140-- Three pages with matrix equations. If $\begin{vmatrix} a & -1 & 3 \\ b & 2 & 4 \\ c & -2 & 1 \end{vmatrix} = 5$, then the three pages intersect. With which of the following lengths do they intersect?
$$\begin{bmatrix} 2 & -1 & 3 \\ 1 & 2 & 4 \\ 3 & -2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} a \\ b \\ c \end{bmatrix}$$
(1) $-\dfrac{1}{3}$ (2) $-\dfrac{1}{2}$ (3) $\dfrac{1}{3}$ (4) $\dfrac{1}{2}$

130- From the matrix relation $\begin{bmatrix} 3 & -1 & 1 \\ 4 & 0 & -2 \\ 1 & 2 & 0 \end{bmatrix} \begin{bmatrix} x \\ 2x \\ -1 \end{bmatrix} = 0$, $\begin{bmatrix} x & 2x & -1 \end{bmatrix}$, the nonzero value of $x$ is which of the following?
(1) $\dfrac{2}{9}$ (2) $\dfrac{3}{8}$ (3) $\dfrac{4}{9}$ (4) $\dfrac{3}{5}$

131- If $A = \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}$, from the relation $AX = A - 2I$, matrix $X$ is which of the following?
(1) $\begin{bmatrix} -2 & 2 \\ 3 & -1 \end{bmatrix}$ (2) $\begin{bmatrix} -2 & 1 \\ 4 & -1 \end{bmatrix}$ (3) $\begin{bmatrix} 2 & -1 \\ 4 & 2 \end{bmatrix}$ (4) $\begin{bmatrix} 1 & 2 \\ 4 & -1 \end{bmatrix}$

139- If $A = \begin{bmatrix} 1 & -1 & -3 \\ 4 & 1 & 2 \\ 2 & 1 & 3 \end{bmatrix}$ and matrix $X$ satisfies the matrix equation $X = \begin{bmatrix} 3 & 0 \\ -2 & 1 \end{bmatrix}$, $$\begin{bmatrix} 2|A| & |A| \\ 1 & \dfrac{2}{|A|} \end{bmatrix} X = \begin{bmatrix} 3 & 0 \\ -2 & 1 \end{bmatrix}$$ holds. The smallest main diagonal entry of matrix $X$ is which?
\[ \text{(1)}\ -15 \qquad \text{(2)}\ -3 \qquad \text{(3)}\ 6 \qquad \text{(4)}\ 8 \]

