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

1) If \displaystyle A \text{ and } B are two matrices of order \displaystyle 3 \times 2 \text{ and } 2 \times 3 respectively; then their sum \displaystyle A + B is possible.

2) The Matrices \displaystyle A_{2 \times 3} \text{ and } A_{2 \times 3} are conformable for subtraction.

3) Transpose of a \displaystyle 2 \times 1 matrix is a \displaystyle 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 \displaystyle A is of the Order \displaystyle 3 \times 2 while \displaystyle B is of the order \displaystyle 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 $ and latex B $ are of the same order.

3) False: The transpose of a matrix is obtained by interchanging rows with columns. Hence the Transpose of a \displaystyle 2 \times 1 matrix is a \displaystyle 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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\displaystyle \text{Question 2: Given } \begin{bmatrix} x & y+2 \\ 3 & z-1 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ 3 & 2 \end{bmatrix} \text{ , find } x, \ y \ and \ z

Answer:

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

\displaystyle \Rightarrow x = 3  

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

\displaystyle \text{also } z-1 = 2 \Rightarrow z = 3  

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

\displaystyle \text{1) } \begin{bmatrix} -4 & a+5 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} b+4 & 2 \\ 3 & c-1 \end{bmatrix}  

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

Answer:

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

\displaystyle -4 = b+ 4 \Rightarrow b = -8  

\displaystyle a+5 = 2 \Rightarrow a = -3  

\displaystyle 2 = c-1 \Rightarrow c = 3  

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

\displaystyle a = 3  

\displaystyle a-b=-1 \Rightarrow b = a-1 = 3-1 = 2  

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

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

1) \displaystyle A+B            2) \displaystyle B-A  

Answer:

1) \displaystyle A+B  

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

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

2) \displaystyle B-A  

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

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

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\displaystyle \text{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} \text{ find: }

1) \displaystyle B+C            2) \displaystyle A-C            3) \displaystyle A+B-C            4) \displaystyle A-B+C  

Answer:

\displaystyle \text{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}

\displaystyle \text{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}

\displaystyle \text{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}

\displaystyle \text{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:

\displaystyle \text{1) } \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} -1 & -2 \\ 1 & -7 \end{bmatrix}  

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

\displaystyle \text{3) } \begin{bmatrix} 0 & 1 & 2 \\ 4 & 6 & 7 \end{bmatrix} + \begin{bmatrix} 3 & 4 \\ 6 & 8 \end{bmatrix}  

Answer:

\displaystyle \text{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}

\displaystyle \text{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 \displaystyle x \text{ and } y from the following equations:

\displaystyle \text{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}  

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

Answer:

\displaystyle \text{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}  

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

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

Therefore

\displaystyle 3-x = 7 \Rightarrow x = -4  

\displaystyle y+2 = 2 \Rightarrow y = 0  

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

Therefore

\displaystyle -8+y=-3 \Rightarrow y = 5  

\displaystyle x-2=2 \Rightarrow x = 4  

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

 1) \displaystyle M+M^{t}            2) \displaystyle M^{t}-M  

Answer:

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

\displaystyle \text{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}

\displaystyle \text{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 \displaystyle A = \begin{bmatrix} 6 & -5 \end{bmatrix} \text{ and } B = \begin{bmatrix} -2 & 0 \\ 4 & -1 \end{bmatrix} \text{ and } C = \begin{bmatrix} -2 \\ 4 \end{bmatrix} .

Answer:

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

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

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

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

 1) \displaystyle X+B=C-A            2) \displaystyle A-X=B+C  

Answer:

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

1) \displaystyle X+B=C-A  

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

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

Therefore

\displaystyle a = -3 \ and \ b = 5  

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

 2) \displaystyle A-X=B+C  

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

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

Therefore \displaystyle a = 3 \ and \ b = -9  

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

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

 1) \displaystyle A+X=B           2) \displaystyle A-X=B             3) \displaystyle X-B=A  

Answer:

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

1) \displaystyle A+X=B  

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

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

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

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

 2) \displaystyle A-X=B  

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

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

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

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

 3) \displaystyle X-B=A  

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

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

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

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

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