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# Solved Examples on Matrix Operations

You may verify your answers as applicable with: Matrices Calculators
For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

For JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

Solve all questions.
Show all work.
Use at least two methods to solve applicable questions.

(1.) AQA A-Level FM Three matrices $A$, $B$, and $C$ are given by

$A = \begin{bmatrix} 5 & 2 & -3 \\[3ex] 0 & 7 & 6 \\[3ex] 4 & 1 & 0 \end{bmatrix} \:\:\: B = \begin{bmatrix} 1 & 0 \\[3ex] 3 & -5 \\[3ex] -2 & 6 \end{bmatrix} \;\;\;and\;\;\; C = \begin{bmatrix} 6 & 4 & 3 \\[3ex] 1 & 2 & 0 \end{bmatrix} \\[3ex]$
Which of the following cannot be calculated?
$AB$   $AC$   $BC$   $A^2$

$\underline{AB} \\[3ex] Order\;\;of\;\;A = 3 * 3 \\[3ex] Order\;\;of\;\;B = 3 * 2 \\[3ex] Number\;\;of\;\;columns\;\;of\;\;A = Number\;\;of\;\;rows\;\;of\;\;B \\[3ex] AB \;\;is\;\; feasible \\[3ex] \underline{AC} \\[3ex] Order\;\;of\;\;A = 3 * 3 \\[3ex] Order\;\;of\;\;C = 2 * 3 \\[3ex] Number\;\;of\;\;columns\;\;of\;\;A \ne Number\;\;of\;\;rows\;\;of\;\;C \\[3ex] AC \;\;is\;\; \underline{not} \;\;feasible...\boldsymbol{cannot} \;\;be\;\;calculated \\[3ex] \underline{BC} \\[3ex] Order\;\;of\;\;B = 3 * 2 \\[3ex] Order\;\;of\;\;C = 2 * 3 \\[3ex] Number\;\;of\;\;columns\;\;of\;\;B = Number\;\;of\;\;rows\;\;of\;\;C \\[3ex] BC \;\;is\;\; feasible \\[3ex] \underline{A^2} \\[3ex] A^2 = A * A \\[3ex] Order\;\;of\;\;A = 3 * 3 \\[3ex] Order\;\;of\;\;A = 3 * 3 \\[3ex] Number\;\;of\;\;columns\;\;of\;\;A = Number\;\;of\;\;rows\;\;of\;\;A \\[3ex] A^2 \;\;is\;\; feasible$
(2.) Given the matrices:

$A = \begin{bmatrix} 1 & 3 \\[3ex] -2 & -1 \end{bmatrix} \:\:and\:\: D = \begin{bmatrix} 3 & -4 \\[3ex] 2 & -2 \end{bmatrix} \\[5ex]$ Show that $(A^T)^T = A$

$From \:\:LHS \\[3ex] A^T = \begin{bmatrix} 1 & -2 \\[3ex] 3 & -1 \end{bmatrix} \\[7ex] (A^T)^T = \begin{bmatrix} 1 & 3 \\[3ex] -2 & -1 \end{bmatrix} = A$
(3.) Given the matrices:

$A = \begin{bmatrix} 1 & 3 \\[3ex] -2 & -1 \end{bmatrix} \:\:and\:\: D = \begin{bmatrix} 3 & -4 \\[3ex] 2 & -2 \end{bmatrix} \\[5ex]$ Show that $(A + D)^T = A^T + D^T$

$From \:\:LHS \\[3ex] A + D = \begin{bmatrix} 1 & 3 \\[3ex] -2 & -1 \end{bmatrix} + \begin{bmatrix} 3 & -4 \\[3ex] 2 & -2 \end{bmatrix} \\[7ex] = \begin{bmatrix} 1 + 3 & 3 + (-4) \\[3ex] -2 + 2 & -1 +(-2) \end{bmatrix} = \begin{bmatrix} 4 & -1 \\[3ex] 0 & -3 \end{bmatrix} \\[7ex] (A + D)^T = \begin{bmatrix} 4 & 0 \\[3ex] -1 & -3 \end{bmatrix} \\[7ex] From \:\:RHS \\[3ex] A^T = \begin{bmatrix} 1 & -2 \\[3ex] 3 & -1 \end{bmatrix} \:\:\:\: D^T = \begin{bmatrix} 3 & 2 \\[3ex] -4 & -2 \end{bmatrix} \\[7ex] A^T + D^T = \begin{bmatrix} 1 + 3 & -2 + 2 \\[3ex] 3 + (-4) & -1 +(-2) \end{bmatrix} = \begin{bmatrix} 4 & 0 \\[3ex] -1 & -3 \end{bmatrix} \\[7ex] LHS = RHS \\[3ex] \therefore (A + D)^T = A^T + D^T$
(4.) Given the matrices:

$A = \begin{bmatrix} 1 & 3 \\[3ex] -2 & -1 \end{bmatrix} \:\:and\:\: D = \begin{bmatrix} 3 & -4 \\[3ex] 2 & -2 \end{bmatrix} \\[5ex]$ Show that $(AD)^T = D^T * A^T$

$From \:\:LHS \\[3ex] A * D = \begin{bmatrix} 1 & 3 \\[3ex] -2 & -1 \end{bmatrix} * \begin{bmatrix} 3 & -4 \\[3ex] 2 & -2 \end{bmatrix} \\[7ex] = \begin{bmatrix} 1 * 3 + 3 * 2 & 1 * -4 + 3 * -2 \\[3ex] -2 * 3 + -1 * 2 & -2 * -4 + -1 * -2 \end{bmatrix} \\[7ex] = \begin{bmatrix} 3 + 6 & -4 + -6 \\[3ex] -6 + -2 & 8 + 2 \end{bmatrix} = \begin{bmatrix} 9 & -10 \\[3ex] -8 & 10 \end{bmatrix} \\[7ex] (AD)^T = \begin{bmatrix} 9 & -8 \\[3ex] -10 & 10 \end{bmatrix} \\[7ex] From \:\:RHS \\[3ex] D^T = \begin{bmatrix} 3 & 2 \\[3ex] -4 & -2 \end{bmatrix} \:\:\:\: A^T = \begin{bmatrix} 1 & -2 \\[3ex] 3 & -1 \end{bmatrix} \\[7ex] D^T * A^T = \begin{bmatrix} 3 * 1 + 2 * 3 & 3 * -2 + 2 * -1 \\[3ex] -4 * 1 + -2 * 3 & -4 * -2 + -2 * -1 \end{bmatrix} \\[7ex] = \begin{bmatrix} 3 + 6 & -6 + -2 \\[3ex] -4 + -6 & 8 + 2 \end{bmatrix} = \begin{bmatrix} 9 & -8 \\[3ex] -10 & 10 \end{bmatrix} \\[7ex] LHS = RHS \\[3ex] \therefore (AD)^T = D^T * A^T$
(5.) JAMB If $P = \begin{bmatrix} 2 & 1 \\[3ex] -3 & 0 \end{bmatrix}$ and I is a $2 * 2$ unit matrix, evaluate $P^2 - 2P + 4I$