18. $\mathrm { A } = \left[ \begin{array} { c c c } \mathrm { a } & 0 & 1 \\ 1 & \mathrm { c } & \mathrm { b } \\ 1 & \mathrm {~d} & \mathrm {~b} \end{array} \right] , \mathrm { B } = \left[ \begin{array} { c c c } \mathrm { a } & 1 & 1 \\ 0 & \mathrm {~d} & \mathrm { c } \\ \mathrm { f } & \mathrm { g } & \mathrm { h } \end{array} \right] , \mathrm { U } = \left[ \begin{array} { c } \mathrm { f } \\ \mathrm { g } \\ \mathrm { h } \end{array} \right] , \mathrm { V } = \left[ \begin{array} { c } \mathrm { a } ^ { 2 } \\ 0 \\ 0 \end{array} \right]$. If there is vector matrix X , such that $\mathrm { AX } = \mathrm { U }$ has infinitely many solutions, then prove that $\mathrm { BX } = \mathrm { V }$ cannot have a unique solution. If afd $\neq 0$ then prove that $\mathrm { BX } = \mathrm { V }$ has no solution.
Sol. $\mathrm { AX } = \mathrm { U }$ has infinite solutions $\Rightarrow | \mathrm { A } | = 0$ $\left| \begin{array} { c c c } \mathrm { a } & 0 & 1 \\ 1 & \mathrm { c } & \mathrm { b } \\ 1 & \mathrm {~d} & \mathrm {~b} \end{array} \right| = 0 \Rightarrow \mathrm { ab } = 1$ or $\mathrm { c } = \mathrm { d }$ and $\left| \mathrm { A } _ { 1 } \right| = \left| \begin{array} { c c c } \mathrm { a } & 0 & \mathrm { f } \\ 1 & \mathrm { c } & \mathrm { g } \\ 1 & \mathrm {~d} & \mathrm {~h} \end{array} \right| = 0 \Rightarrow \mathrm {~g} = \mathrm { h } ; \left| \mathrm { A } _ { 2 } \right| = \left| \begin{array} { c c c } \mathrm { a } & \mathrm { f } & 1 \\ 1 & \mathrm {~g} & \mathrm {~b} \\ 1 & \mathrm {~h} & \mathrm {~b} \end{array} \right| = 0 \Rightarrow \mathrm {~g} = \mathrm { h }$ $\left| \mathrm { A } _ { 3 } \right| = \left| \begin{array} { l l l } \mathrm { f } & 0 & 1 \\ \mathrm {~g} & \mathrm { c } & \mathrm { b } \\ \mathrm { h } & \mathrm { d } & \mathrm { b } \end{array} \right| = 0 \Rightarrow \mathrm {~g} = \mathrm { h } , \mathrm { c } = \mathrm { d } \Rightarrow \mathrm { c } = \mathrm { d }$ and $\mathrm { g } = \mathrm { h }$ $\mathrm { BX } = \mathrm { V }$ $| \mathrm { B } | = \left| \begin{array} { l l l } \mathrm { a } & 1 & 1 \\ 0 & \mathrm {~d} & \mathrm { c } \\ \mathrm { f } & \mathrm { g } & \mathrm { h } \end{array} \right| = 0 \quad$ (since $\mathrm { C } _ { 2 }$ and $\mathrm { C } _ { 3 }$ are equal) $\quad \Rightarrow \mathrm { BX } = \mathrm { V }$ has no unique solution. and $\left| \mathrm { B } _ { 1 } \right| = \left| \begin{array} { l l l } \mathrm { a } ^ { 2 } & 1 & 1 \\ 0 & \mathrm {~d} & \mathrm { c } \\ 0 & \mathrm {~g} & \mathrm {~h} \end{array} \right| = 0 ($ since $\mathrm { c } = \mathrm { d } , \mathrm { g } = \mathrm { h } )$ $\left| \mathrm { B } _ { 2 } \right| = \left| \begin{array} { c c c } \mathrm { a } & \mathrm { a } ^ { 2 } & 1 \\ 0 & 0 & \mathrm { c } \\ \mathrm { f } & 0 & \mathrm {~h} \end{array} \right| = \mathrm { a } ^ { 2 } \mathrm { cf } = \mathrm { a } ^ { 2 } \mathrm { df } \quad ($ since $\mathrm { c } = \mathrm { d } )$
$$\left| \mathrm { B } _ { 3 } \right| = \left| \begin{array} { c c c } \mathrm { a } & 1 & \mathrm { a } ^ { 2 } \\ 0 & \mathrm {~d} & 0 \\ \mathrm { f } & \mathrm {~g} & 0 \end{array} \right| = \mathrm { a } ^ { 2 } \mathrm { df }$$
since if $\operatorname { adf } \neq 0$ then $\left| \mathrm { B } _ { 2 } \right| = \left| \mathrm { B } _ { 3 } \right| \neq 0$. Hence no solution exist.

