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It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it.
- Samuel Dominic Chukwuemeka

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Dominic Chukwuemeka



Solved Examples on Matrices (All Topics)

Samuel Dominic Chukwuemeka (SamDom For Peace) 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.
Use at least two methods for each question as applicable.
Show all work.

(1.) CSEC $A,\:B\:and\:C$ are three $2$ x $2$ matrices such that

$ A = \begin{bmatrix} a & b \\[3ex] c & d \end{bmatrix} \:\: B = \begin{bmatrix} 5 & 3 \\[3ex] 3 & 2 \end{bmatrix} \:\:and\:\: C = \begin{bmatrix} 14 & 0 \\[3ex] -9 & 5 \end{bmatrix} \\[5ex] $ Find
(i) $3A$

(ii) $B^{-1}$

(iii) $3A + B^{-1}$

(iv) the value of $a$, $b$, $c$, and $d$ given that $3A + B^{-1} = C$


$ (i)\:\: 3A = 3\begin{bmatrix} a & b \\[3ex] c & d \end{bmatrix} = \begin{bmatrix} 3(a) & 3(b) \\[3ex] 3(c) & 3(d) \end{bmatrix} = \begin{bmatrix} 3a & 3b \\[3ex] 3c & 3d \end{bmatrix} \\[7ex] (ii)\:\: B^{-1} \\[3ex] $ Let us find the inverse in two ways.
Use any method you like.

$ \underline{First\:\:Method} \\[3ex] For\:\: A = \begin{bmatrix} a & b \\[3ex] c & d \end{bmatrix} A^{-1} = \dfrac{ \begin{bmatrix} d & -b \\[3ex] -c & a \end{bmatrix}}{ad - cb} \\[7ex] Compare\:\:to\:\:Matrix\:B \\[3ex] a = 5 \\[3ex] b = 3 \\[3ex] c = 3 \\[3ex] d = 2 \\[3ex] -b = -3 \\[3ex] -c = -3 \\[3ex] ad - bc = 5(2) - 3(3) = 10 - 9 = 1 \\[3ex] B^{-1} = \dfrac{ \begin{bmatrix} 2 & -3 \\[3ex] -3 & 5 \end{bmatrix}}{1} = \begin{bmatrix} \dfrac{2}{1} & -\dfrac{3}{1} \\[5ex] -\dfrac{3}{1} & \dfrac{5}{1} \end{bmatrix} = \begin{bmatrix} 2 & -3 \\[3ex] -3 & 5 \end{bmatrix} \\[10ex] \underline{Second\:\:Method:\;\; Row\:\:Reduction\:\:Method} \\[3ex] B \ \ | \ \ I = I \ \ | \ \ B^{-1} \\[3ex] \left[ \begin{array}{cc|cc} 5 & 3 & 1 & 0 \\[3ex] 3 & 2 & 0 & 1 \end{array} \right] \underrightarrow{-3R_1 + 5R_2} \left[ \begin{array}{cc|cc} 5 & 3 & 1 & 0 \\[3ex] 0 & 1 & -3 & 5 \end{array} \right] \underrightarrow{-3R_2 + R_1} \left[ \begin{array}{cc|cc} 5 & 0 & 10 & -15 \\[3ex] 0 & 1 & -3 & 5 \end{array} \right] \begin{matrix} \underrightarrow{R_1 \div 5}\end{matrix} \left[ \begin{array}{cc|cc} 1 & 0 & 2 & -3 \\[3ex] 0 & 1 & -3 & 5 \end{array} \right] \\[10ex] B^{-1} = \begin{bmatrix} 2 & -3 \\[3ex] -3 & 5 \end{bmatrix} \\[10ex] (iii)\:\: 3A + B^{-1} = \begin{bmatrix} 3a & 3b \\[3ex] 3c & 3d \end{bmatrix} + \begin{bmatrix} 2 & -3 \\[3ex] -3 & 5 \end{bmatrix} \\[10ex] 3A + B^{-1} = \begin{bmatrix} 3a + 2 & 3b - 3 \\[3ex] 3c - 3 & 3d + 5 \end{bmatrix} \\[10ex] (iv)\:\: 3A + B^{-1} = C \\[3ex] \begin{bmatrix} 3a + 2 & 3b - 3 \\[3ex] 3c - 3 & 3d + 5 \end{bmatrix} = \begin{bmatrix} 14 & 0 \\[3ex] -9 & 5 \end{bmatrix} \\[10ex] \rightarrow 3a + 2 = 14 \\[3ex] 3a = 14 - 2 \\[3ex] 3a = 12 \\[3ex] a = \dfrac{12}{3} \\[5ex] a = 4 \\[3ex] \rightarrow 3b - 3 = 0 \\[3ex] 3b = 0 + 3 \\[3ex] 3b = 3 \\[3ex] b = \dfrac{3}{3} \\[5ex] b = 1 \\[3ex] \rightarrow 3c - 3 = -9 \\[3ex] 3c = -9 + 3 \\[3ex] 3c = -6 \\[3ex] c = -\dfrac{6}{3} \\[5ex] c = -2 \\[3ex] \rightarrow 3d + 5 = 5 \\[3ex] 3d = 5 - 5 \\[3ex] 3d = 0 \\[3ex] d = \dfrac{0}{3} \\[5ex] d = 0 $
(2.)


