Solved Examples on Matrices (All Topics)



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

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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.) Which of the following matrices is in reduced row echelon​ form?

$ A.\;\;\left[ \begin{array}{cc|c} 1 & 9 & 2 \\[3ex] 0 & 1 & 6 \end{array} \right] \hspace{7em} B.\;\;\left[ \begin{array}{cc|c} 1 & 9 & 9 \\[3ex] 7 & -1 & 2 \end{array} \right] \\[10ex] C.\;\;\left[ \begin{array}{cc|c} 1 & 0 & 5 \\[3ex] 0 & 1 & 9 \end{array} \right] \hspace{7em} D.\;\;\left[ \begin{array}{cc|c} 1 & 0 & 9 \\[3ex] 0 & 7 & 4 \end{array} \right] \\[10ex] $

These are the properties of reduced row echelon matrices.
Based on the properties:
Number (1.): The leading entry in any non-zero row is 1
Matrix D is out (7 in the second row).

Number (2.): All entries in the column above or below a leading 1 is zero
Matrices A (9 in the second column) and B (7 in the first column) are out.

Matrix C is the answer.
(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.) Determine whether these statements are true or false.
The matrix $ \left[ \begin{array}{cc|c} 1 & 3 & -2 \\[3ex] 0 & 1 & 5 \\[3ex] 0 & 0 & 0 \end{array} \right]$ is in row echelon form.


These are the properties of row echelon matrices.
A matrix is said to be in row echelon form (ref) or echelon form if each row has more leading zeros that the rows before it.
Based on this definition, the matrix is in row echelon form.
(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.) Which of the following statements accurately describes the system represented by the matrix​ below?

$ \left[ \begin{array}{ccc|c} 1 & 4 & -2 & 3 \\[3ex] 0 & 1 & 3 & -2 \\[3ex] 0 & 0 & 0 & 4 \end{array} \right] $

A. The system has no solution.
B. The system has infinitely many solutions.
C. The number of solutions cannot be determined.
D. The system has one solution.


Equation (3.) from the augmented matrix can be written as:
0x + 0y + 0z = 4
This means that: 0 = 4
This is contradiction.
The system has no solution.
(7.)


(8.) Write the augmented matrix of the following systems of equations.

$ (a.)\;\;\begin{cases} -8x + 7y + 4 = 0 \\[3ex] -x + 9y - 1 = 0 \end{cases} \\[7ex] (b.)\;\;\begin{cases} 0.07x - 0.09y = 0.04 \\[3ex] 0.18x + 0.30y = 0.4 \end{cases} \\[7ex] (c.)\;\;\begin{cases} x - y + z = 12 \\[3ex] 2x + 2y = 4 \\[3ex] x + y + 9z = 3 \end{cases} \\[10ex] (d.)\;\;\begin{cases} x - y - z = -2 \\[3ex] -7x + y - 3z = -1 \\[3ex] 2x + 8y = -5 \\[3ex] 3x + 9y + z = 0 \end{cases} $


System of Equation: Variables on the LHS; constants on the RHS
Augmented Matrix: Coefficients of the variables on the LHS; constants on the RHS

The augmented matrices are:

$ (a.) \\[3ex] -8x + 7y + 4 = 0...eqn.(1) \\[3ex] -8x + 7y = -4 \\[5ex] -x + 9y - 1 = 0...eqn.(2) \\[3ex] -x + 9y = 1 \\[5ex] \left[ \begin{array}{cc|c} -8 & 7 & -4 \\[3ex] -1 & 9 & 1 \end{array} \right] \\[7ex] (b.) \\[3ex] \left[ \begin{array}{cc|c} 0.07 & -0.09 & 0.04 \\[3ex] 0.18 & 0.3 & 0.4 \end{array} \right] \\[7ex] (c.) \\[3ex] 2x + 2y = 4...eqn.(2) \\[3ex] 2x + 2y + 0z = 4 \\[5ex] \left[ \begin{array}{ccc|c} 1 & -1 & 1 & 12 \\[3ex] 2 & 2 & 0 & 4 \\[3ex] 1 & 1 & 9 & 3 \end{array} \right] \\[10ex] (d.) \\[3ex] 2x + 8y = -5...eqn.(3) \\[3ex] 2x + 8y + 0z = -5 \\[5ex] \left[ \begin{array}{ccc|c} 1 & -1 & -1 & -2 \\[3ex] -7 & 1 & -3 & -1 \\[3ex] 2 & 8 & 0 & -5 \\[3ex] 3 & 9 & 1 & 0 \end{array} \right] $
(9.)


