Addition and Subtraction MCQ Quiz in తెలుగు - Objective Question with Answer for Addition and Subtraction - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Given:

\(\rm 2\begin{bmatrix} 1 & 3\\ 0& \rm x \end{bmatrix}+\begin{bmatrix}\rm y& 0\\ 1&2 \end{bmatrix}=\begin{bmatrix} 5& 6\\ 1& 8\end{bmatrix}\)

Concept:

Matrices: If two matrices are equal, then their corresponding elements must also be equal.

Calculation:

\(\rm 2\begin{bmatrix} 1 & 3\\ 0& \rm x \end{bmatrix}+\begin{bmatrix}\rm y& 0\\ 1&2 \end{bmatrix}=\begin{bmatrix} 5& 6\\ 1& 8\end{bmatrix}\)

⇒ \(\rm \begin{bmatrix} 2 & 6\\ 0& \rm 2x \end{bmatrix}+\begin{bmatrix}\rm y& 0\\ 1&2 \end{bmatrix}=\begin{bmatrix} 5& 6\\ 1& 8\end{bmatrix}\)

⇒ \(\rm \begin{bmatrix} 2+\rm y & 6+0\\ 0+1& \rm 2x+2 \end{bmatrix}=\begin{bmatrix} 5& 6\\ 1& 8\end{bmatrix}\)

Since the matrices are equal, their corresponding elements must also be equal.

⇒ 2 + y = 5 and 2x + 2 = 8.

On solving the above equations, we get 

⇒ y = 3 and x = 3

∴ x = 3 and y = 3

A +  I = \(\begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} \)

⇒A = \(\begin{bmatrix} 0 & a & 1 \\ 2 & 0 & 0 \\ a & 1 & 1 \end{bmatrix} \)

$Given\; det(A)=-4$.

det(A)= 0⋅(0. 1 − 0. 1)−a. (2. 1−0. a)+1. (2. 1−0. a)

= - a. (2)+1. (2) =−2a + 2

Set equal to -4 -4" id="MathJax-Element-180-Frame" role="presentation" style="position: relative;" tabindex="0">−4$-4$ $-4$

⇒ 2a + 2 =  -4 ⟹ −2a = −6 ⟹a = 3:

Also

det((a+1) adj ((a−1)A)) =2^m3ⁿ

• For a 3 × 3 Matrix, det(adj)(B) = (det(B))²
• For a scalar k det(kB) = k³det(B)

Let’s plug in values:

⇒ det((a−1) A) = (a−1)³.det(A) = (3−1)³.(−4)=2³.(−4) = 8.(−4) = −32

⇒ det(adj(a - 1)A)) = (-32)² = 1024

⇒ (a +1)³ = (3 + 1)³ = 64

⇒ det((a+1)adj((a−1)A)) = 64⋅1024 = 65536

Thus, 65536 = 2¹⁶3⁰

⇒ Then m = 16,n = 0

⇒ m + n  = 16 + 0 = 16

Calculation:

Given:

x + 2y = \(\begin{bmatrix} 2 & -3\\ 1 & 5 \end{bmatrix}\)                    .... (1)

2x + 5y = \(\begin{bmatrix} 7 & 5\\ 2 & 3 \end{bmatrix}\)                     .... (2)

Multiplying by 2 in the equation (1), we get

⇒ 2x + 4y = \(\begin{bmatrix} 4 & -6\\ 2 & 10 \end{bmatrix}\)             .... (3)

Subtracting equation (3) from equation (2), we get

⇒ (2x + 5y) - (2x + 4y) = \(\begin{bmatrix} 7 & 5\\ 2 & 3 \end{bmatrix}-\begin{bmatrix} 4 & -6\\ 2 & 10 \end{bmatrix}\)

∴ y = \( \begin{bmatrix} 3 & 11\\ 0 & -7 \end{bmatrix}\)

\(P^TP= \begin{bmatrix} \cfrac { \sqrt { 3 } }{ 2 } & \cfrac { -1 }{ 2 } \ \cfrac { 1 }{ 2 } & \cfrac { \sqrt { 3 } }{ 2 } \end{bmatrix}\begin{bmatrix} \cfrac { \sqrt { 3 } }{ 2 } & \cfrac { 1 }{ 2 } \ \cfrac { -1 }{ 2 } & \cfrac { \sqrt { 3 } }{ 2 } \end{bmatrix}=\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}=I\)

