Question 11 Mark
Given $A=\left[\begin{array}{lll}0 & 4 & 6 \\ 3 & 0 & 1\end{array}\right]\left[\begin{array}{cc}0 & 1 \\ -1 & 2 \\ -5 & -6\end{array}\right]$ Find if possible $A^2$
AnswerProduct $AA (=A^2)$ is not possible as the number of columns of matrix A is not equal to its number of rows
View full question & answer→Question 21 Mark
If $A=\left[\begin{array}{cc}0 & 2 \\ 5 & -2\end{array}\right], B=\left[\begin{array}{cc}1 & -1 \\ 3 & 2\end{array}\right]$and $I$ is a unit matrix of order $2 \times 2$ findb $BA$
Answer$B A=\left[\begin{array}{cc}1 & -1 \\ 3 & 2\end{array}\right]\left[\begin{array}{cc}0 & 2 \\ 5 & -2\end{array}\right] $
$ =\left[\begin{array}{cc}0-5 & 2+2 \\ 0+10 & 6-4\end{array}\right] $
$ =\left[\begin{array}{cc}-5 & 4 \\ 10 & 2\end{array}\right]$
View full question & answer→Question 31 Mark
If $A=\left[\begin{array}{cc}0 & 2 \\ 5 & -2\end{array}\right], B=\left[\begin{array}{cc}1 & -1 \\ 3 & 2\end{array}\right]$ and $I$ is a unit matrix of order $2 \times 2$ find $AB$
Answer$A B=\left[\begin{array}{cc}0 & 2 \\ 5 & -2\end{array}\right]\left[\begin{array}{cc}1 & -1 \\ 3 & 2\end{array}\right] $
$ =\left[\begin{array}{cc}0+6 & 0+4 \\ 5-6 & -5-4\end{array}\right] $
$ =\left[\begin{array}{cc}6 & 4 \\ -1 & -9\end{array}\right]$
View full question & answer→Question 41 Mark
Evaluate if possible $\left[\begin{array}{ll}6 & 4 \\ 3 & 1\end{array}\right][-1,3]$
Answer$\left[\begin{array}{ll}6 & 4 \\ 3 & 1\end{array}\right][-1,3]$
The number of columns in the first matrix is not equal to the number of rows in the second matrix. Thus, the product is not possible.
View full question & answer→Question 51 Mark
Evaluate if possible $\left[\begin{array}{cc}6 & 4 \\ 3 & -1\end{array}\right]\left[\begin{array}{c}-1 \\ 3\end{array}\right]$
Answer$\left[\begin{array}{cc}6 & 4 \\ 3 & -1\end{array}\right]\left[\begin{array}{c}-1 \\ 3\end{array}\right]=\left[\begin{array}{c}-6+12 \\ -3-3\end{array}\right]=\left[\begin{array}{c}6 \\ -6\end{array}\right]$
View full question & answer→Question 61 Mark
Evaluate if possible $\left[\begin{array}{ll}1 & -2\end{array}\right]\left[\begin{array}{ll}-2 & 3 \\ -1 & 4\end{array}\right]$
Answer$[1-2]\left[\begin{array}{ll}-2 & 3 \\ -1 & 4\end{array}\right]=[-2+2,3-8]=\left[\begin{array}{ll}0 & -5\end{array}\right]$
View full question & answer→Question 71 Mark
Evaluate if possible $[3 \ 2 ]\left[\begin{array}{l}2 \\ 0\end{array}\right]$
Answer$[3 \ 2 ]\left[\begin{array}{l}2 \\ 0\end{array}\right]$
$=[6+0]$
$=[6]$
View full question & answer→Question 81 Mark
Given $A=\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right]$ Find $A B$
Answer$A B=\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right]\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right] $
$=\left[\begin{array}{ll}4-2 & 0+1 \\ 2-6 & 0+3\end{array}\right] $
$ =\left[\begin{array}{cc}2 & 1 \\ -4 & 3\end{array}\right]$
View full question & answer→Question 91 Mark
Given $A=\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right]$ Find $A^2$
Answer$A^2=\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right]\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right] $
$ =\left[\begin{array}{cc}16+2 & 4+3 \\ 8+6 & 2+9\end{array}\right] $
$=\left[\begin{array}{cc}18 & 7 \\ 14 & 11\end{array}\right]$
View full question & answer→Question 101 Mark
Given $A=\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right]$ Find A - B
Answer$A-B=\left[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\right]-\left[\begin{array}{cc}1 & 0 \\ -2 & 1\end{array}\right]=\left[\begin{array}{ll}3 & 1 \\ 4 & 2\end{array}\right]$
View full question & answer→Question 111 Mark
Evaluate $7\left[\begin{array}{cc}-1 & 2 \\ 0 & 1\end{array}\right]$
Answer$7\left[\begin{array}{cc}-1 & 2 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}-7 & 14 \\ 0 & 7\end{array}\right]$
