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# Solved Examples on the Determinant of a Matrix 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.

Calculate the determinants of these matrices.
For the $3 * 3$ and $4 * 4$ matrices, use at least two methods as applicable.
Show all work.

(1.) $\begin{vmatrix} -3 & 3 \\[3ex] -1 & -2 \end{vmatrix}$

$\begin{vmatrix} -3 & 3 \\[3ex] -1 & -2 \end{vmatrix} \\[3ex]$
$= (-3)(-2) - (-1)(3) \\[3ex] = 6 - (-3) \\[3ex] = 6 + 3 \\[3ex] = 9$
(2.) $\begin{vmatrix} 7 & \sqrt{7} \\[3ex] \sqrt{7} & 7 \end{vmatrix}$

$\begin{vmatrix} 7 & \sqrt{7} \\[3ex] \sqrt{7} & 7 \end{vmatrix} \\[3ex]$
$= (7)(7) - (\sqrt{7})(\sqrt{7}) \\[3ex] = 49 - 7 \\[3ex] = 42$
(3.) $\begin{vmatrix} 3x^7 & -12 \\[3ex] x^7 & x^5 \end{vmatrix}$

$\begin{vmatrix} 3x^7 & -12 \\[3ex] x^7 & x^5 \end{vmatrix} \\[3ex]$
$= (3x^7)(x^5) - (x^7)(-12) \\[3ex] = 3x^{12} - (-12x^7) \\[3ex] = 3x^{12} + 12x^7$
(4.) $\begin{vmatrix} p & c \\[3ex] -d & -e \end{vmatrix}$

$\begin{vmatrix} p & c \\[3ex] -d & -e \end{vmatrix} \\[3ex]$
$= (p)(-e) - (-d)(c) \\[3ex] = -ep - (-cd) \\[3ex] = -ep + cd \\[3ex] = cd - ep$
(5.) $\begin{vmatrix} 2 & -1 & 6 \\[3ex] 4 & -6 & 0 \\[3ex] 5 & -2 & 6 \end{vmatrix}$

First Method: Formula
To solve it faster, we shall use the second row because it contains a $0$
However, you may use any row that you prefer.

$\begin{vmatrix} + & - & + \\[3ex] - & + & - \\[3ex] + & - & + \end{vmatrix} \hspace{3em} \begin{vmatrix} 2 & -1 & 6 \\[3ex] 4 & -6 & 0 \\[3ex] 5 & -2 & 6 \end{vmatrix} \\[3ex]$
$\underline{2nd\:\: Row} \\[3ex] = -4 \begin{vmatrix} -1 & 6 \\[3ex] -2 & 6 \end{vmatrix} + -6 \begin{vmatrix} 2 & 6 \\[3ex] 5 & 6 \end{vmatrix} \\[5ex] = -4(-6 + 12) - 6(12 - 30) \\[3ex] = -4(6) - 6(-18) \\[3ex] = -24 + 108 \\[3ex] = 84 \\[3ex]$ Second Method: "Diagonal" Method

$\begin{vmatrix} 2 & -1 & 6 & 2 & -1 \\[3ex] 4 & -6 & 0 & 4 & -6 \\[3ex] 5 & -2 & 6 & 5 & -2 \end{vmatrix} \\[3ex]$
$\underline{UpDown} \\[3ex] 1st:\:\: 2(-6)(6) = -72 \\[3ex] 2nd:\:\: -1(0)(5) = 0 \\[3ex] 3rd:\:\: 6(4)(-2) = -48 \\[3ex] UpDown\;\;Sum: \\[3ex] = -72 + 0 + -48 \\[3ex] = -120 \\[3ex] \underline{DownUp} \\[3ex] 1st:\:\: 5(-6)(6) = -180 \\[3ex] 2nd:\:\: -2(0)(2) = 0 \\[3ex] 3rd:\:\: 6(4)(-1) = -24 \\[3ex] DownUp\;\;Sum: \\[3ex] = -180 + 0 + -24 \\[3ex] = -204 \\[3ex] det = UpDown\;\;Sum - DownUp\;\;Sum \\[3ex] = -120 - (-204) \\[3ex] = -120 + 204 \\[3ex] = 84$
(6.) JAMB Evaluate $\begin{vmatrix} -1 & -1 & -1 \\[3ex] 3 & 1 & -1 \\[3ex] 1 & 2 & 1 \end{vmatrix}$

JAMB is a timed exam without the use of a calculator.
To solve it faster, we can use any row because they all have small numbers
But, let us use Row $1$ because it has smallest numbers ($1's$)
First Method: Formula

