & -5 & -3 \\ 9 & 6 & 3 \end{array}\right]\\ &\text { Sum of elements of } \mathrm{U}^{-1}=0 \end{aligned} $The value of $\left[\begin{array}{lll}3 & 2 & 0\end{array}\right] U\left[\begin{array}{l}3 \\ 2 \\ 0\end{array}\right]$ is :
5
$5 / 2$
4
$3 / 2$
Let $S=\left\{A=\left(\begin{array}{lll}0 & 1 & c \\ 1 & a & d \\ 1 & b & e\end{array}\right): a, b, c, d, e \in\{0,1\}\right.$ and $\left.|A| \in\{-1,1\}\right\}$, where $|A|$ denotes the determinant of $A$. Then the number of elements in $S$ is __________.
Explanation:
$\begin{aligned} & |A|=0(a e-b d)-1(e-d)+c(b-a) \\ & =c(b-a)+(d-e) \end{aligned}$
$|\mathrm{A}| \in\{-1,1\}$ and $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e} \in\{0,1\}$
Then the number of invertible matrices in $R$ is :
Explanation:
$\begin{gathered}R=\left[\begin{array}{lll}a & 3 & b \\ c & 2 & d \\ 0 & 5 & 0\end{array}\right] \\\\ a, b, c, d, \in\{0,3,5,7,11,13,17,19\}\end{gathered}$
Number of invertible matrices $=$ (Total matrices $)-$ (Non Invertible matrices)
$\begin{aligned} & \text { Total matrices }=\begin{array}{cccc}a, & b, & c, & d \\ \downarrow& \downarrow & \downarrow & \downarrow \\ 8 & 8 & 8 & 8\end{array} \\\\ & =8 \times 8 \times 8 \times 8=8^4=4096 \\ & \end{aligned}$
For Non-invertible matrices,
$ \begin{aligned} & |R|=0 \\\\ & |R|=-5(a d-b c)=0 \end{aligned} $
Cases when both side are zero.
(i) All four $a, b, c, d$ are zero.
$ a d=b c=0 \quad 1 \text { ways } $
(ii) Three zero and one different digit used for $a, b$, $c, d$.
$ \Rightarrow a d=b c $
Select three from four $a, b, c, d$ assign them zero.
$ \text { i.e., }{ }^4 C_3 \times 1 \times 7=28 \text { ways } $
(iii) Two zero and two different digits
Hence $2 \times 7 \times 2 \times 7=196$ ways
Case II: When both side are same but non zero number.
$ a d=b c \neq 0 $
(i) All four $a, b, c, d$ are same.
i.e., $a d=b c$ ( 7 ways)
(ii) Two alike & two alike of another.
$ a d=b c $
$ { }^7 \mathrm{C}_1 \times{ }^6 \mathrm{C}_1 \times 2 !=84 \text { ways } $
Total number of non invertible matrices are
$ \begin{aligned} & =1+28+196+7+84 \\\\ & =316 \end{aligned} $
Hence number of invertible matric
$ \begin{aligned} & =8^4-316 \\\\ & =3780 \end{aligned} $
$ A=\left(\begin{array}{ccc} \beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2 \end{array}\right) $
If $A^{7}-(\beta-1) A^{6}-\beta A^{5}$ is a singular matrix, then the value of $9 \beta$ is _________.
Explanation:
${A^7} - (\beta - 1){A^6} - \beta {A^5}$ is a singular matrix. So determinant of this matrix equal to zero.
