Solution: As per the given question, Now we have to find \[{{\left( AB \right)}^{T}}\] So,

Solution: As per the given question, Consider, \[L.H.S\text{ }=\text{ }R.H.S\] So,

Solution: As per the given question, Consider, \[L.H.S\text{ }=\text{ }R.H.S\] So,

Solution: (i) As per the given question, Consider, \[L.H.S\text{ }=\text{ }R.H.S\] So, (ii) As per the given question, Consider, \[L.H.S\text{ }=\text{ }R.H.S\]...

Solution: As per the given question, \[L.H.S\text{ }=\text{ }R.H.S\] So,

Solution: (i) As per the given question, Consider, \[L.H.S\text{ }=\text{ }R.H.S\] (ii) As per the given question, So,  

Solution: (i) As per the given question, Consider, Put the value of \[A\] \[L.H.S\text{ }=\text{ }R.H.S\] (ii) As per the given question, Consider, \[L.H.S\text{ }=\text{ }R.H.S\] Hence...

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{...

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...

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...

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...

Solution: As per the given question,

Solution: As per the given question,

Solution: As per the given question,

Solution: As per the given question,

Solution: As per the given question,

Solution: As per the given question, Hence the proof.

Solution: As per the given question, Hence the proof.

Solution: As per the given question,

Solution; Now we have to prove \[{{A}^{2}}-\text{ }A\text{ }+\text{ }2\text{ }I\text{ }=\text{ }0\]

Solution: As per the given question, By multiplying we get,  

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)...

Solution: \[=\text{ }\left[ 2x\text{ }+\text{ }1\text{ }+\text{ }2\text{ }+\text{ }x\text{ }+\text{ }3 \right]\text{ }=\text{ }0\] \[=\text{ }\left[ 3x\text{ }+\text{ }6 \right]\text{ }=\text{ }0\]...

Solution: Consider \[{{A}^{2}},\] Hence \[{{A}^{2}}~=\text{ }{{I}_{3}}\]

Solution: As per the question given Consider \[{{A}^{2}}\] Therefore \[{{A}^{2}}~=\text{ }A\]

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{...

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...

Solution: Given From the above matrix, \[{{a}_{43}}~=\text{ }8and\text{ }{{a}_{22}}~=\text{ }0\]

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{...

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,...

Solution: As per the given question, Consider, Now again consider, \[{{B}^{2}}\] Now by subtracting equation (2) from equation (1) we get,

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.

Solution: Consider, Again consider, From equation (1) and (2) AB = BA = 03×3

Solution: As per the given question, Consider, Again consider, From equation (1) and (2) \[AB\text{ }=\text{ }BA\text{ }=\text{ }{{0}_{3\times 3}}\]

Solution: As per the given question, Consider, We know that, Again we have,

Solution: As per the given question, Consider, Hence the proof.

Solution: As per the given question, Consider, Again consider, Hence the proof.

Solution: Consider, Hence the proof.

Solution: As per the given question, Consider, Now we have to find,

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{...

Solution: As per the given question, First we have subtract the matrix which is inside the bracket,  

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{...

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, \[AB\text{ }=\text{ }\left[ 0+\left( -1 \right)+6+6 \right]AB=\left[ 0+\left( -1 \right)+6+6 \right]\] \[AB\text{ }=\text{ }11\]...

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...

Solution: As per the given question, Consider, Now again consider, From equation (1) and (2), it is clear that AB ≠ BA

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...

Solution: As per the given question, Suppose On simplification we get,

Solution: As per the given question, (i) Suppose On simplification we get, (ii) Suppose On simplification we get,

Solution: As per the given question, Consider Now, again consider Therefore, And

Solution: As per the given question, Given Now by multiplying equation \[\left( 1 \right)\text{ }and\text{ }\left( 2 \right)\] we get, Now by adding equation \[\left( 2 \right)\text{ }and\text{...

Solution: As per the given question, Given Now by transposing, we get Therefore,

Solution: As per the given question, Consider, By simplifying we get, Hence, Again suppose, Now by simplifying we get, Therefore,

Solution: Given Now we have to verify \[\left( A\text{ }+\text{ }B \right)\text{ }+\text{ }C\text{ }=\text{ }A\text{ }+\text{ }\left( B\text{ }+\text{ }C \right)\] First consider \[LHS,\text{...

Given \[A\text{ }=\text{ }diag\text{ }\left( 2\text{ }-5\text{ }9 \right),\text{ }\] \[B\text{ }=\text{ }diag\text{ }\left( 1\text{ }1\text{ }-4 \right)\text{ }\] \[and\text{ }C\text{ }=\text{...

(i) Given \[A\text{ }=\text{ }diag\text{ }\left( 2\text{ }-5\text{ }9 \right),\text{ }\] \[B\text{ }=\text{ }diag\text{ }\left( 1\text{ }1\text{ }-4 \right)\text{ }\] \[and\text{ }C\text{ }=\text{...