\(Q = PAP^{T}\)

\(P^{T}Q^{2015} P=P^T\cdot Q \cdot Q \cdot Q \cdot Q........Q\cdot P\)

\(=P^T(PAP^T)(PAP^T)........P\)

\((P^TP)A(P^TP)A(P^TP)........P=A^{2015}\)

\(A^2=\begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix}\)

\(A^3=\begin{bmatrix} 1 & 3 \ 0 & 1 \end{bmatrix}\)

\(A^{2015}=\begin{bmatrix} 1 & 2015 \ 0 & 1 \end{bmatrix}\)

Since, \(A\) and \(I\) are of order \(2\times 2\). So, \(x\) will be a matrix of order \(2\times 2\)

Let \(x=\begin{bmatrix} a & b \ c & d \end{bmatrix}\)

From \((i)\), we get

\(Ax=I\)

\(\implies \begin{bmatrix} 1 & 2 \ 4 & 3 \end{bmatrix} \begin{bmatrix} a & b \ c & d \end{bmatrix}=\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\)

\(\implies \begin{bmatrix} a+2c & b+2d \ 4a+3c & 4b+3d \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\)

\(\implies a+2c=1\) ...... \((ii)\), \(4a+3c=0\) ........ \((iii)\)

\(b+2d=0\) ....... \((iv)\) and \(4b+3d=1\) ....... \((v)\)

Solving \((ii)\) and \((iii)\) simultaneously we get

\(a=-\dfrac{3}{5}\) and \(c=\dfrac{4}{5}\)

Then solving \((iv)\) and \((v)\) simultaneously we get

\(b=\dfrac{2}{5}\) and \(d=-\dfrac{1}{5}\)

Substituting all these values in \(x\), we get

\(x=\begin{bmatrix} -\dfrac{3}{5} & \dfrac{2}{5} \ \dfrac{4}{5} & -\dfrac{1}{5} \end{bmatrix}\)

\(\therefore x=\dfrac{1}{5} \begin{bmatrix} -3 & 2 \ 4 & -1 \end{bmatrix}\)

Calculation

Here, \(a_{ij}\) stands for element of \(A\) with \(i^{th}\) row and \(j^{th}\) column. Also, \(A_{ij}\) stands for co-factor of \(a_{ij}\) of matrix \(A\)

So, we get \(a_{11}=1, a_{12}=1\) and \(a_{13}=0\)

To find the co-factor

\(A_{21}=(-1)^{2+1}\begin{vmatrix} 1 & 0 \\ 2 & 1\end{vmatrix}=(-1)(1-0)=-1\)

\(A_{22}=(-1)^{2+2}\begin{vmatrix} 1 & 0 \\ 1 & 1\end{vmatrix}=(-1)^{4}(1-0)=1\)

\(A_{23}=(-1)^{2+3}\begin{vmatrix} 1 & 1 \\ 1 & 2\end{vmatrix}=(-1)^{5}(2-1)=-1\)

\(\therefore a_{11}A_{21} + a_{12}A_{22} +a_{13}A_{23}=1(-1)+1(1)+0(-1)=0\)

Hence option 2 is correct

Concept:

Conditions for the subtraction of matrices:

Two matrices should be of the same order (number of rows = number of columns).

Add the corresponding element of other matrices.

Calculation:

Given:

\(f(x) = \left[ {\begin{array}{*{20}{c}} x&λ \\ {2λ }&x \end{array}} \right]\)

\(\Rightarrow f(λ x) = \left[ {\begin{array}{*{20}{c}} {λ x}&λ \\ {2λ }&{λ x} \end{array}} \right]\)

\(f(λ x) - f(x) = \left[ {\begin{array}{*{20}{c}} λ x &λ \\ {2λ }&{λ x} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} x&λ \\ {2λ }&x \end{array}} \right]\)

\(f(λ x) - f(x) = \left[ {\begin{array}{*{20}{c}} λ x -x &0\\ {0 }&{λ x - x} \end{array}} \right] \)

|f(λx) - f(x)| = (λx - x)²

|f(λx) - f(x)| = x²(λ - 1)²

Concept:

• Here we have to use the multiplication properties of the matrixes.

