Question 1: State True or False. If False, please state the reason.

  1. If A   and B   are two matrices of order 3 \times 2   and 2 \times 3   respectively; then their sum A + B   is possible.
  2. The Matrices A_{2 \times 3}   and  A_{2 \times 3}   are conformable for subtraction.
  3. Transpose of a 2 \times 1   matrix is a 2\times 1   matrix.
  4. Transpose of a square matrix is a square matrix.
  5. A column matrix has many columns and only one row.

Answer:

  1. False: Two matrices can be added together if they are of the same order. Here A   is of the Order 3 \times 2   while B   is of the order 2 \times 3  . Hence they cannot be added.
  2. True: Two matrices can be subtracted together if they are of the same order. Here both latex A  &s=0$ and latex B  &s=0$ are of the same order.
  3. False: The transpose of a matrix is obtained by interchanging rows with columns. Hence the Transpose of a 2 \times 1   matrix is a 1 \times 2   matrix.
  4. True: Yes Transpose of a square matrix is a square matrix. Here the number of rows is equal to the number of columns. Hence even on transposing, the matrix would remain as a square matrix.
  5. False: A Column matrix has one column and many rows.

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Question 2: Given \begin{bmatrix}  x & y+2  \\ 3 & z-1 \end{bmatrix} = \begin{bmatrix}  3 & 1  \\ 3 & 2 \end{bmatrix} , find x, \ y \ and \  z .

Answer:

\begin{bmatrix}  x & y+2  \\ 3 & z-1 \end{bmatrix} = \begin{bmatrix}  3 & 1  \\ 3 & 2 \end{bmatrix}

\Rightarrow x = 3

y+2 = 1 \ \Rightarrow y = -1

also  z-1 = 2  \Rightarrow z = 3 

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Question 3: Solve for a, \ b \ and \  c if;

  1. \begin{bmatrix}  -4 & a+5  \\ 3 & 2 \end{bmatrix} = \begin{bmatrix}  b+4 & 2  \\ 3 & c-1 \end{bmatrix}
  2. \begin{bmatrix}  a & a-b  \\ b+c & 0 \end{bmatrix} = \begin{bmatrix}  3 & -1  \\ 2 & 0 \end{bmatrix}

Answer:

1)     Given \begin{bmatrix}  -4 & a+5  \\ 3 & 2 \end{bmatrix} = \begin{bmatrix}  b+4 & 2  \\ 3 & c-1 \end{bmatrix} ; Therefore

-4 = b+ 4 \Rightarrow b = -8 

a+5 = 2 \Rightarrow a = -3 

2 = c-1 \Rightarrow c = 3 

2)    \begin{bmatrix}  a & a-b  \\ b+c & 0 \end{bmatrix} = \begin{bmatrix}  3 & -1  \\ 2 & 0 \end{bmatrix}

a = 3

a-b=-1 \Rightarrow b = a-1 = 3-1 = 2 

b+c = 2 \Rightarrow c = 2-b = 2-2 = 0 

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Question 4: If  A =  \begin{bmatrix} 8 & -3   \end{bmatrix} and  B =  \begin{bmatrix} 4 & -5   \end{bmatrix}  find

  1. A+B 
  2. B-A 

Answer:

1)    A+B 

= \begin{bmatrix} 8 & -3   \end{bmatrix} +  \begin{bmatrix} 4 & -5   \end{bmatrix}

=\begin{bmatrix} 8+4 & -3-5 \end{bmatrix} = \begin{bmatrix} 12 & -8 \end{bmatrix} 

2)    B-A 

= \begin{bmatrix} 4 & -5   \end{bmatrix} - \begin{bmatrix} 8 & -3   \end{bmatrix}

=\begin{bmatrix} 4-8 & -5-(-3) \end{bmatrix} = \begin{bmatrix} -4 & -2 \end{bmatrix} 

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Question 5: If A = \begin{bmatrix}  2  \\ 5 \end{bmatrix}, \ B=\begin{bmatrix}  1  \\ 4 \end{bmatrix} \ and \ C=\begin{bmatrix}  6  \\ -2 \end{bmatrix} find:

  1. B+C
  2. A-C
  3. A+B-C
  4. A-B+C

Answer:

1)    B+C

= \begin{bmatrix}  1  \\ 4 \end{bmatrix} + \begin{bmatrix}  6  \\ -2 \end{bmatrix} 