$ \begin{vmatrix} 7 & \sqrt{7} \\ \sqrt{7} & 7 \end{vmatrix} \\[3ex] = (7)(7) - (\sqrt{7})(\sqrt{7}) \\[2ex] = 49 - 7 \\[2ex] = 42 $
(3.) CSEC
(a) (i) a)
Find the matrix product

$ \begin{bmatrix} -1 & 3 \\[2ex] 4 & h \end{bmatrix} \begin{bmatrix} k \\[2ex] 5 \end{bmatrix} $

b) Hence, find the values of $h$ and $k$ that satisfy the matrix equation.

$ \begin{bmatrix} -1 & 3 \\[2ex] 4 & h \end{bmatrix} \begin{bmatrix} k \\[2ex] 5 \end{bmatrix} = \begin{bmatrix} 0 \\[2ex] 0 \end{bmatrix} $

(ii) Using a matrix method, solve the simultaneous equations.

$ 2x + 3y = 5 \\[3ex] -5x + y = 13 \\[3ex] $

(i) a)
$ Order:\: (2\:by\:2) * (2\:by\:1) \rightarrow (2\:by\:1) \\[3ex] \begin{bmatrix} -1 & 3 \\[3ex] 4 & h \end{bmatrix} \begin{bmatrix} k \\[3ex] 5 \end{bmatrix} \\[7ex] = \begin{bmatrix} -1(k) + 3(5) \\[3ex] 4(k) + h(5) \end{bmatrix} = \begin{bmatrix} -k + 15 \\[3ex] 4k + 5h \end{bmatrix} \\[7ex] b)\:\: \begin{bmatrix} -1 & 3 \\[2ex] 4 & h \end{bmatrix} \begin{bmatrix} k \\[2ex] 5 \end{bmatrix} = \begin{bmatrix} 0 \\[2ex] 0 \end{bmatrix} \rightarrow \begin{bmatrix} -k + 15 \\[3ex] 4k + 5h \end{bmatrix} = \begin{bmatrix} 0 \\[2ex] 0 \end{bmatrix} \\[7ex] \rightarrow -k + 15 = 0 \\[3ex] -k = -15 \\[3ex] k = \dfrac{-15}{-1} \\[5ex] k = 15 \\[3ex] \rightarrow 4k + 5h = 0 \\[3ex] 4(15) + 5h = 0 \\[3ex] 60 + 5h = 0 \\[3ex] 5h = 0 - 60 \\[3ex] 5h = - 60 \\[3ex] h = -\dfrac{60}{5} \\[5ex] h = -12 \\[3ex] (ii)\:\: \\[3ex] 2x + 3y = 5 ...eqn.(1) \\[3ex] -5x + y = 13 ...eqn.(2) \\[3ex] $ Prerequisite Topic: Cramer's Rule