Solution: Given Now we have to compute \[2A\text{ }\text{ }3B\text{ }+\text{ }4C\]

Solution: (i) Consider \[A\text{ }+\text{ }B,\] A + B is not possible because matrix A is an order of \[2\text{ }x\text{ }2\] and Matrix B is an order of \[2\text{ }x\text{ }3\] so the Sum of the...

Solution: As per the given question (i) First we have to compute 3A, Now, Therefore,   (ii) First we have to compute 3A Now we have to compute 2B By computing 3C we get,...

Solution: As per the given question, (i) First we have to compute 2A Now by computing 3B we get, Now by we have to compute 2A – 3B we get Therefore (ii) Given First we have to compute 4C, Now,...

Solution: As per the given question, Corresponding elements of two matrices should be added Therefore, we get Therefore, Now, As per the given question,...

Solution: We know that if two matrices are equal then the elements of each matrices are also equal. Given that two matrices are equal. Therefore by equating them we get, \[2a\text{ }+\text{ }b\text{...

Solution: We know that if two matrices are equal then the elements of each matrices are also equal. Given that two matrices are equal. Therefore by equating them we get, \[2a\text{ }+\text{ }b\text{...

Solution: 15. Given that two matrices are equal. We know that if two matrices are equal then the elements of each matrices are also equal. Therefore by equating them we get, \[3x\text{ }+\text{...

Given \[{{a}_{i\text{ }j}}~=\text{ }i\] Let \[A\text{ }=\text{ }{{[{{a}_{i\text{ }j}}]}_{4\times 3}}\] So, the elements in a \[4\text{ }\times \text{ }3\] matrix are...

(i) Given \[{{a}_{i\text{ }j}}~=\text{ }2i\text{ }+\text{ }i/j\] Let \[A\text{ }=\text{ }{{[{{a}_{i\text{ }j}}]}_{4\times 3}}\] So, the elements in a \[4\text{ }\times \text{ }3\] matrix are...

Given \[{{a}_{i\text{ }j}}~=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }\left| -3i\text{ }+\text{ }j \right|\] Let \[A\text{ }=\text{ }{{[{{a}_{i\text{ }j}}]}_{2\times 3}}\] So, the elements in...

(i) Given \[{{a}_{i\text{ }j}}~=\text{ }2i\] Let \[A\text{ }=\text{ }{{[{{a}_{i\text{ }j}}]}_{2\times 3}}\] So, the elements in a \[3\times 4\] matrix are \[{{a}_{11}},\text{ }{{a}_{12}},\text{...

(i) Given \[{{a}_{i\text{ }j}}~=\text{ }i\text{ }+\text{ }j\] Let \[A\text{ }=\text{ }{{[{{a}_{i\text{ }j}}]}_{2\times 3}}\] So, the elements in a \[3\text{ }\times \text{ }4\]matrix are...

(i) Given \[{{a}_{i\text{ }j}}~=\text{ }{{\left( i\text{ }\text{ }2j \right)}^{2~}}/2\] Let \[A\text{ }=\text{ }{{[{{a}_{i\text{ }j}}]}_{2\times 2}}\] So, the elements in a \[2\text{ }\times \text{...

(i) Given \[{{\left( i\text{ }+\text{ }j \right)}^{2~}}/2\] Let \[A\text{ }=\text{ }{{[{{a}_{i\text{ }j}}]}_{2\times 2}}\] So, the elements in a \[2\text{ }\times \text{ }2\] matrix are...

(i) Given \[{{a}_{i\text{ }j~}}=\text{ }i\text{ }+\text{ }j\] Let \[A\text{ }=\text{ }[a{{~}_{i\text{ }j}}]{{~}_{2\times 3}}\] So, the elements in a \[2\text{ }\times \text{ }3\] matrix are...

(i) Given \[{{a}_{i\text{ }j}}~=\text{ }i\text{ }\times \text{ }j\] Let \[A\text{ }=\text{ }{{[{{a}_{i\text{ }j}}]}_{2\text{ }\times \text{ }3}}\] So, the elements in a \[2\text{ }\times \text{...

Given A be a matrix of order \[3\text{ }\times \text{ }4\] So, \[A\text{ }=\text{ }[{{a}_{i\text{ }j}}]{{~}_{3\times 4}}\] \[{{R}_{1}}~=\text{ }first\text{ }row\text{ }of\text{ }A\text{ }=\text{...

Solution: (i) Now, Comparing with equation (1) and (2) \[{{a}_{22}}~=\text{ }4\text{ }and\text{ }{{b}_{21}}~=\text{ }\text{ }3\] \[{{a}_{22}}~+\text{ }{{b}_{21}}~=\text{ }4\text{ }+\text{ }\left(...

If a matrix is of order \[m\text{ }\times \text{ }n\]elements, it has \[m\text{ }n\] elements. So, if the matrix has \[8\]elements, we will find the ordered pairs m and n. \[m\text{ }n\text{...