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The following system of linear equations $7x+6y-2z=0$ $3x+4+2z=0$ $x-2y-6z=0,$ has
Option: 1 infinitely many solutions, $(x,y,z)$ satisfying $y=2z.$
Option: 2 infinitely many solutions, $(x,y,z)$ satisfying $x=2z.$
Option: 3 no solution
Option: 4 only the trivial solution.

System of Homogeneous linear equations -

$\\\mathrm{Let,} \\\mathrm{a_1x+b_1y +c_1z=0\;\;\; ...(i)} \\\mathrm{a_2x+b_2y +c_2z=0\;\;\; ...(ii)} \\\mathrm{a_3x+b_3y +c_3z=0\;\;\; ...(iii)} \\\mathrm{be \; three\; homogeneous\; equation} \\\mathrm{and \; let\; \Delta = \begin{vmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{vmatrix}}$

If ? ≠ 0, then x= 0, y = 0, z = 0 is the only solution of the above system. This solution is also known as a trivial solution.

If ? = 0, at least one of x, y and z are non-zero. This solution is called a non-trivial solution.

Explanation: using equation (ii) and (iii), we have

$\\\mathrm{\frac{x}{b_2c_3-b_3c_2} = \frac{y}{c_2a_3-c_3a_2}=\frac{z}{a_2b_3-a_3b_2}} \\\\\mathrm{or \;\; \frac{x}{\begin{vmatrix} b_2 &c_2 \\ b_3 & c_3 \end{vmatrix}}=\frac{y}{\begin{vmatrix} c_2 &a_2 \\ c_3 & a_3 \end{vmatrix}} = \frac{z}{\begin{vmatrix} a_2 & b_2\\ a_3 & b_3 \end{vmatrix}} = k (say \neq 0)} \\\mathrm{\therefore x = k\begin{vmatrix} b_2 &c_2 \\ b_3 & c_3 \end{vmatrix}, y = k\begin{vmatrix} c_2& a_2\\ c_3 & a_3 \end{vmatrix} \; and \; z = k\begin{vmatrix} a_2 & b_2\\ a_3 & b_3 \end{vmatrix}} \\\mathrm{putting\; these\; value\; in \; equation \; (i), we\; have} \\\mathrm{a_1\left \{ k\begin{vmatrix} b_2 & c_2\\ b_3 & c_3 \end{vmatrix} \right \}+b_1\left \{ k\begin{vmatrix} c_2 & a_2\\ c_3 & a_3 \end{vmatrix} \right \}+c_1\left \{ \begin{vmatrix} a_2 & b_2\\ a_3 & b_3 \end{vmatrix} \right \}=0}$

$\\\mathrm{\Rightarrow a_1\begin{vmatrix} b_2 &c_2 \\ b_3 & c_2 \end{vmatrix}-b_1\begin{vmatrix} a_2 & c_2\\ a_3 &c_3 \end{vmatrix}+c_1\begin{vmatrix} a_2 &b_2 \\ a_3 &b_3 \end{vmatrix} = 0 \;\;\;[\because \; k \neq 0]} \\\mathrm{or \;\; \begin{vmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{vmatrix} = 0 \; or \; \Delta = 0}$

This is the condition for a system have Non-trivial solution.

$\begin{aligned} &(1)\;\;7 x+6 y-2 z=0\\ &(2)\;\;3 x+4 y+2 z=0\\ &(3)\;\;x-2 y-6 z=0 \end{aligned}$

$\left|\begin{array}{ccc}{7} & {6} & {-2} \\ {3} & {4} & {2} \\ {1} & {-2} & {-6}\end{array}\right| =7(-20)-6(-20)-2(-10)=-140+120+20=0$

so infinite non-trivial solution exist

now equation (1) + 3 equation (3)

10x - 20z = 0

x = 2z

Correct Option 2

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avinash.dongre

Let $a-2b+c=1.$ If $f\left ( x \right )= \begin{vmatrix} x+a & x+2 &x+1 \\ x+b &x+3 &x+2 \\ x+c &x+4 & x+3 \end{vmatrix},$ then :
Option: 1 $f(-50)=501$
Option: 2 $f(-50)=-1$
Option: 3 $f(50)=1$
Option: 4 $f(50)=-501$

Elementary row operations -

Elementary row operations

Row transformation: Following three types of operation (Transformation) on the rows of a given matrix are known as elementary row operation (transformation).

i) Interchange of i^th row with j^th row, this operation is denoted by

$\\\mathrm{R_i \leftrightarrow R_j \;\;or\; R_{ij}}$

ii) The multiplication of ith row by a constant k (k≠0) is denoted by

$\\\mathrm{R_i \leftrightarrow kR_i \;\; or \; k\cdot R_i}$

iii) The addition of ith row to the elements of jth row multiplied by constant k (k≠0) is denoted by

$\\\mathrm{R_i \leftrightarrow R_i + kR_j \;\; or \; k\cdot R_{ij}}$

In the same way, three-column operations can also be defined too.