$P = \begin{bmatrix} 2 & 1 \\[3ex] -3 & 0 \end{bmatrix} \\[7ex] P^2 = \begin{bmatrix} 2 & 1 \\[3ex] -3 & 0 \end{bmatrix} * \begin{bmatrix} 2 & 1 \\[3ex] -3 & 0 \end{bmatrix} \\[7ex] = \begin{bmatrix} 2(2) + 1(-3) & 2(1) + 1(0) \\[3ex] -3(2) + 0(-3) & -3(1) + 0(0) \end{bmatrix} \\[7ex] = \begin{bmatrix} 4 + (-3) & 2 + 0 \\[3ex] -6 + 0 & -3 + 0 \end{bmatrix} \\[7ex] = \begin{bmatrix} 4 - 3 & 2 \\[3ex] -6 & -3 \end{bmatrix} \\[7ex] P^2 = \begin{bmatrix} 1 & 2 \\[3ex] -6 & -3 \end{bmatrix} \\[5ex] 2P = \begin{bmatrix} 2(2) & 2(1) \\[3ex] 2(-3) & 2(0) \end{bmatrix} \\[7ex] 2P = \begin{bmatrix} 4 & 2 \\[3ex] -6 & 0 \end{bmatrix} \\[7ex] I = \begin{bmatrix} 1 & 0 \\[3ex] 0 & 1 \end{bmatrix} \\[7ex] 4I = \begin{bmatrix} 4(1) & 4(0) \\[3ex] 4(0) & 4(1) \end{bmatrix} \\[7ex] 4I = \begin{bmatrix} 4 & 0 \\[3ex] 0 & 4 \end{bmatrix} \\[7ex] P^2 - 2P + 4I = \begin{bmatrix} 1 & 2 \\[3ex] -6 & -3 \end{bmatrix} - \begin{bmatrix} 4 & 2 \\[3ex] -6 & 0 \end{bmatrix} + \begin{bmatrix} 4 & 0 \\[3ex] 0 & 4 \end{bmatrix} \\[7ex] = \begin{bmatrix} 1 - 4 + 4 & 2 - 2 + 0 \\[3ex] -6 - (-6)+ 0 & -3 - 0 + 4 \end{bmatrix} \\[7ex] = \begin{bmatrix} 1 & 0 \\[3ex] -6 + 6 + 0 & 1 \end{bmatrix} \\[7ex] = \begin{bmatrix} 1 & 0 \\[3ex] 0 & 1 \end{bmatrix}$
(6.) ACT

$If\:\: A = \begin{bmatrix} 2 & -4 \\[3ex] 6 & 0 \end{bmatrix} \:\:and\:\: B = \begin{bmatrix} -2 & 4 \\[3ex] -6 & 0 \end{bmatrix} \\[5ex]$ then $A - B$ = ?

$A - B \\[3ex] = \begin{bmatrix} 2 & -4 \\[3ex] 6 & 0 \end{bmatrix} - \begin{bmatrix} -2 & 4 \\[3ex] -6 & 0 \end{bmatrix} \\[7ex] = \begin{bmatrix} 2 -(-2) & -4 - 4 \\[3ex] 6 -(-6) & 0 - 0 \end{bmatrix} \\[7ex] = \begin{bmatrix} 2 + 2 & -8 \\[3ex] 6 + 6 & 0 \end{bmatrix} \\[7ex] = \begin{bmatrix} 4 & -8 \\[3ex] 12 & 0 \end{bmatrix}$
(7.) JAMB Given the matrix:

$K = \begin{bmatrix} 2 & 1 \\[3ex] 3 & 4 \end{bmatrix}$, the matrix $K^2 + K + I$ where I is the $2 * 2$ identity matrix, is

$K = \begin{bmatrix} 2 & 1 \\[3ex] 3 & 4 \end{bmatrix} \\[7ex] K^2 = \begin{bmatrix} 2 & 1 \\[3ex] 3 & 4 \end{bmatrix} * \begin{bmatrix} 2 & 1 \\[3ex] 3 & 4 \end{bmatrix} \\[7ex] = \begin{bmatrix} 2(2) + 1(3) & 2(1) + 1(4) \\[3ex] 3(2) + 4(3) & 3(1) + 4(4) \end{bmatrix} \\[7ex] = \begin{bmatrix} 4 + 3 & 2 + 4 \\[3ex] 6 + 12 & 3 + 16 \end{bmatrix} \\[7ex] = \begin{bmatrix} 7 & 6 \\[3ex] 18 & 19 \end{bmatrix} \\[7ex] K^2 + K + I = \begin{bmatrix} 7 & 6 \\[3ex] 18 & 19 \end{bmatrix} + \begin{bmatrix} 2 & 1 \\[3ex] 3 & 4 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\[3ex] 0 & 1 \end{bmatrix} \\[7ex] = \begin{bmatrix} 7 + 2 + 1 & 6 + 1 + 0 \\[3ex] 18 + 3 + 0 & 19 + 4 + 1 \end{bmatrix} \\[7ex] = \begin{bmatrix} 10 & 7 \\[3ex] 21 & 24 \end{bmatrix}$
(8.) WASSCE:FM Given that

$M = \begin{bmatrix} 0 & 2 \\[3ex] 2 & 1 \end{bmatrix} \:\:and\:\: N = \begin{bmatrix} 0 & 2 \\[3ex] 2 & -1 \end{bmatrix} \\[5ex]$ find $(2M - N)$

