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.

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

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

$\displaystyle \Rightarrow x = 3$

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

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

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

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

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

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

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

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

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

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

$\displaystyle a = 3$

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

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

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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) $\displaystyle B-A$

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) $\displaystyle A-B+C$

$\displaystyle \text{1) } B+C = \begin{bmatrix} 1 \\ 4 \end{bmatrix} + \begin{bmatrix} 6 \\ -2 \end{bmatrix} = \begin{bmatrix} 1+6 \\ 4-2 \end{bmatrix} = \begin{bmatrix} 7 \\ 2 \end{bmatrix}$

$\displaystyle \text{2) }A-C = \begin{bmatrix} 2 \\ 5 \end{bmatrix} - \begin{bmatrix} 6 \\ -2 \end{bmatrix} = \begin{bmatrix} 2-6 \\ 5-(-2) \end{bmatrix} = \begin{bmatrix} -4 \\ 7 \end{bmatrix}$

$\displaystyle \text{3) }A+B-C = \begin{bmatrix} 2 \\ 5 \end{bmatrix} + \begin{bmatrix} 1 \\ 4 \end{bmatrix} - \begin{bmatrix} 6 \\ -2 \end{bmatrix} = \begin{bmatrix} 2+1-6 \\ 5+4-(-2) \end{bmatrix} = \begin{bmatrix} -3 \\ 11 \end{bmatrix}$

$\displaystyle \text{4) }A-B+C = \begin{bmatrix} 2 \\ 5 \end{bmatrix} - \begin{bmatrix} 1 \\ 4 \end{bmatrix} + \begin{bmatrix} 6 \\ -2 \end{bmatrix} = \begin{bmatrix} 2-1+6 \\ 5-4+(-2) \end{bmatrix} = \begin{bmatrix} 7 \\ -1 \end{bmatrix}$

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

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

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

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

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

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

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:

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

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

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

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

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

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) $\displaystyle M^{t}-M$

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

$\displaystyle \text{1) } M+M^{t} = \begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix} + \begin{bmatrix} 5 & -2 \\ -3 & 4 \end{bmatrix} = \begin{bmatrix} 10 & -5 \\ -5 & 8 \end{bmatrix}$

$\displaystyle \text{2) } M^{t}-M = \begin{bmatrix} 5 & -2 \\ -3 & 4 \end{bmatrix} - \begin{bmatrix} 5 & -3 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$

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

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

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

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

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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) $\displaystyle A-X=B+C$

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) $\displaystyle X-B=A$

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

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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