$\\\text { Apply } \mathrm{R}_{1}=\mathrm{R}_{1}+\mathrm{R}_{3}-2 \mathrm{R}_{2}\\\Rightarrow f(x)=\left|\begin{array}{ccc}{1} & {0} & {0} \\ {x+b} & {x+3} & {x+2} \\ {x+c} & {x+4} & {x+3}\end{array}\right| \quad \Rightarrow f(x)=1 \\\quad \Rightarrow f(50)=1$

Correct Option (3)

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If for some $\alpha \: and\: \beta$ in R, the intersection of the following three planes $x+4y-2z=1$ $x+7y-5z=\beta$ $x+5y+\alpha z=5$ is aline in $R^{3}$ , then $\alpha +\beta$ is equal to :
Option: 1 $0$
Option: 2 $10$
Option: 3 $-10$
Option: 4 $2$

Cramer’s law -

Cramer’s law for the system of equations in two variables :

$\\\mathrm{Let \; a_1x +b_1y = c_1\; and \; a_2x + b_2y = c_2, where} \\\mathrm{\frac{a_1}{a_2}\neq\frac{b_1}{b_2}} \\\mathrm{on \; solving \; this \; equation \; by \; cross \; multiplication, we \; get} \\\mathrm{\frac{x}{b_1c_2-b_2c_1}=\frac{y}{a_1c_2-a_2c_1}=\frac{1}{a_1b_2-a_2b_1}} \\\mathrm{or \; \frac{x}{\begin{vmatrix} b_1 & c_1\\ b_2 & c_2 \end{vmatrix}}=\frac{y}{\begin{vmatrix} a_1 & c_1\\ a_2 & c_2 \end{vmatrix}}=\frac{1}{\begin{vmatrix} a_1 & b_1\\ a_2 & b_2 \end{vmatrix}}} \\\mathrm{or \; x=\frac{\begin{vmatrix} b_1 & c_1\\ b_2 & c_2 \end{vmatrix}}{\begin{vmatrix} a_1 & b_1\\ a_2 & b_2 \end{vmatrix}}, y=\frac{\begin{vmatrix} a_1 & c_1\\ a_2 & c_2 \end{vmatrix}}{\begin{vmatrix} a_1 & b_1\\ a_2 & b_2 \end{vmatrix}}}$

We can observe that first row in the numerator of x is of constants and 2nd row in the numerator is of constants, and the denominator is of the coefficient of variables.

We can follow this analogy for the system of equations of 3 variable where third in the numerator of the value of z will be of constant and denominator will be formed by the value of coefficients of the variables.

$\\\mathrm{so\; let \;us\; consider\; the \; system\; of \;equations} \\\mathrm{a_1x+b_1y +c_1z =d_1 \;\;...(i)} \\\mathrm{a_2x+b_2y +c_2z =d_2\;\;\;...(ii)} \\\mathrm{a_3x+b_3y +c_3z =d_3\;\;\;...(iii)} \\\mathrm{then \; \Delta,\; which\; will\; be \;determinant\; of\; coefficient\; of \;variables, will\; be } \\\mathrm{\Delta = \begin{vmatrix} a_1 & b_1 & c_1\\ a_2& b_2 & c_2\\ a_3 & b_3 & c_3 \end{vmatrix}, \Delta_1\; numerator\; of\; x \;is:} \\\mathrm{\Delta_1= \begin{vmatrix} d_1 & b_1 &c_1 \\ d_2 & b_2 & c_2\\ d_3 & b_3 & c_3 \end{vmatrix}, similarly\; \Delta_2 = \begin{vmatrix} a_1 & d_1 & c_1\\ a_2 & d_2 & c_2\\ a_3 & d_3 & c_3 \end{vmatrix}\; and \;} \\\mathrm{\Delta_3 = \begin{vmatrix} a_1 & b_1 & d_1\\ a_2 & b_2 & d_2\\ a_3 & b_3 & d_3 \end{vmatrix}, if \Delta \neq 0\; then\; \;x=\frac{\Delta_1}{\Delta}} \\\mathrm{y=\frac{\Delta_2}{\Delta}, z=\frac{\Delta_3}{\Delta}}$

i) If ? ≠ 0, then the system of equations has a unique finite solution and so equations are consistent, and solutions are $\\\mathrm{x=\frac{\Delta_1}{\Delta}, y=\frac{\Delta_2}{\Delta}, z=\frac{\Delta_3}{\Delta}}$

ii) If ? = 0, and any of $\Delta_1\neq 0 \; or \;\Delta_2\neq 0 \; or \;\Delta_3\neq 0$