$A.\:\: \begin{bmatrix} 0 & 2 \\[3ex] 2 & -3 \end{bmatrix} \\[5ex] B.\:\: \begin{bmatrix} 0 & 2 \\[3ex] 2 & -1 \end{bmatrix} \\[5ex] C.\:\: \begin{bmatrix} 0 & 2 \\[3ex] 2 & 1 \end{bmatrix} \\[5ex] D.\:\: \begin{bmatrix} 0 & 2 \\[3ex] 2 & 3 \end{bmatrix} \\[5ex]$

$2M = 2\begin{bmatrix} 0 & 2 \\[3ex] 2 & 1 \end{bmatrix} = \begin{bmatrix} 2(0) & 2(2) \\[3ex] 2(2) & 2(1) \end{bmatrix} = \begin{bmatrix} 0 & 4 \\[3ex] 4 & 2 \end{bmatrix} \\[7ex] 2M - N = \begin{bmatrix} 0 - 0 & 4 - 2 \\[3ex] 4 - 2 & 2 - (-1) \end{bmatrix} \\[7ex] = \begin{bmatrix} 0 & 2 \\[3ex] 2 & 2 + 1 \end{bmatrix} \\[7ex] = \begin{bmatrix} 0 & 2 \\[3ex] 2 & 3 \end{bmatrix}$
(9.) ACT Four matrices are given below.

$W = \begin{bmatrix} 1 & 2 \\[3ex] 5 & 8 \end{bmatrix} \:\: X = \begin{bmatrix} 3 & 9 \\[3ex] 7 & 4 \end{bmatrix} \:\: Y = \begin{bmatrix} 1 & 3 & 7 \\[3ex] 4 & 2 & 6 \end{bmatrix} \:\: Z = \begin{bmatrix} 5 & 8 \\[3ex] 2 & 9 \\[3ex] 3 & 7 \end{bmatrix} \\[5ex]$ Which of the following matrix products is undefined?

$F.\:\: WX \\[3ex] G.\:\: WY \\[3ex] H.\:\: YZ \\[3ex] J.\:\: XW \\[3ex] K.\:\: XZ \\[3ex]$

Two matrices say $A$ and $B$ is conformable for multiplication (which means that $AB$ exists) if the number of columns of $A$ is equal to the number of rows of $B$

Two matrices say $B$ and $A$ is conformable for multiplication (which means that $BA$ exists) if the number of columns of $B$ is equal to the number of rows of $A$

$Order\:\:of\:\:a\:\:Matrix = row\:\:by\:\:column \\[3ex] Order\:\:of\:\:W = 2\:by\:2 \rightarrow row = 2, \:\: column = 2 \\[3ex] Order\:\:of\:\:X = 2\:by\:2 \rightarrow row = 2, \:\: column = 2 \\[3ex] \underline{For\:\:WX} \\[3ex] 2 = 2 \\[3ex] WX\:\:exists \\[3ex] Order\:\:of\:\:Y = 2\:by\:3 \rightarrow row = 2, \:\: column = 3 \\[3ex] \underline{For\:\:WY} \\[3ex] 2 = 2 \\[3ex] WY\:\:exists \\[3ex] Order\:\:of\:\:Z = 3\:by\:2 \rightarrow row = 3, \:\: column = 2 \\[3ex] \underline{For\:\:YZ} \\[3ex] 3 = 3 \\[3ex] YZ\:\:exists \\[3ex] \underline{For\:\:XW} \\[3ex] 2 = 3 \\[3ex] XW\:\:exists \\[3ex] \underline{For\:\:XZ} \\[3ex] 2 \ne 3 \\[3ex] XZ\:\:DNE$
(10.) CSEC Three matrices are given as follows:

$P = \begin{bmatrix} -1 & 2 \\[3ex] 0 & 5 \end{bmatrix} \:\: Q = \begin{bmatrix} a \\[3ex] b \end{bmatrix} \:\:and\:\: R = \begin{bmatrix} 11 \\[3ex] 15 \end{bmatrix} \\[5ex]$ (i) Using a calculation to support your answer, explain whether matrix $P$ is a singular or a non-singular matrix.