$\begin{vmatrix} + & - & + \\[3ex] - & + & - \\[3ex] + & - & + \end{vmatrix} \hspace{3em} \begin{vmatrix} -1 & -1 & -1 \\[3ex] 3 & 1 & -1 \\[3ex] 1 & 2 & 1 \end{vmatrix} \\[3ex]$
$\underline{1st\:\: Row} \\[3ex] -1 \begin{vmatrix} 1 & -1 \\[3ex] 2 & 1 \end{vmatrix} - -1 \begin{vmatrix} 3 & -1 \\[3ex] 1 & 1 \end{vmatrix} + -1 \begin{vmatrix} 3 & 1 \\[3ex] 1 & 2 \end{vmatrix} \\[5ex] = -1(1 + 2) + 1(3 + 1) - 1(6 - 1) \\[3ex] = -1(3) + 1(4) - 1(5) \\[3ex] = -3 + 4 - 5 \\[3ex] = -4 \\[3ex]$ Second Method: "Diagonal" Method

$\begin{vmatrix} -1 & -1 & -1 & -1 & -1 \\[3ex] 3 & 1 & -1 & 3 & 1 \\[3ex] 1 & 2 & 1 & 1 & 2 \end{vmatrix} \\[3ex]$
$\underline{UpDown} \\[3ex] 1st:\:\: -1(1)(1) = -1 \\[3ex] 2nd:\:\: -1(-1)(1) = 1 \\[3ex] 3rd:\:\: -1(3)(2) = -6 \\[3ex] UpDown\;\;Sum: \\[3ex] = -1 + 1 + -6 \\[3ex] = -6 \\[3ex] \underline{DownUp} \\[3ex] 1st:\:\: 1(1)(-1) = -1 \\[3ex] 2nd:\:\: 2(-1)(-1) = 2 \\[3ex] 3rd:\:\: 1(3)(-1) = -3 \\[3ex] DownUp\;\;Sum: \\[3ex] = -1 + 2 + -3 \\[3ex] = -2 \\[3ex] det = UpDown\;\;Sum - DownUp\;\;Sum \\[3ex] = -6 - (-2) \\[3ex] = -6 + 2 \\[3ex] = -4$
(7.) ACT The determinant of any $2$ x $2$ matrix

$\begin{bmatrix} a & b \\[3ex] c & d \end{bmatrix} \:\:is\:\: ad - bc \\[7ex]$ The determinant of

$\begin{bmatrix} (x + 3) & 7 \\[3ex] 2 & (x - 2) \end{bmatrix} \:\:is\:\:equal\:\:to\:\: 0 \\[7ex]$ What are all possible values of $x$?

$A.\:\: -5\:\:and\:\:4 \\[3ex] B.\:\: -4\:\:and\:\:5 \\[3ex] C.\:\: -3\:\:and\:\:2 \\[3ex] D.\:\: -1\:\:and\:\:9 \\[3ex] E.\:\: -\sqrt{20}\:\:and\:\:\sqrt{20} \\[3ex]$

Compare

$\begin{bmatrix} a & b \\[3ex] c & d \end{bmatrix} \:\:and\:\: \begin{bmatrix} (x + 3) & 7 \\[3ex] 2 & (x - 2) \end{bmatrix} \\[7ex] a = x + 3 \\[3ex] b = 7 \\[3ex] c = 2 \\[3ex] d = x - 2 \\[3ex] determinant = ad - bc \\[3ex] determinant = (x + 3)(x - 2) - 7(2) \\[3ex] determinant = 0...from\:\:the\:\:Question \\[3ex] \rightarrow (x + 3)(x - 2) - 7(2) = 0 \\[3ex] x^2 - 2x + 3x - 6 - 14 = 0 \\[3ex] x^2 + x - 20 = 0 \\[3ex] (x + 5)(x - 4) = 0 \\[3ex] x + 5 = 0 \:\:OR\:\: x - 4 = 0 \\[3ex] x = -5 \:\:OR\:\: x = 4$
(8.) ACT For what value of $b$ will the determinant of the matrix

$\begin{bmatrix} 4 & b \\[3ex] 2 & 3 \end{bmatrix}$ have a value of $18$?

$F.\:\: -\dfrac{10}{3} \\[5ex] G.\:\: -3 \\[3ex] H.\:\: 3 \\[3ex] J.\:\: 6 \\[3ex] K.\;\; 15 \\[3ex]$

$\begin{vmatrix} 4 & b \\[3ex] 2 & 3 \end{vmatrix} = 18 \\[7ex] (4 * 3) - (2 * b) = 18 \\[3ex] 12 - 2b = 18 \\[3ex] 12 - 18 = 2b \\[3ex] -6 = 2b \\[3ex] 2b = -6 \\[3ex] b = -\dfrac{6}{2} \\[5ex] b = -3$