Calculation:

Given:  A and B are square matrices of order n × n.

We know that,

(A - B)2 = (A - B) (A - B)

⇒ (A - B)2 = A2 - AB - BA + B2

∵  AB ≠ BA 

So, option 3 is correct.                              

Concept:

Conditions for the subtraction of matrices:

Two matrices should be of the same order (number of rows = number of columns).

Add the corresponding element of other matrices.

Calculation:

Given:

\(f(x) = \left[ {\begin{array}{*{20}{c}} x&\lambda \\ {2\lambda }&x \end{array}} \right]\)

\(\Rightarrow f(\lambda x) = \left[ {\begin{array}{*{20}{c}} {\lambda x}&\lambda \\ {2\lambda }&{\lambda x} \end{array}} \right]\)

\(f(\lambda x) - f(x) = \left[ {\begin{array}{*{20}{c}} \lambda &\lambda \\ {2\lambda }&{\lambda x} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} x&\lambda \\ {2\lambda }&x \end{array}} \right]\)

\(\begin{array}{l} = ({(\lambda x)^2} - 2{\lambda ^2}) - ({x^2} - 2{\lambda ^2}) = {\lambda ^2}{x^2} - {x^2} = {x^2}({\lambda ^2} - 1)\\ \end{array}\)

Concept:

Diagonal Matrix:

Any square matrix in which all the elements are zero except those in the principal diagonal is called a diagonal matrix.

i.e A = [aij]n × n is a diagonal matrix if aij = 0 for i ≠ j.

Note: If A = diag [a11, a22, a33, ....., ann] is a diagonal matrix of order n then \(A = \;\left[ {\begin{array}{*{20}{c}} {{a_{11}}}& \cdots &0\\ \vdots & \ddots & \vdots \\ 0& \cdots &{{a_{nn}}} \end{array}} \right]\) 

Scalar Matrix:

A diagonal matrix in which all the principal diagonal elements are equal is called a scalar matrix.

Let A and B be any two matrices of same order m × n, then their sum A ± B = [aij ± bij]m × n where A = [aij]m × n and B = [bij]m × n

Calculation:

Given: A = diag [2, - 5, 9], B = diag [- 3, 7, 14] and C = diag [4, - 6, 3]

Here, we have to find the value of 2A + B - 5C

As we know that if, A = diag [a11, a22, a33, ....., ann] is a diagonal matrix of order n then \(A = \;\left[ {\begin{array}{*{20}{c}} {{a_{11}}}& \cdots &0\\ \vdots & \ddots & \vdots \\ 0& \cdots &{{a_{nn}}} \end{array}} \right]\) 

∵  A = diag [2, - 5, 9], B = diag [- 3, 7, 14] and C = diag [4, - 6, 3]

⇒ \(A = \left[ {\begin{array}{*{20}{c}} 2&0&0\\ 0&{ - \;5}&0\\ 0&0&9 \end{array}} \right],\;B = \left[ {\begin{array}{*{20}{c}} { - \;3}&0&0\\ 0&7&0\\ 0&0&{14} \end{array}} \right]\;and\;C = \left[ {\begin{array}{*{20}{c}} 4&0&0\\ 0&{ - \;6}&0\\ 0&0&3 \end{array}} \right]\)

⇒ \(2A = \left[ {\begin{array}{*{20}{c}} 4&0&0\\ 0&{ - \;10}&0\\ 0&0&{18} \end{array}} \right],\;B = \left[ {\begin{array}{*{20}{c}} { - \;3}&0&0\\ 0&7&0\\ 0&0&{14} \end{array}} \right]\;and\;5C = \left[ {\begin{array}{*{20}{c}} {20}&0&0\\ 0&{ - \;30}&0\\ 0&0&{15} \end{array}} \right]\)

⇒ \(2A + B - 5C = \left[ {\begin{array}{*{20}{c}} {\left( {4 - 3 - 20} \right)}&0&0\\ 0&{\left( { - \;10 + 7 + 30} \right)}&0\\ 0&0&{\left( {18 + 14 - 15} \right)} \end{array}} \right]\)

\(= \left[ {\begin{array}{*{20}{c}} { - \;19}&0&0\\ 0&{27}&0\\ 0&0&{17} \end{array}} \right]\)

Hence, 2A + B - 5C = diag [- 19, 27, 17]