Then the system of equations is inconsistent and hence no solution exists.

iii) If all $\Delta =\Delta_1=\Delta_2=\Delta_3= 0$ then

System of equations is consistent and dependent and it has an infinite number of solution.

$\begin{array}{l}{\Delta=0 \Rightarrow\left|\begin{array}{ccc}{1} & {4} & {-2} \\ {1} & {7} & {-5} \\ {1} & {5} & {\alpha}\end{array}\right|=0} \\ {(7 \alpha+25)-(4 \alpha+10)+(-20+14)=0} \\ {3 \alpha+9=0 \Rightarrow \alpha=-3}\end{array}$

$\begin{array}{l}{\text { Also }\quad D_{3}=0 \Rightarrow\left|\begin{array}{lll}{1} & {4} & {1} \\ {1} & {7} & {\beta} \\ {1} & {5} & {5}\end{array}\right|=0} \\ {1(35-5 \beta)-(15)+1(4 \beta-7)=0} \\ {\beta=13}\end{array}$

$\alpha+ \beta=10$

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avinash.dongre

The number of all $3\times 3$ matrices A, with enteries from the set $\left \{ -1,0,1 \right \}$ such that the sum of the diagonal elements of $AA^{T}$ is 3, is
Option: 1 672
Option: 2 512
Option: 31024
Option: 4 256

Let matrix A be

$\\A=\begin{bmatrix} a &b &c \\ d& e &f \\ g &h & i \end{bmatrix}\\\\\\ A^T=\begin{bmatrix} a &d &g \\ b& e &h \\ c &f & i \end{bmatrix} \\\\\\ \text{trace}(AA^T)=a^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2+i^2=3$

So out of 9 elements, 3 elements must be equal to 1 or −1, and the rest elements must be 0.

Possible cases

$\begin{array}{cc} 0,0,0,0,0,0,1,1,1 & \rightarrow \text{Total possibilities}=^9C_6 \\ 0,0,0,0,0,0,-1,-1,-1 & \rightarrow \text{Total possibilities}=^9C_6 \\ 0,0,0,0,0,0,1,1,-1 & \rightarrow \text{Total possibilities}=^9C_6\times 3 \\ 0,0,0,0,0,0,-1,1,-1 & \rightarrow \text{Total possibilities}=^9C_6\times3 \end{array}$

$\\\\\text{Total number of cases}=^{9}{C}_{6}\times 8=672$

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Kuldeep Maurya

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For which of the following ordered pairs $(\mu ,\delta)$ , the system of linear equations $x+2y+3z=1$ $3x+4y+5z=\mu$ $4x+4y+4z=\varrho$ is inconsistent ?
Option: 1 $(4,6)$
Option: 2 $(3,4)$
Option: 3 $(1,0)$
Option: 4 $(4,3)$

Solution of System of Linear Equations Using Matrix Method -

let us consider n linear equations in n unknowns, given as below

$\\\mathrm{a_{11}x_1+a_{12}x_2+...+a_{1n}x_n = b_1} \\\mathrm{a_{21}x_1+a_{22}x_2+...+a_{2n}x_n = b_2} \\\mathrm{...\;\;\;...\;\;\;...\;\;\;...\;\;\;\;\;\;\;\;...\;\;\;\;\;\;...} \\\mathrm{...\;\;\;...\;\;\;...\;\;\;...\;\;\;\;\;\;\;\;...\;\;\;\;\;\;...} \\\mathrm{a_{n1}x_1+a_{n2}x_2+...+a_{nn}x_n = b_n} \\\mathrm{here \; x_{1}, x_2,...x_n \; are \;unknown\; variables} \\\mathrm{if \; b_1=b_2 =...=b_n=0\; then\; the\; system \; of \; equation \; is} \\\mathrm{known\; as \; homogenous\; system \; of \; equation\; and \;if } \\\mathrm{any \; of \;b_1,b_2,...b_n \; is\; not \; known\; then \; it \; is \;called\; } \\\mathrm{non\; homogenous\; system \; of \; equation}$