$\therefore$ $|{A^7} - (\beta - 1){A^6} - \beta {A^5}| = 0$
$ \Rightarrow |{A^5}({A^2} - (\beta - 1)A - \beta I)| = 0$
$ \Rightarrow |{A^5}||({A^2} - \beta A + A - \beta I)| = 0$
$ \Rightarrow |A{|^5}|A(A + I) - \beta (A + I)| = 0$
$ \Rightarrow |A{|^5}|(A - \beta I)(A + I)| = 0$
$ \Rightarrow |A{|^5}|A - \beta I||A + I| = 0$
Now given,
$A = \left[ {\matrix{ \beta & 0 & 1 \cr 2 & 1 & { - 2} \cr 3 & 1 & { - 2} \cr } } \right]$
$\therefore$ $|A| = 2 - 3 = - 1$
$|A + I| = \left| {\matrix{ {\beta + 1} & 0 & 1 \cr 2 & 2 & { - 2} \cr 3 & 1 & { - 1} \cr } } \right|$
$ = (\beta + 1)( - 2 + 2) + 1(2 - 6)$
$ = - 4$
$\therefore$ We get $|A| \ne 0$ and $|A + I| \ne 0$
$\therefore$ $|A{|^5}|A - \beta I||A + I| = 0$ is possible only when $|A - \beta I| = 0$
$\therefore$ $|A - \beta I| = \left| {\matrix{ 0 & 0 & 1 \cr 2 & {1 - \beta } & { - 2} \cr 3 & 1 & { - 2 - \beta } \cr } } \right|$
$ = 2 - 3 - 3\beta $
$\therefore$ $2 - 3 + 3\beta = 0$
$ \Rightarrow 3\beta = 1$
$ \Rightarrow 9\beta = 3$
x + 2y + 3z = $\alpha$
4x + 5y + 6z = $\beta$
7x + 8y + 9z = $\gamma $ $-$ 1
is consistent. Let | M | represent the determinant of the matrix
$M = \left[ {\matrix{ \alpha & 2 & \gamma \cr \beta & 1 & 0 \cr { - 1} & 0 & 1 \cr } } \right]$
Let P be the plane containing all those ($\alpha$, $\beta$, $\gamma$) for which the above system of linear equations is consistent, and D be the square of the distance of the point (0, 1, 0) from the plane P.
The value of | M | is _________.
Explanation:
On equating the coefficients,
4A + B = 7 .... (i)
5A + 2B = 8 .... (ii)
and $-$ ($\gamma$ $-$ 1) = $-$ A$\beta$ $-$ $\alpha$B ..... (iii)
On solving Eqs. (i) and (ii), we get A = 2 and B = $-$1
From Eq. (iii), we get
$-$ $\gamma$ + 1 = $-$ 2$\beta$ $-$ $\alpha$($-$1)
$\Rightarrow$ $\alpha$ $-$ 2$\beta$ + $\gamma$ = 1 ..... (iv)
Now, determinant of
$M = \left| M \right| = \left| {\matrix{ \alpha & 2 & \gamma \cr \beta & 1 & 0 \cr { - 1} & 0 & 1 \cr } } \right| = \alpha - 2\beta + \gamma = 1$ [from Eq. (iv)]
x + 2y + 3z = $\alpha$
4x + 5y + 6z = $\beta$
7x + 8y + 9z = $\gamma $ $-$ 1
is consistent. Let | M | represent the determinant of the matrix
$M = \left[ {\matrix{ \alpha & 2 & \gamma \cr \beta & 1 & 0 \cr { - 1} & 0 & 1 \cr } } \right]$
Let P be the plane containing all those ($\alpha$, $\beta$, $\gamma$) for which the above system of linear equations is consistent, and D be the square of the distance of the point (0, 1, 0) from the plane P.
The value of D is _________.