18. $\mathrm { A } = \left[ \begin{array} { l l l } \mathrm { a } & 0 & 1 \\ 1 & \mathrm { c } & \mathrm { b } \\ 1 & \mathrm {~d} & \mathrm {~b} \end{array} \right] , \mathrm { B } = \left[ \begin{array} { c c c } \mathrm { a } & 1 & 1 \\ 0 & \mathrm {~d} & \mathrm { c } \\ \mathrm { f } & \mathrm { g } & \mathrm { h } \end{array} \right] , \mathrm { U } = \left[ \begin{array} { c } \mathrm { f } \\ \mathrm { g } \\ \mathrm { h } \end{array} \right] , \mathrm { V } = \left[ \begin{array} { c } \mathrm { a } ^ { 2 } \\ 0 \\ 0 \end{array} \right]$. If there is vector matrix X , such that $\mathrm { AX } = \mathrm { U }$ has infinitely many solutions, then prove that $\mathrm { BX } = \mathrm { V }$ cannot have a unique solution. If afd $\neq 0$ then prove that $\mathrm { BX } = \mathrm { V }$ has no solution.
Sol. $\mathrm { AX } = \mathrm { U }$ has infinite solutions $\Rightarrow | \mathrm { A } | = 0$ $\left| \begin{array} { c c c } \mathrm { a } & 0 & 1 \\ 1 & \mathrm { c } & \mathrm { b } \\ 1 & \mathrm {~d} & \mathrm {~b} \end{array} \right| = 0 \Rightarrow \mathrm { ab } = 1$ or $\mathrm { c } = \mathrm { d }$ and $\left| \mathrm { A } _ { 1 } \right| = \left| \begin{array} { c c c } \mathrm { a } & 0 & \mathrm { f } \\ 1 & \mathrm { c } & \mathrm { g } \\ 1 & \mathrm {~d} & \mathrm {~h} \end{array} \right| = 0 \Rightarrow \mathrm {~g} = \mathrm { h } ; \left| \mathrm { A } _ { 2 } \right| = \left| \begin{array} { c c c } \mathrm { a } & \mathrm { f } & 1 \\ 1 & \mathrm {~g} & \mathrm {~b} \\ 1 & \mathrm {~h} & \mathrm {~b} \end{array} \right| = 0 \Rightarrow \mathrm {~g} = \mathrm { h }$ $\left| \mathrm { A } _ { 3 } \right| = \left| \begin{array} { l l l } \mathrm { f } & 0 & 1 \\ \mathrm {~g} & \mathrm { c } & \mathrm { b } \\ \mathrm { h } & \mathrm { d } & \mathrm { b } \end{array} \right| = 0 \Rightarrow \mathrm {~g} = \mathrm { h } , \mathrm { c } = \mathrm { d } \Rightarrow \mathrm { c } = \mathrm { d }$ and $\mathrm { g } = \mathrm { h }$ $\mathrm { BX } = \mathrm { V }$ $| \mathrm { B } | = \left| \begin{array} { l l l } \mathrm { a } & 1 & 1 \\ 0 & \mathrm {~d} & \mathrm { c } \\ \mathrm { f } & \mathrm { g } & \mathrm { h } \end{array} \right| = 0 \quad$ (since $\mathrm { C } _ { 2 }$ and $\mathrm { C } _ { 3 }$ are equal) $\quad \Rightarrow \mathrm { BX } = \mathrm { V }$ has no unique solution. and $\left| \mathrm { B } _ { 1 } \right| = \left| \begin{array} { l l l } \mathrm { a } ^ { 2 } & 1 & 1 \\ 0 & \mathrm {~d} & \mathrm { c } \\ 0 & \mathrm {~g} & \mathrm {~h} \end{array} \right| = 0 ($ since $\mathrm { c } = \mathrm { d } , \mathrm { g } = \mathrm { h } )$ $\left| \mathrm { B } _ { 2 } \right| = \left| \begin{array} { c c c } \mathrm { a } & \mathrm { a } ^ { 2 } & 1 \\ 0 & 0 & \mathrm { c } \\ \mathrm { f } & 0 & \mathrm {~h} \end{array} \right| = \mathrm { a } ^ { 2 } \mathrm { cf } = \mathrm { a } ^ { 2 } \mathrm { df } \quad ($ since $\mathrm { c } = \mathrm { d } )$
$$\left| \mathrm { B } _ { 3 } \right| = \left| \begin{array} { c c c } \mathrm { a } & 1 & \mathrm { a } ^ { 2 } \\ 0 & \mathrm {~d} & 0 \\ \mathrm { f } & \mathrm {~g} & 0 \end{array} \right| = \mathrm { a } ^ { 2 } \mathrm { df }$$
since if $\operatorname { adf } \neq 0$ then $\left| \mathrm { B } _ { 2 } \right| = \left| \mathrm { B } _ { 3 } \right| \neq 0$. Hence no solution exist.

31. The sum of the elements of $\mathrm { U } ^ { - 1 }$ is
(A) - 1
(B) 0
(C) 1
(D) 3
Sol. (B) Moreover adj $\mathrm { U } = \left[ \begin{array} { c c c } - 1 & - 2 & 0 \\ - 7 & - 5 & - 3 \\ 9 & 6 & 3 \end{array} \right]$. Hence $\mathrm { U } ^ { - 1 } = \frac { \operatorname { adj } \mathrm { U } } { 3 }$ and sum of the elements of $\mathrm { U } ^ { - 1 } = 0$.