(ii) Given that $PQ = R$, determine the values of $a$ and $b$.

(iii) State the reason why the matrix product $QP$ is not possible.

A singular matrix is a matrix whose determinant is zero.
A non-singular matrix is a matrix whose determinant is non-zero.

$(i)\:\: \underline{Matrix\:P} \\[3ex] |P| = \begin{vmatrix} -1 & 2 \\[3ex] 0 & 5 \end{vmatrix} = -1(5) - 0(2) = -5 - 0 = -5 \\[5ex] -5 \ne 0 \\[3ex] Matrix\:P\:\:is\:\:a\:\:non-singular\:\:matrix \\[3ex] (ii)\:\: PQ = R \\[3ex] \underline{For PQ} \\[3ex] Order:\: (2\:by\:2) * (2\:by\:1) \rightarrow (2\:by\:1) \\[3ex] PQ = \begin{bmatrix} -1 & 2 \\[3ex] 0 & 5 \end{bmatrix} \begin{bmatrix} a \\[3ex] b \end{bmatrix} = \begin{bmatrix} -1(a) + 2(b) \\[3ex] 0(a) + 5(b) \end{bmatrix} = \begin{bmatrix} -a + 2b \\[3ex] 0 + 5b \end{bmatrix} \\[7ex] \therefore \begin{bmatrix} -a + 2b \\[3ex] 0 + 5b \end{bmatrix} = R \\[7ex] \begin{bmatrix} -a + 2b \\[3ex] 5b \end{bmatrix} = \begin{bmatrix} 11 \\[3ex] 15 \end{bmatrix} \\[7ex] \rightarrow 5b = 15 \\[3ex] b = \dfrac{15}{5} \\[5ex] b = 3 \\[3ex] \rightarrow -a + 2b = 11 \\[3ex] -a + 2(3) = 11 \\[3ex] -a + 6 = 11 \\[3ex] -a = 11 - 6 \\[3ex] -a = 5 \\[3ex] a = \dfrac{5}{-1} \\[5ex] a = -5 \\[3ex]$ Two matrices say $A$ and $B$ is conformable for multiplication (which means that $AB$ exists) if the number of columns of $A$ is equal to the number of rows of $B$

Two matrices say $B$ and $A$ is conformable for multiplication (which means that $BA$ exists) if the number of columns of $B$ is equal to the number of rows of $A$

$Order\:\:of\:\:a\:\:Matrix = row\:\:by\:\:column \\[3ex] Order\:\:of\:\:Q = 2\:by\:1 \rightarrow row = 2, \:\: column = 1 \\[3ex] Order\:\:of\:\:P = 2\:by\:2 \rightarrow row = 2, \:\: column = 2 \\[3ex] \underline{For\:\:WX} \\[3ex] 1 \ne 2 \\[3ex] QP\:\:DNE \\[3ex]$
(11.)

(12.) WASSCE

$If\:\: \begin{bmatrix} 2 & 4 \\[3ex] 3 & z \end{bmatrix} + \begin{bmatrix} x & y \\[3ex] 3 & 4 \end{bmatrix} = \begin{bmatrix} 4 & -4 \\[3ex] w & 0 \end{bmatrix} \\[5ex]$ find $(w, x, y, z)$

$A.\:\: (6, -2, -8, -4) \\[3ex] B.\:\: (6, 2, -8, -4) \\[3ex] C.\:\: (6, 2, -8, 4) \\[3ex] D.\:\: (6, 4, -4, 0) \\[3ex]$