Explanation:
On equating the coefficients,
4A + B = 7 .... (i)
5A + 2B = 8 .... (ii)
and $-$ ($\gamma$ $-$ 1) = $-$ A$\beta$ $-$ $\alpha$B ..... (iii)
On solving Eqs. (i) and (ii), we get A = 2 and B = $-$1
From Eq. (iii), we get
$-$ $\gamma$ + 1 = $-$ 2$\beta$ $-$ $\alpha$($-$1)
$\Rightarrow$ $\alpha$ $-$ 2$\beta$ + $\gamma$ = 1 ..... (iv)
Now, determinant of
$M = \left| M \right| = \left| {\matrix{ \alpha & 2 & \gamma \cr \beta & 1 & 0 \cr { - 1} & 0 & 1 \cr } } \right| = \alpha - 2\beta + \gamma = 1$ [from Eq. (iv)]
Equation of plane P is given by $x - 2y + z = 1$
Hence, perpendicular distance of the point (0, 1, 0) from the plane
$P = {{\left| {0 - 2 \times 1 + 0 - 1} \right|} \over {\sqrt {{1^2} + {{( - 2)}^2} + {1^2}} }} = {{\left| 3 \right|} \over {\sqrt 6 }}$
$ \Rightarrow D = {\left( {{{\left| 3 \right|} \over {\sqrt 6 }}} \right)^2} = {9 \over 6} = 1.5$
Explanation:
$A = \left[ {\matrix{ x & y \cr z & {3 - x} \cr } } \right]$
So, ${A^2} = \left[ {\matrix{ x & y \cr z & {3 - x} \cr } } \right]\left[ {\matrix{ x & y \cr z & {3 - x} \cr } } \right]$
$ = \left[ {\matrix{ {{x^2} + yz} & {xy + 3y - xy} \cr {xz + 3z - xz} & {yz + {{(3 - x)}^2}} \cr } } \right]$
$ \therefore $ ${A^3} = \left[ {\matrix{ {{x^2} + yz} & {3y} \cr { + 3z} & {yz + 9 + {x^2} - 6x} \cr } } \right]\left[ {\matrix{ x & y \cr z & {3 - x} \cr } } \right]$
$ \because $ ${t_r}({A^3}) = {x^3} + xyz + 3yz + 3yz + 3yz - xyz + 27 - 9x + 3{x^2} - {x^3} - 18x + 6{x^2}$
$ = 9yz + 27 - 27x + 9{x^2} = - 18$ (given)
$ \Rightarrow yz + 3 - 3x + {x^2} = - 2$
$ \Rightarrow 3x - {x^2} - yz = 5$
$ \because $ $|A|\, = \,\left[ {\matrix{ x & y \cr z & {3 - x} \cr } } \right]$
$ = 3x - {x^2} - yz = 5$
det$\left| {\matrix{ {\sum\limits_{k = 0}^n k } & {\sum\limits_{k = 0}^n {{}^n{C_k}{k^2}} } \cr {\sum\limits_{k = 0}^n {{}^n{C_k}.k} } & {\sum\limits_{k = 0}^n {{}^n{C_k}{3^k}} } \cr } } \right| = 0$
holds for some positive integer n. Then $\sum\limits_{k = 0}^n {{{{}^n{C_k}} \over {k + 1}}} $ equals ..............
Explanation:
$\left| {\matrix{ {\sum\limits_{k = 0}^n k } & {\sum\limits_{k = 0}^n {{}^n{C_k}{k^2}} } \cr {\sum\limits_{k = 0}^n {{}^n{C_k}.k} } & {\sum\limits_{k = 0}^n {{}^n{C_k}{3^k}} } \cr } } \right| = 0$
$ \Rightarrow \left| {\matrix{ {{{n(n + 1)} \over 2}} & {n(n + 1){2^{n - 2}}} \cr {n{{.2}^{n - 1}}} & {{4^n}} \cr } } \right| = 0$
$ \because $ $\left[ \matrix{ \sum\limits_{k = 0}^n k = {{n(n + 1)} \over 2},\,\sum\limits_{k = 0}^n {{}^n{C_k}k = n{{.2}^{n - 1}}} \hfill \cr \sum\limits_{k = 0}^n {} {}^n{C_k}{k^2} = n(n + 1){2^{n - 2}}\,\,and\,\,\sum\limits_{k = 0}^n {} {}^n{C_k}{3^k} = {4^n} \hfill \cr} \right]$
$ \Rightarrow {{n(n + 1)} \over 2}{4^n} - {n^2}(n + 1)\,{2^{2n - 3}} = 0$
$ \Rightarrow {{{4^n}} \over 2} - n{{{4^{n - 1}}} \over 2} = 0$
$ \Rightarrow n = 4$
$ \therefore $ $\sum\limits_{k = 0}^n {} {{{}^n{C_k}} \over {k + 1}} = \sum\limits_{k = 0}^4 {} {{{}^4{C_k}} \over {k + 1}}$
= ${1 \over 5}\sum\limits_{k = 0}^4 {} {}^5{C_{k + 1}} = {1 \over 5}({2^5} - 1)$
$ = {1 \over 5}(32 - 1) = {{31} \over 5} = 6.20$