The number of distinct real values of $\lambda$ for which the system of linear equations $$x + y + z = 0$$ $$x + \lambda y + z = 0$$ $$x + y + \lambda z = 0$$ has a non-trivial solution is
(A) 0
(B) 1
(C) 2
(D) 3

Let $\mathscr { A }$ be the set of all $3 \times 3$ symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0.
The number of matrices $A$ in $\mathscr { A }$ for which the system of linear equations
$$A \left[ \begin{array} { l } x \\ y \\ z \end{array} \right] = \left[ \begin{array} { l } 1 \\ 0 \\ 0 \end{array} \right]$$
has a unique solution, is
(A) less than 4
(B) at least 4 but less than 7
(C) at least 7 but less than 10
(D) at least 10

Let $\mathscr { A }$ be the set of all $3 \times 3$ symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0.
The number of matrices $A$ in $\mathscr { A }$ for which the system of linear equations
$$A \left[ \begin{array} { l } x \\ y \\ z \end{array} \right] = \left[ \begin{array} { l } 1 \\ 0 \\ 0 \end{array} \right]$$
is inconsistent, is
(A) 0
(B) more than 2
(C) 2
(D) 1

Let $(x,y,z)$ be points with integer coordinates satisfying the system of homogeneous equations: $$\begin{array}{r} 3x-y-z=0\\ -3x+z=0\\ -3x+2y+z=0 \end{array}$$ Then the number of such points for which $x^{2}+y^{2}+z^{2}\leq100$ is

The number of $3 \times 3$ matrices A whose entries are either 0 or 1 and for which the system $A \left[ \begin{array} { l } x \\ y \\ z \end{array} \right] = \left[ \begin{array} { l } 1 \\ 0 \\ 0 \end{array} \right]$ has exactly two distinct solutions, is
A) 0
B) $2 ^ { 9 } - 1$
C) 168
D) 2

Let $P = \left[\begin{array}{ccc} 3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0 \end{array}\right]$, where $\alpha \in \mathbb{R}$. Suppose $Q = [q_{ij}]$ is a matrix such that $PQ = kI$, where $k \in \mathbb{R}, k \neq 0$ and $I$ is the identity matrix of order 3. If $q_{23} = -\frac{k}{8}$ and $\det(Q) = \frac{k^2}{2}$, then
(A) $\alpha = 0, k = 8$
(B) $4\alpha - k + 8 = 0$
(C) $\det(P\operatorname{adj}(Q)) = 2^9$
(D) $\det(Q\operatorname{adj}(P)) = 2^{13}$

Let $\alpha , \beta$ and $\gamma$ be real numbers. Consider the following system of linear equations
$x + 2 y + z = 7$
$x + \alpha z = 11$
$2 x - 3 y + \beta z = \gamma$
Match each entry in List-I to the correct entries in List-II.
List-I
(P) If $\beta = \frac { 1 } { 2 } ( 7 \alpha - 3 )$ and $\gamma = 28$, then the system has
(Q) If $\beta = \frac { 1 } { 2 } ( 7 \alpha - 3 )$ and $\gamma \neq 28$, then the system has
(R) If $\beta \neq \frac { 1 } { 2 } ( 7 \alpha - 3 )$ where $\alpha = 1$ and $\gamma \neq 28$, then the system has
(S) If $\beta \neq \frac { 1 } { 2 } ( 7 \alpha - 3 )$ where $\alpha = 1$ and $\gamma = 28$, then the system has
List-II
(1) a unique solution
(2) no solution
(3) infinitely many solutions
(4) $x = 11 , y = - 2$ and $z = 0$ as a solution
(5) $x = - 15 , y = 4$ and $z = 0$ as a solution
The correct option is:
(A) $( P ) \rightarrow ( 3 ) \quad ( Q ) \rightarrow ( 2 ) \quad ( R ) \rightarrow ( 1 ) \quad ( S ) \rightarrow ( 4 )$
(B) $( P ) \rightarrow ( 3 ) \quad ( Q ) \rightarrow ( 2 ) \quad ( R ) \rightarrow ( 5 ) \quad ( S ) \rightarrow ( 4 )$
(C) $( P ) \rightarrow ( 2 ) \quad ( Q ) \rightarrow ( 1 ) \quad ( R ) \rightarrow ( 4 ) \quad ( S ) \rightarrow ( 5 )$
(D) $( P ) \rightarrow ( 2 ) \quad ( Q ) \rightarrow ( 1 ) \quad ( R ) \rightarrow ( 1 ) \quad ( S ) \rightarrow ( 3 )$

The number of values of $k$ for which the linear equations $4x+ky+2z=0$;\ $kx+4y+z=0$;\ $2x+2y+z=0$ possess a non-zero solution is
(1) 2
(2) 1
(3) 0
(4) 3