$\begin{bmatrix} 2 & 4 \\[3ex] 3 & z \end{bmatrix} + \begin{bmatrix} x & y \\[3ex] 3 & 4 \end{bmatrix} = \begin{bmatrix} 4 & -4 \\[3ex] w & 0 \end{bmatrix} \\[10ex] \begin{bmatrix} 2 + x & 4 + y \\[3ex] 3 + 3 & z + 4 \end{bmatrix} = \begin{bmatrix} 4 & -4 \\[3ex] w & 0 \end{bmatrix} \\[10ex] \rightarrow 2 + x = 4 \\[3ex] x = 4 - 2 \\[3ex] x = 2 \\[3ex] \rightarrow 4 + y = -4 \\[3ex] y = -4 - 4 \\[3ex] y = -8 \\[3ex] \rightarrow 3 + 3 = w \\[3ex] 6 = w \\[3ex] w = 6 \\[3ex] z + 4 = 0 \\[3ex] z = 0 - 4 \\[3ex] z = -4 \\[3ex] (w, x, y, z) = (6, 2, -8, -4)$
(13.) ACT Given $A = \begin{bmatrix} 2 & 0 & 3 \\[3ex] -1 & 5 & -2 \end{bmatrix}, \;\;\; B = \begin{bmatrix} -3 & 1 \\[3ex] 4 & 1 \\[3ex] 1 & 2 \end{bmatrix}, \;\;\;and\;\;\; C = \begin{bmatrix} 0 & -2 \\[3ex] 1 & -4 \end{bmatrix}, \\[3ex]$ if it is possible to calculate $C + AB$, which of the following matrices is the result?

$A.\:\: \begin{bmatrix} -6 & -3 \\[3ex] 1 & 1 \end{bmatrix} \\[5ex] B.\:\: \begin{bmatrix} -3 & 6 \\[3ex] 22 & -4 \end{bmatrix} \\[5ex] C.\:\: \begin{bmatrix} 11 & 6 \\[3ex] 2 & -3 \end{bmatrix} \\[5ex] D.\:\: \begin{bmatrix} -6 & 0 \\[3ex] 8 & 1 \\[3ex] 3 & -4 \end{bmatrix} \\[3ex]$ $E.$ It is not possible to calculate $C + AB$

$A = \begin{bmatrix} 2 & 0 & 3 \\[3ex] -1 & 5 & -2 \end{bmatrix} \;\;\; B = \begin{bmatrix} -3 & 1 \\[3ex] 4 & 1 \\[3ex] 1 & 2 \end{bmatrix} \\[10ex] Order\;\;of\;\;A = 2 * 3 \\[3ex] Order\;\;of\;\;B = 3 * 2 \\[3ex] Number\;\;of\;\;columns\;\;of\;\;A = Number\;\;of\;\;rows\;\;of\;\;B \\[3ex] AB\;\;is\;\;feasible \\[3ex] Order\;\;of\;\;AB = Number\;\;of\;\;rows\;\;of\;\;A * Number\;\;of\;\;columns\;\;of\;\;B \\[3ex] Order\;\;of\;\;AB = 2 * 2 \\[3ex] AB \\[3ex] = \begin{bmatrix} 2 & 0 & 3 \\[3ex] -1 & 5 & -2 \end{bmatrix} \begin{bmatrix} -3 & 1 \\[3ex] 4 & 1 \\[3ex] 1 & 2 \end{bmatrix} \\[10ex] = \begin{bmatrix} 2(-3) + 0(4) + 3(1) & 2(1) + 0(1) + 3(2) \\[3ex] -1(-3) + 5(4) + -2(1) & -1(1) + 5(1) + -2(2) \end{bmatrix} \\[10ex] = \begin{bmatrix} -6 + 0 + 3 & 2 + 0 + 6 \\[3ex] 3 + 20 - 2 & -1 + 5 - 4 \end{bmatrix} \\[10ex] = \begin{bmatrix} -3 & 8 \\[3ex] 21 & 0 \end{bmatrix} \\[10ex] C + AB \\[3ex] = \begin{bmatrix} 0 & -2 \\[3ex] 1 & -4 \end{bmatrix} + \begin{bmatrix} -3 & 8 \\[3ex] 21 & 0 \end{bmatrix} \\[10ex] = \begin{bmatrix} 0 + -3 & -2 + 8 \\[3ex] 1 + 21 & -4 + 0 \end{bmatrix} \\[10ex] = \begin{bmatrix} -3 & 6 \\[3ex] 22 & -4 \end{bmatrix}$
(14.) ACT The $2 * 2$ matrices $A$ and $B$ below are related to matrix $C$ by the equation $C = 2A - 3B$
What is matrix $C$?