Explanation:
$ = {a_1}({b_2}{c_3} - {b_3}{c_2}) - {a_2}({b_1}{c_3} - {b_3}{c_1}) + {a_3}({b_1}{c_2} - {b_2}{c_1})$
Now, maximum value of Det (P) = 6
If ${a_1} = 1$, ${a_2} = - 1$, ${a_3} = 1$, ${b_2}{c_3} = {b_1}{c_3} = {b_1}{c_2} = 1$ and ${b_3}{c_2} = {b_3}{c_1} = {b_2}{c_1} = - 1$
But it is not possible as
$({b_2}{c_3})({b_3}{c_1})({b_1}{c_2})$ = $-$1
and $({b_1}{c_3})({b_3}{c_2})({b_2}{c_1})$ = 1
i.e., ${b_1}{c_2}{b_3}{c_1}{c_2}{c_3}$ = 1 and $-$1
Similar contradiction occurs when
${a_1} = 1$, ${a_2} = 1$, ${a_3} = 1$, ${b_2}{c_1} = {b_3}{c_1} = {b_1}{c_2}$ = 1 and ${b_3}{c_2} = {b_1}{c_3} = {b_1}{c_2} = - 1$
Now, for value to be 5 one of the terms must be zero but that will make 2 terms zero which means answer cannot be 5
Now,
$\left| {\matrix{ 1 & 1 & 1 \cr { - 1} & 1 & 1 \cr 1 & { - 1} & 1 \cr } } \right| = 4$
Hence, maximum value is 4
$\left[ {\matrix{ 1 & \alpha & {{\alpha ^2}} \cr \alpha & 1 & \alpha \cr {{\alpha ^2}} & \alpha & 1 \cr } } \right]\left[ {\matrix{ x \cr y \cr z \cr } } \right] = \left[ {\matrix{ 1 \cr { - 1} \cr 1 \cr } } \right]$
of linear equations, has infinitely many solutions, then 1 + $\alpha $ + $\alpha $2 =
Explanation:
It is given that
$\left[ {\matrix{ 1 & \alpha & {{\alpha ^2}} \cr \alpha & 1 & \alpha \cr {{\alpha ^2}} & \alpha & 1 \cr } } \right]\left[ {\matrix{ x \cr y \cr z \cr } } \right] = \left[ {\matrix{ 1 \cr { - 1} \cr 1 \cr } } \right]$
$\left[ {\matrix{ 1 & \alpha & {{\alpha ^2}} \cr \alpha & 1 & \alpha \cr {{\alpha ^2}} & \alpha & 1 \cr } } \right] = 0$
$ \Rightarrow 1(1 - {\alpha ^2}) - \alpha (\alpha - {\alpha ^3}) + {\alpha ^2}({\alpha ^2} - {\alpha ^2}) = 0$
$ \Rightarrow \alpha (1 - {\alpha ^2}) - {\alpha ^2}(1 - {\alpha ^2}) = 0$
$ \Rightarrow (1 - {\alpha ^2})(1 - {\alpha ^2}) = 0$
$ \Rightarrow {(1 - {\alpha ^2})^2} = 0$
$ \Rightarrow {\alpha ^2} = 1 \Rightarrow \alpha = \pm 1$
For $\alpha$ = 1, the given system of linear equations has no solution.
$\left[ {\matrix{ { + 1} & { + 1} & { + 1} \cr { + 1} & { + 1} & { + 1} \cr { + 1} & { + 1} & { + 1} \cr } } \right]\left[ {\matrix{ x \cr y \cr z \cr } } \right] = \left[ {\matrix{ 1 \cr { - 1} \cr 1 \cr } } \right]$
$x + y + z = 1$
$x + y + z = - 1$
$x + y + z = 1$
Since two planes are parallel. So, $\alpha$ = 1 is rejected for $\alpha$ = $-$1 the given system of linear equations has coincident planes
$\left[ {\matrix{ 1 & { - 1} & 1 \cr { - 1} & 1 & { - 1} \cr 1 & { - 1} & 1 \cr } } \right]\left[ {\matrix{ x \cr y \cr z \cr } } \right] = \left[ {\matrix{ 1 \cr { - 1} \cr 1 \cr } } \right]$
$x - y + z = 1$
$ \Rightarrow - x + y - z = - 1 \Rightarrow x - y + z = 1$
$x - y + 1 = 1$
Therefore, $\alpha$ = $-$1 is accepted. That is,
$1 + \alpha + {\alpha ^2} = 1 + ( - 1) + {( - 1)^2} = 1 - 1 + 1 = 1$
$ \Rightarrow 1 + \alpha + {\alpha ^2} = 1$
The total number of distinct x $\in$ R for which
$\left| {\matrix{ x & {{x^2}} & {1 + {x^3}} \cr {2x} & {4{x^2}} & {1 + 8{x^3}} \cr {3x} & {9{x^2}} & {1 + 27{x^3}} \cr } } \right| = 10$ is ______________.