The above system of equations can be written in matrix form as

$\\\mathrm{\begin{bmatrix} a_{11} & a_{12} & ... & ... & a_{1n}\\ a_{21} & a_{22} & ... & ... & a_{2n}\\ ... & ... & ... & ... & ...\\ ... & ... & ... & ... & ...\\ a_{n1} & a_{n2} & ... & ... & a_{nn} \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ ...\\ ...\\ x_n \end{bmatrix} = \begin{bmatrix} b_1\\ b_2\\ ...\\ ...\\ b_n \end{bmatrix}} \\\\\mathrm{\Rightarrow AX = B, \; where} \\\\\mathrm{A = \begin{bmatrix} a_{11} & a_{12} & ... & ... & a_{1n}\\ a_{21} & a_{22} & ... & ... & a_{2n}\\ ... & ... & ... & ... & ...\\ ... & ... & ... & ... & ...\\ a_{n1} & a_{n2} & ... & ... & a_{nn} \end{bmatrix} , X=\begin{bmatrix} x_1\\ x_2\\ ...\\ ...\\ x_n \end{bmatrix}, B=\begin{bmatrix} b_1\\ b_2\\ ...\\ ...\\ b_n \end{bmatrix}}$

Premultiplying equation AX=B by A^-1, we get

A^-1(AX) = A^-1B ⇒ (A^-1A)X = A^-1B

⇒ IX = A^-1B

⇒ X = A^-1B

⇒ $\mathrm{X=\frac{adj A}{\left | A \right |}B}$

Types of equation :

System of equations is non-homogenous:
1. If |A| ≠ 0, then the system of equations is consistent and has a unique solution X = A^-1B
2. If |A| = 0 and (adj A)·B ≠ 0, then the system of equations is inconsistent and has no solution.
3. If |A| = 0 and (adj A)·B = 0, then the system of equations is consistent and has infinite number of solutions.
System of equations is homogenous:
1. If |A| ≠ 0, then the system of equations has only trivial solution and it has one solution.
2. If |A| = 0 then the system of equations has non-trivial solution and it has an infinite number of solution.
3. If number of equation < number of unknown then it has non-trivial solution.

$\begin{aligned} D &=\left|\begin{array}{lll}{3} & {4} & {5} \\ {1} & {2} & {3} \\ {4} & {4} & {4}\end{array}\right| \\ &=2\left|\begin{array}{lll}{3} & {4} & {5} \\ {1} & {2} & {3} \\ {2} & {2} & {2}\end{array}\right| \end{aligned}$

$\\R_1\rightarrow R_1-(R_2+R_3)\\D=\left|\begin{array}{lll}{3} & {4} & {5} \\ {1} & {2} & {3} \\ {2} & {2} & {2}\end{array}\right|=0$

Hence, it has infinitely many solutions

$\mathrm{P}_{3} \equiv \alpha \mathrm{P}_{1}+\beta \mathrm{P}_{2}$

$3 \alpha+\beta=4\;\; \&\;\; 4 \alpha+2 \beta=4 \Rightarrow \alpha=2 \;\; \&\;\; \beta=-2$

$2 \mu-2=\delta$

For non infinite solution $2 \mu-2\neq\delta$

Correct Option (4)

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Kuldeep Maurya

If the system of linear equations 2x+2ay+az=0 2x+3by+bz=0, 2x+4cy+cz=0, where $a,b,c\epsilon R$ are non-zero and distinct ; has a non-zero solution, then :
Option: 1 $a+b+c=0$
Option: 2 $a,b,c$ are in AP
Option: 3 $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.
Option: 4 $a,b,c$ are in G.P.

Cramer’s law -

Cramer’s law for the system of equations in two variables :

We can observe that first row in the numerator of x is of constants and 2nd row in the numerator is of constants, and the denominator is of the coefficient of variables.

ii) If ? = 0, and any of $\Delta_1\neq 0 \; or \;\Delta_2\neq 0 \; or \;\Delta_3\neq 0$

Then the system of equations is inconsistent and hence no solution exists.

iii) If all $\Delta =\Delta_1=\Delta_2=\Delta_3= 0$ then

System of equations is consistent and dependent and it has an infinite number of solution.