= \begin{bmatrix}  1+6  \\ 4-2 \end{bmatrix} 

= \begin{bmatrix}  7  \\ 2 \end{bmatrix} 

2)    A-C

= \begin{bmatrix}  2  \\ 5 \end{bmatrix} - \begin{bmatrix}  6  \\ -2 \end{bmatrix} 

= \begin{bmatrix}  2-6  \\ 5-(-2) \end{bmatrix}  

= \begin{bmatrix}  -4  \\ 7 \end{bmatrix}  

3)    A+B-C

= \begin{bmatrix}  2  \\ 5 \end{bmatrix} + \begin{bmatrix}  1  \\ 4 \end{bmatrix} - \begin{bmatrix}  6  \\ -2 \end{bmatrix} 

= \begin{bmatrix}  2+1-6  \\ 5+4-(-2) \end{bmatrix}  

= \begin{bmatrix}  -3  \\ 11 \end{bmatrix}  

4)    A-B+C

= \begin{bmatrix}  2  \\ 5 \end{bmatrix} - \begin{bmatrix}  1  \\ 4 \end{bmatrix} + \begin{bmatrix}  6  \\ -2 \end{bmatrix} 

= \begin{bmatrix}  2-1+6  \\ 5-4+(-2) \end{bmatrix}  

= \begin{bmatrix}  7  \\ -1 \end{bmatrix}  

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Question 6: Wherever possible, write each of the following in a single matrix:

  1. \begin{bmatrix}  1 & 2  \\ 3 & 4 \end{bmatrix} + \begin{bmatrix}  -1 & -2  \\ 1 & -7 \end{bmatrix}  
  2. \begin{bmatrix}  2 &3 & 4  \\ 5 & 6 & 7 \end{bmatrix} - \begin{bmatrix}  0 &2 & 3  \\ 6 & -1 & 0 \end{bmatrix}  
  3. \begin{bmatrix}  0 & 1 & 2  \\ 4 & 6 & 7 \end{bmatrix} + \begin{bmatrix}  3 & 4   \\ 6 & 8 \end{bmatrix}

Answer:

1) \begin{bmatrix}  1 & 2  \\ 3 & 4 \end{bmatrix} + \begin{bmatrix}  -1 & -2  \\ 1 & -7 \end{bmatrix} = \begin{bmatrix}  0 & 0  \\ 4 & -3 \end{bmatrix} 

2) \begin{bmatrix}  2 &3 & 4  \\ 5 & 6 & 7 \end{bmatrix} - \begin{bmatrix}  0 &2 & 3  \\ 6 & -1 & 0 \end{bmatrix} =  \begin{bmatrix}  2 &5 & 7  \\ 11 & 5 & 7 \end{bmatrix}

3) Adding this is is not possible as the order of the metrices are not the same.

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Question 7: Find x and y from the following equations:

  1. \begin{bmatrix}  5 & 2  \\ -1 & y-1 \end{bmatrix} - \begin{bmatrix}  1 & x-1  \\ 2 & -3 \end{bmatrix} = \begin{bmatrix}  4 & 7  \\ -3 & 2 \end{bmatrix}
  2. \begin{bmatrix} -8 & x \end{bmatrix} + \begin{bmatrix} y & -2 \end{bmatrix} = \begin{bmatrix} -3 & 2 \end{bmatrix} 

Answer:

1)     \begin{bmatrix}  5 & 2  \\ -1 & y-1 \end{bmatrix} - \begin{bmatrix}  1 & x-1  \\ 2 & -3 \end{bmatrix} = \begin{bmatrix}  4 & 7  \\ -3 & 2 \end{bmatrix}

\begin{bmatrix}  5-1 & 2-(x-1)  \\ (-1-2) & y-1-(-3) \end{bmatrix} =\begin{bmatrix}  4 & 7  \\ -3 & 2 \end{bmatrix}

= \begin{bmatrix}  4 & 3-x  \\ -3 & y+2 \end{bmatrix} =\begin{bmatrix}  4 & 7  \\ -3 & 2 \end{bmatrix}

Therefore

3-x = 7 \Rightarrow x = -4

y+2 = 2 \Rightarrow y = 0 

2)    \begin{bmatrix} -8 & x \end{bmatrix} + \begin{bmatrix} y & -2 \end{bmatrix} = \begin{bmatrix} -3 & 2 \end{bmatrix} 

Therefore

-8+y=-3 \Rightarrow y = 5  

x-2=2 \Rightarrow x = 4  

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Question 8: Given M = \begin{bmatrix}  5 & -3  \\ -2 & 4 \end{bmatrix} , find its transpose matrix M^{t} . If possible find:

  1.  M+M^{t}
  2.  M^{t}-M

Answer:

M = \begin{bmatrix}  5 & -3  \\ -2 & 4 \end{bmatrix}

M^{t} = \begin{bmatrix}  5 & -2  \\ -3 & 4 \end{bmatrix}

1) M+M^{t}

= \begin{bmatrix}  5 & -3  \\ -2 & 4 \end{bmatrix} + \begin{bmatrix}  5 & -2  \\ -3 & 4 \end{bmatrix} = \begin{bmatrix}  10 & -5  \\ -5 & 8 \end{bmatrix} 

2) M^{t}-M

= \begin{bmatrix}  5 & -2  \\ -3 & 4 \end{bmatrix} - \begin{bmatrix}  5 & -3  \\ -2 & 4 \end{bmatrix} = \begin{bmatrix}  0 & 1  \\ -1 & 0 \end{bmatrix} 

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Question 9:  Write the additive  inverse of matrices A, B and C where A = \begin{bmatrix}  6 & -5   \end{bmatrix} and B = \begin{bmatrix}  -2 & 0  \\ 4 & -1 \end{bmatrix} and C = \begin{bmatrix}  -2  \\  4 \end{bmatrix} .

Answer:

Additive Inverse of  A = \begin{bmatrix}  6 & -5   \end{bmatrix} is = \begin{bmatrix}  -6 & 5   \end{bmatrix}

Additive Inverse of B = \begin{bmatrix}  -2 & 0  \\ 4 & -1 \end{bmatrix} is = \begin{bmatrix}  2 & 0 \\ -4 & 1   \end{bmatrix}

Additive Inverse of C = \begin{bmatrix}  -2  \\  4 \end{bmatrix}   is = \begin{bmatrix}  7 & -4   \end{bmatrix}

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Question 10: Given A = \begin{bmatrix}  2 & -3   \end{bmatrix}, \ B= \begin{bmatrix}  0 & 2   \end{bmatrix}, \ C= \begin{bmatrix}  -1 & 4   \end{bmatrix} . Find matrix X  in each of the following:

  1.  X+B=C-A
  2.  A-X=B+C

Answer:

Let X = \begin{bmatrix}  a & b   \end{bmatrix} 

1) X+B=C-A

\begin{bmatrix}  a & b   \end{bmatrix} +\begin{bmatrix}  0 & 2   \end{bmatrix}=\begin{bmatrix}  -1 & 4   \end{bmatrix}-\begin{bmatrix}  2 & -3   \end{bmatrix}

\begin{bmatrix}  a & b+2   \end{bmatrix} = \begin{bmatrix}  -3 & 7   \end{bmatrix}

Therefore

a = -3 \ and \ b = 5

Hence  X = \begin{bmatrix}  -3 & 5   \end{bmatrix} 

 2) A-X=B+C

\begin{bmatrix}  2 & -3   \end{bmatrix} - \begin{bmatrix}  a & b   \end{bmatrix} = \begin{bmatrix}  0 & 2   \end{bmatrix} + \begin{bmatrix}  -1 & 4   \end{bmatrix}

\begin{bmatrix} 2-a & -3-b \end{bmatrix} = \begin{bmatrix}  -1 & 6   \end{bmatrix}

Therefore a = 3 \ and \ b = -9

Hence X = \begin{bmatrix}  3 & -9   \end{bmatrix} 

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Question 11: Given A = \begin{bmatrix}  -1 & 0  \\ 2 & -4 \end{bmatrix} and B = \begin{bmatrix}  3 & -3  \\ -2 & 0 \end{bmatrix} . FInd the matrix X  in each of the following:

  1.  A+X=B
  2.  A-X=B
  3.  X-B=A

Answer:

Let X = \begin{bmatrix}  a & b \\ c & d   \end{bmatrix} 

1) A+X=B

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

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

Therefore a = 4, \ b = -3, \ c = -4 \ and \ d = 4

Hence X = \begin{bmatrix}  4 & -3 \\ -4 & 4   \end{bmatrix} 

 2) A-X=B

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

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

Therefore a = -4, \ b = 3, \ c = 4 \ and \ d = -4

Hence X = \begin{bmatrix}  -4 & 3 \\ 4 & -4   \end{bmatrix} 

 3) X-B=A

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

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

Therefore a = 2, \ b = -3, \ c = 0  \ and \ d = -4

Hence X = \begin{bmatrix}  2 & -3 \\ 0 & -4   \end{bmatrix} 

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