Explanation:
Given, $\left| {\matrix{ x & {{x^2}} & {1 + {x^3}} \cr {2x} & {4{x^2}} & {1 + 8{x^3}} \cr {3x} & {9{x^2}} & {1 + 27{x^3}} \cr } } \right| = 10$
$ \Rightarrow x\,.\,{x^2}\left| {\matrix{ 1 & 1 & {1 + {x^3}} \cr 2 & 4 & {1 + 8{x^3}} \cr 3 & 9 & {1 + 27{x^3}} \cr } } \right| = 10$
Apply R2 $\to$ R2 $-$ 2R1 and R3 $\to$ R3 $-$ 3R1, we get
${x^3}\left| {\matrix{ 1 & 1 & {1 + {x^3}} \cr 0 & 2 & { - 1 + 6{x^3}} \cr 0 & 6 & { - 2 + 24{x^3}} \cr } } \right| = 10$
$ \Rightarrow {x^3}\,.\,\left| {\matrix{ 2 & {6{x^2} - 1} \cr 6 & {24{x^3} - 2} \cr } } \right| = 10$
$ \Rightarrow {x^3}(48{x^3} - 4 - 36{x^3} + 6) = 10$
$ \Rightarrow 12{x^6} + 2{x^3} = 10$
$ \Rightarrow 6{x^6} + {x^3} - 5 = 0$
$ \Rightarrow 6{({x^3})^2} + {x^3} - 5 = 6$
$ \Rightarrow 6{({x^3})^2} + 6{x^3} - 5{x^3} - 5 = 0$
$ \Rightarrow 6{x^3}({x^3} + 1) - 5({x^3} + 1) = 0$
$ \Rightarrow (6{x^3} - 5)({x^2} - x + 1)(x + 1) = 0$
$\therefore$ $x = {\left( {{5 \over 6}} \right)^{1/3}}, - 1$
Hence, the number of real solutions is 2.
Let $z = {{ - 1 + \sqrt 3 i} \over 2}$, where $i = \sqrt { - 1} $, and r, s $\in$ {1, 2, 3}. Let $P = \left[ {\matrix{ {{{( - z)}^r}} & {{z^{2s}}} \cr {{z^{2s}}} & {{z^r}} \cr } } \right]$ and I be the identity matrix of order 2. Then the total number of ordered pairs (r, s) for which P2 = $-$I is ____________.
Explanation:
Here, $z = {{ - 1 + i\sqrt 3 } \over 2} = \omega $
$\because$ $P = \left[ {\matrix{ {{{( - \omega )}^r}} & {{\omega ^{2s}}} \cr {{\omega ^{2s}}} & {{\omega ^r}} \cr } } \right]$
${P^2} = \left[ {\matrix{ {{{( - \omega )}^r}} & {{\omega ^{2s}}} \cr {{\omega ^{2s}}} & {{\omega ^r}} \cr } } \right]\left[ {\matrix{ {{{( - \omega )}^r}} & {{\omega ^{2s}}} \cr {{\omega ^{2s}}} & {{\omega ^r}} \cr } } \right]$$ = \left[ {\matrix{ {{\omega ^{2r}} + {\omega ^{4s}}} & {{\omega ^{r + 2s}}[{{( - 1)}^r} + 1]} \cr {{\omega ^{r + 2s}}[{{( - 1)}^r} + 1]} & {{\omega ^{4s}} + {\omega ^{2r}}} \cr } } \right]$
Given, ${P^2} = - I$
$\therefore$ ${\omega ^{2r}} + {\omega ^{4s}} = - 1$
and ${\omega ^{r + 2s}}[{( - 1)^r} + 1] = 0$
Since, r $\in$ {1, 2, 3} and ($-$1)r + 1 = 0
$\Rightarrow$ r = {1, 3}
Also, ${\omega ^{2r}} + {\omega ^{4s}} = - 1$
If r = 1, then ${\omega ^2} + {\omega ^{4s}} = - 1$
which is only possible, when s = 1.
