Solution: We know, is identity matrix of size \[3\] So according to the given criteria Now we will multiply the two matrices on LHS using the formula \[{{c}_{ij}}~=\text{...
Solve:
Solution: So Now, we will find the matrix for \[{{A}^{2}}\] we get Now, we will find the matrix for \[\lambda \text{ }A\] we get But given, \[{{A}^{2}}~=\text{ }\lambda \text{ }A\text{ }+\text{ }\mu...
Solve:
Solution: As per the given question, To show that \[f\text{ }\left( A \right)\text{ }=\text{ }0\] Substitute \[x\text{ }=\text{ }A~in~f\left( x \right)\] we get I is identity matrix, so Now, we will...
Solve:
Solution: As per the given question, I is identity matrix, so Also given, Now, we have to find \[{{A}^{2}}\], we get Now, we will find the matrix for \[8A\], we get So, Substitute corresponding...
Solve:
Solution: As per the given question,
Solve:
Solution: As per the given question,
Solve:
Solution: As per the given question,
Solve:
Solution: As per the given question,
Solve:
Solution: As per the given question,
Solve:
Solution: As per the given question, Hence the proof.
Solve:
Solution: As per the given question, Hence the proof.
Solve:
Solution: As per the given question,
Solve:
Solution; Now we have to prove \[{{A}^{2}}-\text{ }A\text{ }+\text{ }2\text{ }I\text{ }=\text{ }0\]
If given matrix, then find x
Solution: As per the given question, By multiplying we get,
Solve:
Solution: \[\Rightarrow ~\left[ \left( 2x\text{ }+\text{ }4 \right)\text{ }x\text{ }+\text{ }4\text{ }\left( x\text{ }+\text{ }2 \right)\text{ }-\text{ }1\left( 2x\text{ }+\text{ }4 \right)...
Solve:
Solution: \[=\text{ }\left[ 2x\text{ }+\text{ }1\text{ }+\text{ }2\text{ }+\text{ }x\text{ }+\text{ }3 \right]\text{ }=\text{ }0\] \[=\text{ }\left[ 3x\text{ }+\text{ }6 \right]\text{ }=\text{ }0\]...
Solve:
Solution: Consider \[{{A}^{2}},\] Hence \[{{A}^{2}}~=\text{ }{{I}_{3}}\]
Solve:
Solution: As per the question given Consider \[{{A}^{2}}\] Therefore \[{{A}^{2}}~=\text{ }A\]
If ω is a complex cube root of unity, show that
Solution: Given It is also given that \[\omega \] is a complex cube root of unity, Consider the LHS, We know that \[1\text{ }+\text{ }\omega \text{ }+\text{ }{{\omega }^{2}}~=\text{ }0\text{...
Solve:
Solution: Consider, Again consider, Now, consider the RHS Therefore, \[{{A}^{3}}~=\text{ }p\text{ }I\text{ }+\text{ }q\text{ }A\text{ }+\text{ }r{{A}^{2}}\] Hence the...
Compute the elements a43 and a22 of the matrix:
Solution: Given From the above matrix, \[{{a}_{43}}~=\text{ }8and\text{ }{{a}_{22}}~=\text{ }0\]
Solve:
Solution: As per the given question, Given, Consider the LHS, Now consider RHS From the above equations \[LHS\text{ }=\text{ }RHS\] Therefore, \[A\text{ }\left( B\text{ }\text{ }C \right)\text{...
For the following matrices verify the distributivity of matrix multiplication over matrix addition i.e. A (B + C) = AB + AC.
Solution: As per the given question, Consider LHS, Now consider RHS, From equation (1) and (2), it is clear that \[A\text{ }\left( B\text{ }+\text{ }C \right)\text{ }=\text{ }AB\text{ }+\text{ }AC\]...
For the following matrices verify the associativity of matrix multiplication i.e. (AB) C = A (BC)
Solution: Consider, Now consider RHS, From equation (1) and (2), it is clear that \[\left( AB \right)\text{ }C\text{ }=\text{ }A\text{ }\left( BC \right)\] Now, Consider the LHS, Now consider RHS,...
Solve:
Solution: As per the given question, Consider, Now again consider, \[{{B}^{2}}\] Now by subtracting equation (2) from equation (1) we get,
Solve:
Solution: As per the given question, Consider, Therefore \[AB\text{ }=\text{ }A\] Again consider, \[BA\] we get, Hence \[BA\text{ }=\text{ }B\] Hence the proof.
Solve:
Solution: Consider, Again consider, From equation (1) and (2) AB = BA = 03×3
Solve:
Solution: As per the given question, Consider, Again consider, From equation (1) and (2) \[AB\text{ }=\text{ }BA\text{ }=\text{ }{{0}_{3\times 3}}\]
Solve:
Solution: As per the given question, Consider, We know that, Again we have,
Solve:
Solution: As per the given question, Consider, Hence the proof.
Solve:
Solution: As per the given question, Consider, Again consider, Hence the proof.
Solve:
Solution: Consider, Hence the proof.
Solve:
Solution: As per the given question, Consider, Now we have to find,
Solve:
Solution: As per the given question, We know that, Again we know that, Now, consider, We have, Now, from equation (1), (2), (3) and (4), it is clear that \[{{A}^{2~}}=\text{ }{{B}^{2}}=\text{...
Evaluate the following:
Solution: As per the given question, First we have subtract the matrix which is inside the bracket,
Evaluate the following:
Solution: As per the given question, First we have to add first two matrix, On simplifying, we get Now, First we have to multiply first two given matrix, \[=\text{...
Show that AB ≠ BA in each of the following cases:
Solution: (i) As per the given question, Consider, Again consider, From equation (1) and (2), it is clear that \[AB\text{ }\ne \text{ }BA\] (ii) As per the given question, Consider, Again consider,...
(i) (ii) Solution: (i) As per the given question, Consider, \[AB\text{ }=\text{ }\left[ 0+\left( -1 \right)+6+6 \right]AB=\left[ 0+\left( -1 \right)+6+6 \right]\] \[AB\text{ }=\text{ }11\]...
Compute the products AB and BA whichever exists in each of the following cases:
Solution: (i) As per the given question, Consider, BA does not exist Because the number of columns in B is greater than the rows in A (ii) As per the given question, Consider, Again...
Show that AB ≠ BA in each of the following cases:
Solution: As per the given question, Consider, Now again consider, From equation (1) and (2), it is clear that AB ≠ BA
Show that AB ≠ BA in each of the following cases:
Solution: (i) As per the given question, Consider, Again consider, From equation (1) and (2), it is clear that \[AB\text{ }\ne \text{ }BA\] (ii) As per the given question, Consider Now again...
Compute the indicated products:
Solution: As per the given question, Suppose On simplification we get,
Compute the indicated products:
Solution: As per the given question, (i) Suppose On simplification we get, (ii) Suppose On simplification we get,
Find $\frac{d y}{d x}$ in the following exercise $y=\cos ^{-1}\left(\frac{2 x}{1+x^{2}}\right),-1<x<1$
Solution: The provided function is: $y=\cos ^{-1}\left(\frac{2 x}{1+x^{2}}\right),-1<x<1$ Let's take $x=\tan \theta$, we obtain $y=\cos^{-1}\quad[(2 \tan\ \theta) /(1+\tan^ 2$$\theta)]$...