Solved Examples on Matrix Operations



Samuel Dominic Chukwuemeka (SamDom For Peace) You may verify your answers as applicable with: Matrices Calculators

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For JAMB and CMAT Students
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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?
Circle your answer.
  $AB$   $AC$   $BC$   $A^2$


$ \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} $




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(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} $