If the system of equations $$\begin{aligned} & x + y + z = 6 \\ & x + 2 y + 3 z = 10 \\ & x + 2 y + \lambda z = 0 \end{aligned}$$ has a unique solution, then $\lambda$ is not equal to
(1) 1
(2) 0
(3) 2
(4) 3

Statement 1: If the system of equations $x + k y + 3 z = 0, 3 x + k y - 2 z = 0, 2 x + 3 y - 4 z = 0$ has a nontrivial solution, then the value of $k$ is $\frac { 31 } { 2 }$. Statement 2: A system of three homogeneous equations in three variables has a non trivial solution if the determinant of the coefficient matrix is zero.
(1) Statement 1 is false, Statement 2 is true.
(2) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.
(3) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.
(4) Statement 1 is true, Statement 2 is false.

Let $A = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix}$. If $u_{1}$ and $u_{2}$ are column matrices such that $Au_{1} = \begin{pmatrix}1\\0\\0\end{pmatrix}$ and $Au_{2} = \begin{pmatrix}0\\1\\0\end{pmatrix}$, then $u_{1}+u_{2}$ is equal to
(1) $\begin{pmatrix}-1\\1\\0\end{pmatrix}$
(2) $\begin{pmatrix}-1\\1\\-1\end{pmatrix}$
(3) $\begin{pmatrix}-1\\-1\\0\end{pmatrix}$
(4) $\begin{pmatrix}1\\-1\\-1\end{pmatrix}$

The number of values of $k$ for which the linear equations $4x+ky+2z=0$, $kx+4y+z=0$, $2x+2y+z=0$ possess a non-zero solution is
(1) 2
(2) 1
(3) zero
(4) 3

The system of linear equations \begin{align*} x + \lambda y - z &= 0 \lambda x - y - z &= 0 x + y - \lambda z &= 0 \end{align*} has a non-trivial solution for:
(1) infinitely many values of $\lambda$
(2) exactly one value of $\lambda$
(3) exactly two values of $\lambda$
(4) exactly three values of $\lambda$

The system of linear equations \begin{align*} x + \lambda y - z &= 0 \lambda x - y - z &= 0 x + y - \lambda z &= 0 \end{align*} has a non-trivial solution for: (1) infinitely many values of $\lambda$ (2) exactly one value of $\lambda$ (3) exactly two values of $\lambda$ (4) exactly three values of $\lambda$

The set of all values of $\lambda$ for which the system of linear equations $$x - 2 y - 2 z = \lambda x$$ $$x + 2 y + z = \lambda y$$ $$- x - y = \lambda z$$ has a non-trivial solution:
(1) is an empty set
(2) is a singleton
(3) contains two elements
(4) contains more than two elements

If the system of linear equations $x + k y + 3 z = 0$ $3 x + k y - 2 z = 0$ $2 x + 4 y - 3 z = 0$ has a non-zero solution $( x , y , z )$, then $\frac { x z } { y ^ { 2 } }$ is equal to:
(1) 30
(2) - 10
(3) 10
(4) - 30

If $\left[ \begin{array} { l l } 1 & 1 \\ 0 & 1 \end{array} \right] \left[ \begin{array} { l l } 1 & 2 \\ 0 & 1 \end{array} \right] \left[ \begin{array} { l l } 1 & 3 \\ 0 & 1 \end{array} \right] \ldots \left[ \begin{array} { c c } 1 & n - 1 \\ 0 & 1 \end{array} \right] = \left[ \begin{array} { c c } 1 & 78 \\ 0 & 1 \end{array} \right]$, then the inverse of $\left[ \begin{array} { l l } 1 & n \\ 0 & 1 \end{array} \right]$ is:
(1) $\left[ \begin{array} { c c } 1 & - 12 \\ 0 & 1 \end{array} \right]$
(2) $\left[ \begin{array} { c c } 1 & 0 \\ 12 & 1 \end{array} \right]$
(3) $\left[ \begin{array} { c c } 1 & 0 \\ 13 & 1 \end{array} \right]$
(4) $\left[ \begin{array} { c c } 1 & - 13 \\ 0 & 1 \end{array} \right]$

If the system of equations $2 x + 3 y - z = 0 , x + k y - 2 z = 0$ and $2 x - y + z = 0$ has a non-trivial solution $( x , y , z )$, then $\frac { x } { y } + \frac { y } { z } + \frac { z } { x } + k$ is equal to
(1) $- \frac { 1 } { 4 }$
(2) $\frac { 1 } { 2 }$
(3) $- 4$
(4) $\frac { 3 } { 4 }$