$ \underline{Cramer's Rule} \\[3ex] \begin{bmatrix} 2 & 3 \\[3ex] -5 & 1 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \begin{bmatrix} 5 \\[3ex] 13 \end{bmatrix} \\[7ex] x = \dfrac{\begin{vmatrix} 5 & 3 \\[3ex] 13 & 1 \end{vmatrix}} {\begin{vmatrix} 2 & 3 \\[3ex] -5 & 1 \end{vmatrix}} = \dfrac{5(1) - 13(3)}{2(1) - (-5)(3)} = \dfrac{5 - 39}{2 - -15} = \dfrac{26 + 25}{2 + 15} = -\dfrac{34}{17} = -2 \\[10ex] y = \dfrac{\begin{vmatrix} 2 & 5 \\[3ex] -5 & 13 \end{vmatrix}} {\begin{vmatrix} 2 & 3 \\[3ex] -5 & 1 \end{vmatrix}} = \dfrac{2(13) - (-5)(5)}{2(1) - (-5)(3)} = \dfrac{26 - -25}{2 - -15} = \dfrac{26 + 25}{2 + 15} = \dfrac{51}{17} = 3 \\[10ex] $ Check
$x = -2,\:y = 3$
$LHS$ $RHS$
$ 2x + 3y \\[3ex] 2(-2) + 3(3) \\[3ex] -4 + 9 \\[3ex] 5 $ $5$
$ -5x + y \\[3ex] -5(-2) + 3 \\[3ex] 10 + 3 \\[3ex] 13 $ $13$
(4.) $ \begin{vmatrix} p & c \\ -d & -e \end{vmatrix} $


$ \begin{vmatrix} p & c \\ -d & -e \end{vmatrix} \\[3ex] = (p)(-e) - (-d)(c) \\[2ex] = -ep - (-cd) \\[2ex] = -ep + cd \\[2ex] = cd - ep $
(5.) CSEC (a) Given that

$ D = \begin{bmatrix} 1 & 9p \\[3ex] p & 4 \end{bmatrix} $ is a singular matrix, determine the value(s) of $p$.

(b) Given the linear equations

$ 2x + 5y = 6 \\[3ex] 3x + 4y = 8 \\[3ex] $ (i) Write the equations in the form $AX = B$ where $A, X,\:\:and\:\:B$ are matrices.

(ii) a) Calculate the determinant of the matrix $A$

b) Show that

$ A^{-1} = \begin{bmatrix} -\dfrac{4}{7} & \dfrac{5}{7} \\[5ex] \dfrac{3}{7} & -\dfrac{2}{7} \end{bmatrix} \\[3ex] $
c) Use the matrix $A^{-1}$ to solve for $x$ and $y$


A singular matrix is a matrix whose determinant is zero.