$A = \begin{bmatrix} 3 & 5 \\[3ex] -2 & 1 \end{bmatrix} \;\;\; B = \begin{bmatrix} -4 & 5 \\[3ex] 2 & 1 \end{bmatrix} \\[3ex]$
$F.\:\: \begin{bmatrix} 18 & -5 \\[3ex] -10 & -1 \end{bmatrix} \\[5ex] G.\:\: \begin{bmatrix} 13 & -10 \\[3ex] -8 & -2 \end{bmatrix} \\[5ex] H.\:\: \begin{bmatrix} 10 & 5 \\[3ex] -6 & 1 \end{bmatrix} \\[5ex] J.\:\: \begin{bmatrix} 6 & -1 \\[3ex] -5 & -1 \end{bmatrix} \\[5ex] K.\:\: \begin{bmatrix} -6 & 25 \\[3ex] 2 & 5 \end{bmatrix} \\[5ex]$

$2A = 2\begin{bmatrix} 3 & 5 \\[3ex] -2 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 10 \\[3ex] -4 & 2 \end{bmatrix} \\[3ex]$
$3B = 3\begin{bmatrix} -4 & 5 \\[3ex] 2 & 1 \end{bmatrix} = \begin{bmatrix} -12 & 15 \\[3ex] 6 & 3 \end{bmatrix} \\[3ex]$
$2A - 3B \\[3ex] = \begin{bmatrix} 6 & 10 \\[3ex] -4 & 2 \end{bmatrix} - \begin{bmatrix} -12 & 15 \\[3ex] 6 & 3 \end{bmatrix} \\[5ex] = \begin{bmatrix} 6 -(-12) & 10 - 15 \\[3ex] -4 - 6 & 2 - 3 \end{bmatrix} \\[5ex] = \begin{bmatrix} 6 + 12 & -5 \\[3ex] -10 & -1 \end{bmatrix} \\[5ex] = \begin{bmatrix} 18 & -5 \\[3ex] -10 & -1 \end{bmatrix}$
(15.)

(16.) ACT What is the matrix product

$\begin{bmatrix} 2 & 4 \\[3ex] 6 & 5 \end{bmatrix} \begin{bmatrix} a & b \\[3ex] c & d \end{bmatrix} \\[3ex]$
$F.\:\: \begin{bmatrix} 2a & 4b \\[3ex] 6c & 5d \end{bmatrix} \\[5ex] G.\:\: \begin{bmatrix} (2a + 4b) \\[3ex] (6c + 5d) \end{bmatrix} \\[5ex] H.\:\: \begin{bmatrix} (2a + 6c) & (4b + 5d) \end{bmatrix} \\[5ex] J.\:\: \begin{bmatrix} (2a + 6b) & (4a + 5b) \\[3ex] (2c + 6d) & (4c + 5d) \end{bmatrix} \\[5ex] K.\:\: \begin{bmatrix} (2a + 4c) & (2b + 4d) \\[3ex] (6a + 5c) & (6b + 5d) \end{bmatrix} \\[5ex]$

$Order:\;\; (2 * 2) * (2 * 2) \rightarrow (2 * 2) \\[3ex] Eliminate\;\;Options\;\;F,\;\;G,\;\;H \\[3ex] \begin{bmatrix} 2 & 4 \\[3ex] 6 & 5 \end{bmatrix} \begin{bmatrix} a & b \\[3ex] c & d \end{bmatrix} \\[3ex]$
$= \begin{bmatrix} (2a + 4c) & (2b + 4d) \\[3ex] (6a + 5c) & (6b + 5d) \end{bmatrix} \\[3ex]$
Option $K$
(17.)

(18.) ACT Matrices $A$ and $B$ are given below.

$A = \begin{bmatrix} 3 & -5 \\[3ex] -2 & 9 \end{bmatrix} \:\:\; B = \begin{bmatrix} -7 & 6 \\[3ex] 4 & 5 \end{bmatrix} \\[3ex]$
Which of the following matrices is $A - B$?