For non zero solutions D = 0

$\\D=\begin{vmatrix} 2 & 2a &a \\ 2& 3b &b \\ 2& 4c &c \end{vmatrix}=0\\C_2\rightarrow C_2-2C_3\\D=\begin{vmatrix} 1 &0 &a \\ 1& b &b \\ 1 & 2c & c \end{vmatrix}=0\\$

on solving

$\\-bc+2ac-ab=0\\\\\frac{1}{a}+\frac{1}{c}=\frac{2}{b}$

Correct Option (3)

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Kuldeep Maurya

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The system of linear equations $\lambda x+2y+2z=5$ $2\lambda x+3y+5z=8$ $4 x+\lambda y+6z=10$ has :
Option: 1 no solution when $\lambda =2$
Option: 2 infinitely many solutions when $\lambda =2$
Option: 3 no solution when $\lambda =8$
Option: 4 a unique solution when $\lambda =-8$

Cramer’s law -

Cramer’s law for the system of equations in two variables :

We can observe that first row in the numerator of x is of constants and 2nd row in the numerator is of constants, and the denominator is of the coefficient of variables.

ii) If ? = 0, and any of $\Delta_1\neq 0 \; or \;\Delta_2\neq 0 \; or \;\Delta_3\neq 0$

Then the system of equations is inconsistent and hence no solution exists.

iii) If all $\Delta =\Delta_1=\Delta_2=\Delta_3= 0$ then

System of equations is consistent and dependent and it has an infinite number of solution.

$D=\begin{vmatrix} \lambda &2 &2 \\ 2\lambda &3 & 5\\ 4 &\lambda &6 \end{vmatrix}$

D = ( $\lambda$ + 8) ( 2 – $\lambda$ ) for $\lambda$ = 2

$D_1=\begin{vmatrix} 5 &2 &2 \\ 8 &3 & 5\\ 10 &2 &6 \end{vmatrix}$

= 5[18 – 10] – 2 [48 – 50] + 2 (16 – 30]

= 40 + 4 – 28 $\neq$ 0 No solutions for $\lambda$ = 2

Correct Option (1)

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vishal kumar

If $A=\begin{pmatrix} 2 &2 \\ 9 &4 \end{pmatrix}$ and $I=\begin{pmatrix} 1 &0 \\ 0 &1 \end{pmatrix}$ , then $10A^{-1}$ is equal to :
Option: 1 $6I-A$
Option: 5 $A-6I$
Option: 9 $4I-A$
Option: 13 $A-4I$

Inverse of a Matrix -

A non-singular square matrix “A” is said to be invertible if there exists a non-singular square matrix B such that AB = I = BA, (all matrix are of the same order, they must be for this), then B is called inverse of matrix A.

$\\\mathrm{A^{-1} = B \Leftrightarrow AB = \mathbb{I}_n = BA} \\\mathrm{we \; have, } \\ \\\mathrm{A(adj A)=A\mathbb{I}_n} \\\mathrm{\Rightarrow A^{-1}A (adj A)=A^{-1}\mathbb{I}_n|A|} \\\mathrm{\mathbb{I}_n(adj A)=A^{-1}|A|\mathbb{I}_n} \\\mathrm{A^{-1}=\frac{adj A}{\left | A \right |}}$

Inverse of 2 x 2 matrix

$\\\mathrm{Let\;A\;is\;a\;square\;matrix\;of\;order\;2}\\\mathrm{A=\begin{bmatrix} a &b \\ c & d \end{bmatrix}}\\\mathrm{Then,}\\\mathrm{A^{-1}=\begin{bmatrix} a &b \\ c & d \end{bmatrix}^{-1}=\frac{1}{ad-bc}\begin{bmatrix} d &-b \\- c & a \end{bmatrix}}$

Multiplication of two matrices -

Matrix multiplication:

Two matrices A and B are conformable for the product AB if the number of columns in A and the number of rows in B is equal. Otherwise, these two matrices will be non-conformable for matrix multiplication. So on that basis,

i) AB is defined only if col(A) = row(B)

ii) BA is defined only if col(B) = row(A)

$\\\mathrm{A = \left [ a_{ij} \right ]_{m\times n}} \\\mathrm{\\B=\left [ b_{ij} \right ]_{n\times p}}$

$\\\mathrm{C = AB = \left [ c_{ij} \right ]_{m\times p}} \\\mathrm{Where\;\; c_{ij} = \sum_{j=1}^{n}a_{ij}b_{jk}, 1\leq i\leq m,1\leq k\leq p} \\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=a_{i1}b_{1k} + a_{i2}b_{2k} + a_{i3}b_{3k}+ ... + a_{in}b_{nk}}$