$ (a)\:\: \begin{vmatrix} 1 & 9p \\[3ex] p & 4 \end{vmatrix} = 0 \\[3ex] $
$ 1(4) - p(9p) = 0 \\[3ex] 4 - 9p^2 = 0 \\[3ex] 4 = 9p^2 \\[3ex] 9p^2 = 4 \\[3ex] p^2 = \dfrac{4}{9} \\[5ex] p = \pm\sqrt{\dfrac{4}{9}} \\[5ex] p = \pm\dfrac{2}{3} \\[5ex] (b)\:\: 2x + 5y = 6 \\[3ex] 3x + 4y = 8 \\[3ex] AX = B \\[3ex] (i)\:\: \begin{bmatrix} 2 & 5 \\[3ex] 3 & 4 \end{bmatrix} \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \begin{bmatrix} 6 \\[3ex] 8 \end{bmatrix} \\[7ex] A = \begin{bmatrix} 2 & 5 \\[3ex] 3 & 4 \end{bmatrix}\:\:\:\: X = \begin{bmatrix} x \\[3ex] y \end{bmatrix}\:\:\:\: B = \begin{bmatrix} 6 \\[3ex] 8 \end{bmatrix}\\[5ex] (ii)a)\:\: |A| = \begin{vmatrix} 2 & 5 \\[3ex] 3 & 4 \end{vmatrix} = 2(4) - 3(5) = 8 - 15 = -7 \\[7ex] b)\:\: For\:\:A = \begin{bmatrix} a & b \\[3ex] c & d \end{bmatrix}\:\:\:\:\: adj\:\: A = \begin{bmatrix} d & -b \\[3ex] -c & a \end{bmatrix} \\[7ex] Compare\:\:Matrix\:A \:\:to\:\: Matrix\:A \\[3ex] a = 2 \\[3ex] b = 5 \\[3ex] -b = -5 \\[3ex] c = 3 \\[3ex] -c = -3 \\[3ex] d = 4 \\[3ex] adj\:\: A = \begin{bmatrix} 4 & -5 \\[3ex] -3 & 2 \end{bmatrix} \\[7ex] A^{-1} = \dfrac{adj\:A}{det\:A} \\[5ex] A^{-1} = \dfrac{ \begin{bmatrix} 4 & -5 \\[3ex] -3 & 2 \end{bmatrix}}{-7} \\[7ex] A^{-1} = \begin{bmatrix} \dfrac{4}{-7} & \dfrac{-5}{-7} \\[5ex] \dfrac{-3}{-7} & \dfrac{2}{-7} \end{bmatrix} \\[10ex] \therefore A^{-1} = \begin{bmatrix} -\dfrac{4}{7} & \dfrac{5}{7} \\[5ex] \dfrac{3}{7} & -\dfrac{2}{7} \end{bmatrix} \\[10ex] c)\:\: X = A^{-1} * B \\[3ex] A^{-1} * B = \begin{bmatrix} -\dfrac{4}{7} & \dfrac{5}{7} \\[5ex] \dfrac{3}{7} & -\dfrac{2}{7} \end{bmatrix} * \begin{bmatrix} 6 \\[3ex] 8 \end{bmatrix} = \\[10ex] \begin{bmatrix} -\dfrac{4}{7}(6) + \dfrac{5}{7}(8) \\[5ex] \dfrac{3}{7}(6) + -\dfrac{2}{7}(8) \end{bmatrix} = \begin{bmatrix} -\dfrac{24}{7} + \dfrac{40}{7} \\[5ex] \dfrac{18}{7} -\dfrac{16}{7} \end{bmatrix} = \begin{bmatrix} \dfrac{-24 + 40}{7} \\[5ex] \dfrac{18 - 16}{7} \end{bmatrix} = \begin{bmatrix} \dfrac{16}{7} \\[5ex] \dfrac{2}{7} \end{bmatrix} \\[10ex] \rightarrow \begin{bmatrix} x \\[3ex] y \end{bmatrix} = \begin{bmatrix} \dfrac{16}{7} \\[5ex] \dfrac{2}{7} \end{bmatrix} \\[10ex] x = \dfrac{16}{7} \\[5ex] y = \dfrac{2}{7} \\[5ex] \underline{Check} \\[3ex] 2x + 5y = 6 \\[3ex] 3x + 4y = 8 \\[3ex] x = \dfrac{16}{7},\:y = \dfrac{2}{7} \\[5ex] $
$LHS$ $RHS$
$ 2x + 5y \\[3ex] 2\left(\dfrac{16}{7}\right) + 5\left(\dfrac{2}{7}\right) \\[5ex] \dfrac{32}{7} + \dfrac{10}{7} \\[5ex] \dfrac{32 + 10}{7} \\[5ex] \dfrac{42}{7} \\[5ex] 6 $ $6$
$ 3x + 4y \\[3ex] 3\left(\dfrac{16}{7}\right) + 4\left(\dfrac{2}{7}\right) \\[5ex] \dfrac{48}{7} + \dfrac{8}{7} \\[5ex] \dfrac{48 + 8}{7} \\[5ex] \dfrac{56}{7} \\[5ex] 8 $ $8$
(6.) $ \begin{vmatrix} p & c \\ -d & -e \end{vmatrix} $


$ \begin{vmatrix} p & c \\ -d & -e \end{vmatrix} \\[3ex] = (p)(-e) - (-d)(c) \\[2ex] = -ep - (-cd) \\[2ex] = -ep + cd \\[2ex] = cd - ep $