$F.\:\: \begin{bmatrix} -10 & 11 \\[3ex] 6 & -4 \end{bmatrix} \\[5ex] G.\:\: \begin{bmatrix} -4 & 2 \\[3ex] 1 & 14 \end{bmatrix} \\[5ex] H.\:\: \begin{bmatrix} -4 & 1 \\[3ex] 2 & 14 \end{bmatrix} \\[5ex] J.\:\: \begin{bmatrix} 10 & -6 \\[3ex] -11 & 4 \end{bmatrix} \\[5ex] K.\:\: \begin{bmatrix} 10 & -11 \\[3ex] -6 & 4 \end{bmatrix} \\[5ex]$

$A - B \\[3ex] = \begin{bmatrix} 3 & -5 \\[3ex] -2 & 9 \end{bmatrix} - \begin{bmatrix} -7 & 6 \\[3ex] 4 & 5 \end{bmatrix} \\[7ex] = \begin{bmatrix} 3 -(-7) & -5 - 6 \\[3ex] -2 - 4 & 9 - 5 \end{bmatrix} \\[7ex] = \begin{bmatrix} 3 + 7 & -11 \\[3ex] -6 & 4 \end{bmatrix} \\[7ex] = \begin{bmatrix} 10 & -11 \\[3ex] -6 & 4 \end{bmatrix}$
(19.)

(20.) ACT Which of the following matrices is equal to

$\begin{bmatrix} 5 & 7 \\[3ex] -4 & 4 \end{bmatrix} + \begin{bmatrix} -6 & 3 \\[3ex] 6 & 8 \end{bmatrix} ? \\[7ex] F.\:\: \begin{bmatrix} -1 & 10 \\[3ex] 2 & 12 \end{bmatrix} \\[5ex] G.\:\: \begin{bmatrix} -1 & 10 \\[3ex] 10 & 12 \end{bmatrix} \\[5ex] H.\:\: \begin{bmatrix} 11 & 10 \\[3ex] 10 & 12 \end{bmatrix} \\[5ex] J.\:\: \begin{bmatrix} 12 & -3 \\[3ex] 0 & 14 \end{bmatrix} \\[5ex] K.\:\: \begin{bmatrix} 12 & 71 \\[3ex] 48 & 20 \end{bmatrix} \\[5ex]$

$\begin{bmatrix} 5 & 7 \\[3ex] -4 & 4 \end{bmatrix} + \begin{bmatrix} -6 & 3 \\[3ex] 6 & 8 \end{bmatrix} \\[7ex] = \begin{bmatrix} 5 + (-6) & 7 + 3 \\[3ex] -4 + 6 & 4 + 8 \end{bmatrix} \\[7ex] = \begin{bmatrix} 5 - 6 & 10 \\[3ex] 2 & 12 \end{bmatrix} \\[7ex] = \begin{bmatrix} -1 & 10 \\[3ex] 2 & 12 \end{bmatrix}$

(21.)

(22.) ACT Which of the following matrices is equal to

$\begin{bmatrix} 9 & 8 \\[3ex] -4 & 7 \end{bmatrix} + \begin{bmatrix} -6 & 6 \\[3ex] 5 & 4 \end{bmatrix} ? \\[7ex] A.\:\: \begin{bmatrix} 3 & 14 \\[3ex] 1 & 11 \end{bmatrix} \\[5ex] B.\:\: \begin{bmatrix} 3 & 14 \\[3ex] 9 & 11 \end{bmatrix} \\[5ex] C.\:\: \begin{bmatrix} 15 & 14 \\[3ex] 9 & 11 \end{bmatrix} \\[5ex] D.\:\: \begin{bmatrix} 17 & 0 \\[3ex] 3 & 9 \end{bmatrix} \\[5ex] E.\:\: \begin{bmatrix} -14 & 86 \\[3ex] 59 & 4 \end{bmatrix} \\[5ex]$

$\begin{bmatrix} 9 & 8 \\[3ex] -4 & 7 \end{bmatrix} + \begin{bmatrix} -6 & 6 \\[3ex] 5 & 4 \end{bmatrix} \\[7ex] = \begin{bmatrix} 9 + (-6) & 8 + 6 \\[3ex] -4 + 5 & 7 + 4 \end{bmatrix} \\[7ex] = \begin{bmatrix} 9 - 6 & 14 \\[3ex] 1 & 11 \end{bmatrix} \\[7ex] = \begin{bmatrix} 3 & 14 \\[3ex] 1 & 11 \end{bmatrix}$