For examples

$\\\mathrm{Suppose,\;two\;matrices\;are\;given}\\\mathrm{A=\begin{bmatrix} a_{11} &a_{12} &a_{13} \\ a_{21} &a_{22} & a_{33} \end{bmatrix}_{2\times3}\;\;\;and\;\;\;B=\begin{bmatrix} b_{11}& b_{12} &b_{13} \\b_{21} &b_{22} &b_{23} \\b_{31} &b_{32} &b_{33} \end{bmatrix}_{3\times3}}\\\\\mathrm{To\:obtain\:the\:entries\:in\:row\:\mathit{i}\:of\:AB,\:we\:multiply\:the\:entries\:in\:row\:\mathit{i}\:of\:A\:by\:}\\\mathrm{column\:\mathit{j}\:in\:B\:and\:add.}\\\mathrm{given\:matrices\:A\:and\:B,\:where\:the\:order\:of\:A\:are\:2\times3\:and\:the\:order\:of\:B\:are\:3\times3,}\\\mathrm{the\:product\:of\:AB\:will\:be\:a\:2\times3\:matrix.}\\\\\mathrm{To\:obtain\:the\:entry\:in\:row\:1,\:column\:1\:of\:AB,\:multiply\:the\:}\\\mathrm{first\:row\:in\:A\:by\:the\:first\:column\:in\:B,and\:add.}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\begin{bmatrix} a_{11} &a_{12} &a_{13} \end{bmatrix}\begin{bmatrix} b_{11}\\b_{21} \\b_{31} \end{bmatrix}=a_{11}\cdot b_{11}+a_{12}\cdot b_{21}+a_{13}\cdot b_{31}}$

$\\\mathrm{To\:obtain\:the\:entry\:in\:row\:1,\:column\:2\:of\:AB,\:multiply\:the\:}\\\mathrm{first\:row\:in\:A\:by\:the\:second\:column\:in\:B,and\:add.}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\begin{bmatrix} a_{11} &a_{12} &a_{13} \end{bmatrix}\begin{bmatrix} b_{12}\\b_{22} \\b_{32} \end{bmatrix}=a_{11}\cdot b_{12}+a_{12}\cdot b_{22}+a_{13}\cdot b_{32}}\\\\\mathrm{To\:obtain\:the\:entry\:in\:row\:1,\:column\:3\:of\:AB,\:multiply\:the\:}\\\mathrm{first\:row\:in\:A\:by\:the\:thired\:column\:in\:B,and\:add.}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\begin{bmatrix} a_{11} &a_{12} &a_{13} \end{bmatrix}\begin{bmatrix} b_{13}\\b_{23} \\b_{33} \end{bmatrix}=a_{11}\cdot b_{13}+a_{12}\cdot b_{23}+a_{13}\cdot b_{33}}\\\\\mathrm{We\:proceed\:the\:same\:way\:to\:obtain\:the\:second\:row\:of\:AB.\:In\:other\:words,\:}\\\mathrm{row\:2\:of\:A\:times\:column\:1\:of\:B;}\\\mathrm{row\:2\:of\:A\:times\:column\:2\:of\:B;}\\\mathrm{row\:2\:of\:A\;times\:column\:3\:of\:B.}$

$\\\mathrm{When\:complete,\:the\:product\:matrix\:will\:be}\\\\\mathrm{AB=\begin{bmatrix} a_{11}\cdot b_{11}+a_{12}\cdot b_{21}+a_{13}\cdot b_{31}\;\;& a_{11}\cdot b_{12}+a_{12}\cdot b_{22}+a_{13}\cdot b_{32}\;\; &a_{11}\cdot b_{13}+a_{12}\cdot b_{23}+a_{13}\cdot b_{33} \\ a_{21}\cdot b_{11}+a_{22}\cdot b_{21}+a_{23}\cdot b_{31} \;\;& a_{21}\cdot b_{12}+a_{22}\cdot b_{22}+a_{23}\cdot b_{32} \;\;& a_{21}\cdot b_{13}+a_{22}\cdot b_{23}+a_{23}\cdot b_{33} \end{bmatrix}}$

$A=\left[\begin{array}{ll}{2} & {2} \\ {9} & {4}\end{array}\right] \text { and } A^{-1}=\left[\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right]$

$AA^{-1}=\left[\begin{array}{ll}{2a+2c} & {2b+2d} \\ {9a+4c} & {9b+4d}\end{array}\right] =\left[\begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right]$

On comparing we get

$a=-\frac{2}{5}, \;\; b=\frac{1}{5},\;\;c=\frac{9}{10}\;\&\;d=-\frac{1}{5}$

$\Rightarrow \quad 10 A^{-1}=A-61$

Correct Option (2)