View full question & answer→Question 121 Mark
View full question & answer→Question 131 Mark
Given $M=\left[\begin{array}{cc}5 & -3 \\ -2 & 4\end{array}\right]$$M^t$ find its transpose matrix $M^t$ if possible find $M^t-M$
Answer$M^t-M=\left[\begin{array}{cc}5 & -2 \\ -3 & 4\end{array}\right]-\left[\begin{array}{cc}5 & -3 \\ -2 & 4\end{array}\right]=\left[\begin{array}{cc}5-5 & -2+3 \\ -3+2 & 4-4\end{array}\right]=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]$
View full question & answer→Question 141 Mark
Wherever possible write of the following as a single matrix
$\left[\begin{array}{lll}0 & 1 & 2 \\ 4 & 6 & 7\end{array}\right]+\left[\begin{array}{ll}3 & 4 \\ 6 & 8\end{array}\right]$
AnswerAn addition is not possible because of both matrices are not of the same order
View full question & answer→Question 151 Mark
Wherever possible write of the following as a single matrix
$\left[\begin{array}{lll}2 & 3 & 4 \\ 5 & 6 & 7\end{array}\right]-\left[\begin{array}{ccc}0 & 2 & 3 \\ 6 & -1 & 0\end{array}\right]$
Answer$\left[\begin{array}{ccc}2 & 3 & 4 \\ 5 & 6 & 7\end{array}\right]-\left[\begin{array}{ccc}0 & 2 & 3 \\ 6 & -1 & 0\end{array}\right]=\left[\begin{array}{ccc}2-0 & 3-2 & 4-3 \\ 5-6 & 6+1 & 7-0\end{array}\right]=\left[\begin{array}{ccc}2 & 1 & 1 \\ -1 & 7 & 7\end{array}\right]$
View full question & answer→Question 161 Mark
Wherever possible write of the following as a single matrix
$\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]+\left[\begin{array}{cc}-1 & -2 \\ 1 & -7\end{array}\right]$
Answer$\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]+\left[\begin{array}{cc}-1 & -2 \\ 1 & -7\end{array}\right]=\left[\begin{array}{ll}1-1 & 2-2 \\ 3+1 & 4-7\end{array}\right]=\left[\begin{array}{cc}0 & 0 \\ 4 & -3\end{array}\right]$
View full question & answer→Question 171 Mark
If $A=\left[\frac{2}{5}\right], B=\left[\frac{1}{4}\right]$ and $c=\left[\frac{6}{-2}\right]$ Find $A-C$
Answer$A-C=\left[\begin{array}{l}2 \\ 5\end{array}\right]-\left[\begin{array}{c}6 \\ -2\end{array}\right]=\left[\begin{array}{l}2-6 \\ 5+2\end{array}\right]=\left[\begin{array}{c}-4 \\ 7\end{array}\right]$
View full question & answer→Question 181 Mark
If $A=\left[\frac{2}{5}\right], B=\left[\frac{1}{4}\right]$ and $C=\left[\frac{6}{-2}\right]$ Find $B+C$
Answer$B+C=\left[\begin{array}{l}1 \\ 4\end{array}\right]+\left[\begin{array}{c}6 \\ -2\end{array}\right]=\left[\begin{array}{l}1+6 \\ 4-2\end{array}\right]=\left[\begin{array}{l}7 \\ 2\end{array}\right]$
View full question & answer→Question 191 Mark
If A = [8 -3] and B = [4 -5]; find B - A
AnswerB - A = [4 -5] - [8 -3]
= [4 -8 -5 + 3]
= [-4 - 2]
View full question & answer→Question 201 Mark
If A = [8 -3] and B = [4 -5]; find A + B
AnswerA + B = [8 -3] + [4 -5] = [8 + 4 -3-5] = [12 -8]
View full question & answer→Question 211 Mark
Solve for a, b and c if $\left[\begin{array}{cc}a & a-b \\ b+c & 0\end{array}\right]=\left[\begin{array}{cc}3 & -1 \\ 2 & 0\end{array}\right]$
AnswerIf two matrices are equal, then their corresponding elements are also equal.
a= 3
a – b = -1
⇒ b = a + 1 = 4
b + c = 2
⇒ c = 2 – b = 2 – 4 = -2
View full question & answer→Question 221 Mark
Solve for a, b and c if $\left[\begin{array}{cc}-4 & a+5 \\ 3 & 2\end{array}\right]=\left[\begin{array}{cc}b+4 & 2 \\ 3 & c-1\end{array}\right]$
AnswerIf two matrices are equal, then their corresponding elements are also equal.
a + 5 = 2 ⇒ a = -3
-4 = b + 4 ⇒ b = -8
2 = c – 1 ⇒ c = 3
View full question & answer→Question 231 Mark
Given $\left[\begin{array}{ll}x & y+2 \\ 3 & z-1\end{array}\right]=\left[\begin{array}{ll}3 & 1 \\ 3 & 2\end{array}\right]$ , Find x, y, z
AnswerIf two matrices are equal, then their corresponding elements are also equal. Therefore, we have:
x = 3,
y + 2 = 1 ⇒ y = -1
z - 1 = 2 ⇒ z = 3
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