As, ${\omega ^2} + {\omega ^4} = - 1$
$\therefore$ r = 1, s = 1
Again, if r = 3, then
${\omega ^6} + {\omega ^{4s}} = - 1$
$ \Rightarrow {\omega ^{4s}} = - 2$ [never possible]
$\therefore$ r $\ne$ 3
$\Rightarrow$ (r, s) = (1, 1) is the only solution.
Hence, the total number of ordered pairs is 1.
Let M be a 3 $\times$ 3 matrix satisfying $M\left[ {\matrix{ 0 \cr 1 \cr 0 \cr } } \right] = \left[ {\matrix{ { - 1} \cr 2 \cr 3 \cr } } \right]$, $M\left[ {\matrix{ 1 \cr { - 1} \cr 0 \cr } } \right] = \left[ {\matrix{ 1 \cr 1 \cr { - 1} \cr } } \right]$ and $M\left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr {12} \cr } } \right]$. Then the sum of the diagonal entries of M is ___________.
Explanation:
Let $M = \left[ {\matrix{ a & b & c \cr d & e & f \cr g & h & i \cr } } \right]$
$M = \left[ {\matrix{ 0 \cr 1 \cr 0 \cr } } \right] = \left[ {\matrix{ { - 1} \cr 2 \cr 3 \cr } } \right] \Rightarrow b = - 1,\,e = 2,\,h = 3$
$M = \left[ {\matrix{ 1 \cr { - 1} \cr 0 \cr } } \right] = \left[ {\matrix{ 1 \cr 1 \cr { - 1} \cr } } \right] \Rightarrow a = 0,\,d = 3,\,g = 2$
$M = \left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr {12} \cr } } \right] \Rightarrow g + h + i = 12 \Rightarrow i = 7$
Hence, the sum of diagonal elements is 9.
Let $k$ be a positive real number and let
$ \begin{aligned} A & =\left[\begin{array}{ccc} 2 k-1 & 2 \sqrt{k} & 2 \sqrt{k} \\ 2 \sqrt{k} & 1 & -2 k \\ -2 \sqrt{k} & 2 k & -1 \end{array}\right] \text { and } \\\\ \mathbf{B} & =\left[\begin{array}{ccc} 0 & 2 k-1 & \sqrt{k} \\ 1-2 k & 0 & 2 \sqrt{k} \\ -\sqrt{k} & -2 \sqrt{k} & 0 \end{array}\right] . \end{aligned} $
If $\operatorname{det}(\operatorname{adj} A)+\operatorname{det}(\operatorname{adj} B)=10^6$, then $[k]$
is equal to _________.
[ Note : adj M denotes the adjoint of a square matrix M and $[k]$ denotes the largest integer less than or equal to $k$ ].
Explanation:
$ |A|=(2 k+1)^3,|B|=0 $
( $\therefore$ B is a skew - symmetric matrix of order 3 )
$ \begin{aligned} \text { Let }(\operatorname{adj} \mathrm{A}) & =|\mathrm{A}|^{n-1} \\\\ \left((2 k+1)^3\right)^2 & =10^6 \\\\ (2 k+1)^6 & =10^6 \Rightarrow 2 k+1=10 \\\\ 2 k=9 \Rightarrow[k] & =4 \end{aligned} $
$\overrightarrow A = \left( {1,a,{a^2}} \right),\,\,\overrightarrow B = \left( {1,b,{b^2}} \right),\,\,\overrightarrow C = \left( {1,c,{c^2}} \right),$ are non-coplannar, then the product $abc=$ .......