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vishal kumar

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If the system of linear equations, x+y+z=6 x+2y+3z=10 $3x+2y+\lambda z=\mu$ has more than two solutions, then $\mu-\lambda ^{2}$ is equal to _______.
Option: 1 13
Option: 2 17
Option: 3 21
Option: 4 25

Cramer’s law -

Cramer’s law for the system of equations in two variables :

We can observe that first row in the numerator of x is of constants and 2nd row in the numerator is of constants, and the denominator is of the coefficient of variables.

ii) If ? = 0, and any of $\Delta_1\neq 0 \; or \;\Delta_2\neq 0 \; or \;\Delta_3\neq 0$

Then the system of equations is inconsistent and hence no solution exists.

iii) If all $\Delta =\Delta_1=\Delta_2=\Delta_3= 0$ then

System of equations is consistent and dependent and it has an infinite number of solution.

x + y + z = 6 …….. (1)

x + 2y + 3z = 10 …….. (2)

3x + 2y + $\lambda$ z = $\mu$ …….. (3)

from (1) and (2)

if z = 0 $\Rightarrow$ x + y = 6 and x + 2y = 10

$\Rightarrow$ y = 4, x = 2

(2, 4, 0)

if y = 0 $\Rightarrow$ x + z = 6 and x + 3z = 10

$\Rightarrow$ z = 2 and x = 4

(4, 0, 2)

so, 3x + 2y + $\lambda$ z = $\mu$ must pass through (2, 4, 0) and (4, 0, 2)

$\mu$ = 14

and 12 + 2 $\lambda$ = $\mu$

$\lambda$ = 1

$(\mu-\lambda^2)$ = 13

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Posted by

Ritika Jonwal

Let $A=\left [ a_{ij} \right ]$ and $B=\left [ b_{ij} \right ]$ be two $3\times 3$ real matrices such that $b_{ij}=(3)^{(i+j-2)}a_{ji},$ where, i, j=1,2,3. if the determinant of B is 81, then the determinant of A is :
Option: 1 $1/9$
Option: 2 $1/81$
Option: 3 $1/3$
Option: 4 $3$

Matrices, order of matrices, row and column matrix -

A set of $\mathrm{m\times n}$ numbers (real or complex) or objects or symbols arranged in form of a rectangular array having m rows and n columns and bounded by brackets [?] is called matrix of order m × n, read as m by n matrix.

E.g for m = 2 and n =3, we have $\begin{bmatrix} 2 & 4 & -3\\ 5 & 4 & 6 \end{bmatrix}$ order of this matrix is 2×3

The m by n matrix is represented as :

$\begin{bmatrix} a_{11} &a_{12} &... & a_{1n}\\ a_{21}&a_{22} &... &a_{2n} \\ ...& ...& ... & ...\\ a_{m1}&a_{m2} &... &a_{mn} \end{bmatrix}$

This representation can be represented in a more compact form as

$\left [ a_{ij} \right ]_{m\times n}$

Where $a_{ij}$ represents element of i^th row and j^th column and i = 1,2,...,m; j = 1,2,...,n.

For example, to locate the entry in matrix A identified as a_ij, we look for the entry in row i, column j. In matrix A, shown below, the entry in row 2, column 3 is a₂₃.

$A=\left[\begin{array}{lll}{a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}}\end{array}\right]$

Matrix is only a representation of the symbol, number or object. It does not have any value. Usually, a matrix denoted by capital letters.

Properties of Determinants - Part 2 -

Property 5

If each element of a row (or a column) of a determinant is multiplied by a constant k, then the value of the determinant is multiplied by k.

For example

$\\\mathrm{Let,\;\;\Delta=\begin{vmatrix} a_1 &a_2 &a_3 \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}}\\\\\mathrm{and\:\Delta'\:be\:the\:determinant\:obtained\:by\:multiplying\:the\:elements\:of\:}\\\mathrm{the\:first\:row\:by\:k.}\\\\\mathrm{\Delta'=\begin{vmatrix} ka_1 &ka_2 &ka_3 \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}}\\\mathrm{Expanding\:along\:first\:row,\:we\:get}\\\\\mathrm{\Delta'=ka_1(b_2c_3-b_3c_2)-ka_2(b_1c_3-b_3c_1)+ka_3(b_1c_2-b_2c_1)}\\\mathrm{\;\;\;\;=k[a_1(b_2c_3-b_3c_2)-a_2(b_1c_3-b_3c_1)+a_3(b_1c_2-b_2c_1)]}\\\mathrm{\Delta'=k\Delta}\\\mathrm{Hence,}\\\mathrm{\begin{vmatrix} ka_1 &ka_2 &ka_3 \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}=k\begin{vmatrix} a_1 &a_2 &a_3 \\b_1 &b_2 &b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}}$

Note:

By this property, we can take out any common factor from any one row or any one column of a given determinant.
If corresponding elements of any two rows (or columns) of a determinant are proportional (in the same ratio), then the determinant value is zero.

$|B|=\left|\begin{array}{lll}{b_{11}} & {b_{12}} & {b_{13}} \\ {b_{21}} & {b_{22}} & {b_{23}} \\ {b_{31}} & {b_{32}} & {b_{33}}\end{array}\right|$

$|B|=\left|\begin{array}{ccc}{3^{0} a_{11}} & {3^{1} a_{12}} & {3^{2} a_{13}} \\ {3^{1} a_{21}} & {3^{2} a_{22}} & {3^{3} a_{23}} \\ {3^{2} a_{31}} & {3^{3} a_{32}} & {3^{4} a_{33}}\end{array}\right|$

Taking Common $3^2$ from $R_3$ and $3$ from $R_2$

$|B|=3^3\left|\begin{array}{ccc}{3^{0} a_{11}} & {3^{1} a_{12}} & {3^{2} a_{13}} \\ {3^{0} a_{21}} & {3^{1} a_{22}} & {3^{2} a_{23}} \\ {3^{0} a_{31}} & {3^{1} a_{32}} & {3^{2} a_{33}}\end{array}\right|$

Taking Common $3^2$ from $C_3$ and $3$ from $C_2$

$\Rightarrow 81=3^{3} \cdot 3 \cdot 3^{2}|\mathrm{A}| \Rightarrow 3^{4}=3^{6}|\mathrm{A}| \Rightarrow|\mathrm{A}|=\frac{1}{9}$

Correct Option (A)

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Posted by

Ritika Jonwal

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The following system of linear equations has Option: 1 infinitely many solutions, satisfying Option: 2 infinitely many solutions, satisfying Option: 3 no solution Option: 4 only the trivial solution.

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JEE Main high-scoring chapters and topics

For which of the following ordered pairs , the system of linear equations is inconsistent ? Option: 1 Option: 2 Option: 3 Option: 4

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Option: 2 infinitely many solutions, $(x,y,z)$ satisfying $x=2z.$
Option: 3 no solution
Option: 4 only the trivial solution.

Let $a-2b+c=1.$ If $f\left ( x \right )= \begin{vmatrix} x+a & x+2 &x+1 \\ x+b &x+3 &x+2 \\ x+c &x+4 & x+3 \end{vmatrix},$ then :
Option: 1 $f(-50)=501$
Option: 2 $f(-50)=-1$
Option: 3 $f(50)=1$
Option: 4 $f(50)=-501$

If for some $\alpha \: and\: \beta$ in R, the intersection of the following three planes $x+4y-2z=1$ $x+7y-5z=\beta$ $x+5y+\alpha z=5$ is aline in $R^{3}$ , then $\alpha +\beta$ is equal to :
Option: 1 $0$
Option: 2 $10$
Option: 3 $-10$
Option: 4 $2$

The number of all $3\times 3$ matrices A, with enteries from the set $\left \{ -1,0,1 \right \}$ such that the sum of the diagonal elements of $AA^{T}$ is 3, is
Option: 1 672
Option: 2 512
Option: 31024
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Option: 1 $(4,6)$
Option: 2 $(3,4)$
Option: 3 $(1,0)$
Option: 4 $(4,3)$

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Option: 1 $a+b+c=0$
Option: 2 $a,b,c$ are in AP
Option: 3 $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.
Option: 4 $a,b,c$ are in G.P.

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Option: 1 $6I-A$
Option: 5 $A-6I$
Option: 9 $4I-A$
Option: 13 $A-4I$

If the system of linear equations, x+y+z=6 x+2y+3z=10 $3x+2y+\lambda z=\mu$ has more than two solutions, then $\mu-\lambda ^{2}$ is equal to _______.
Option: 1 13
Option: 2 17
Option: 3 21
Option: 4 25

Let $A=\left [ a_{ij} \right ]$ and $B=\left [ b_{ij} \right ]$ be two $3\times 3$ real matrices such that $b_{ij}=(3)^{(i+j-2)}a_{ji},$ where, i, j=1,2,3. if the determinant of B is 81, then the determinant of A is :
Option: 1 $1/9$
Option: 2 $1/81$
Option: 3 $1/3$
Option: 4 $3$