(10.) (I.) Write the system of equations for the following augmented matrices.
(II.) Then, perform the indicated row operations on the augmented matrices.

$ (a.)\;\;\left[ \begin{array}{cc|c} 1 & 9 & -4 \\[3ex] -7 & 4 & 7 \end{array} \right] \\[10ex] R_2 = 7r_1 + r_2 \\[5ex] (b.)\;\;\left[ \begin{array}{ccc|c} 1 & -4 & 3 & 4 \\[3ex] 4 & -3 & 6 & 3 \\[3ex] -3 & 5 & 3 & 3 \end{array} \right] \\[10ex] R_2 = -4r_1 + r_2 \\[3ex] R_3 = 3r_1 + r_3 \\[5ex] (c.)\;\;\left[ \begin{array}{ccc|c} 1 & -5 & 2 & -2 \\[3ex] 2 & -7 & 8 & -5 \\[3ex] -5 & -6 & 4 & 2 \end{array} \right] \\[10ex] R_2 = -2r_1 + r_2 \\[3ex] R_3 = 5r_1 + r_3 \\[5ex] (d.)\;\;\left[ \begin{array}{ccc|c} 9 & -4 & 1 & -7 \\[3ex] 2 & -4 & 6 & -7 \\[3ex] -4 & 1 & 3 & 8 \end{array} \right] \\[10ex] R_1 = -4r_2 + r_1 \\[3ex] R_3 = 2r_2 + r_3 \\[3ex] $

System of Equation: Variables on the LHS; constants on the RHS
Augmented Matrix: Coefficients of the variables on the LHS; constants on the RHS

$ (a.)\;\;\begin{cases} x + 9y = -4 \\[3ex] -7x + 4y = 7 \end{cases} \\[10ex] R_2 = 7r_1 + r_2 \\[5ex] \left[ \begin{array}{cc|c} 1 & 9 & -4 \\[3ex] 7(1) + -7 & 7(9) + 4 & 7(-4) + 7 \end{array} \right] \\[10ex] \left[ \begin{array}{cc|c} 1 & 9 & -4 \\[3ex] 7 - 7 & 63 + 4 & -28 + 7 \end{array} \right] \\[10ex] \left[ \begin{array}{cc|c} 1 & 9 & -4 \\[3ex] 0 & 67 & -21 \end{array} \right] \\[10ex] (b.)\;\;\begin{cases} x - 4y + 3z = 4 \\[3ex] 4x - 3y + 6z = 3 \\[3ex] -3x + 5y + 3z = 3 \end{cases} \\[10ex] R_2 = -4r_1 + r_2 \\[3ex] R_3 = 3r_1 + r_3 \\[5ex] \left[ \begin{array}{ccc|c} 1 & -4 & 3 & 4 \\[3ex] -4(1) + 4 & -4(-4) + -3 & -4(3) + 6 & -4(4) + 3 \\[3ex] 3(1) + -3 & 3(-4) + 5 & 3(3) + 3 & 3(4) + 3 \end{array} \right] \\[10ex] \left[ \begin{array}{ccc|c} 1 & -4 & 3 & 4 \\[3ex] -4 + 4 & 16 - 3 & -12 + 6 & -16 + 3 \\[3ex] 3 - 3 & -12 + 5 & 9 + 3 & 12 + 3 \end{array} \right] \\[10ex] \left[ \begin{array}{ccc|c} 1 & -4 & 3 & 4 \\[3ex] 0 & 13 & -6 & -13 \\[3ex] 0 & -7 & 12 & 15 \end{array} \right] \\[10ex] (c.)\;\;\begin{cases} x - 5y + 2z = -2 \\[3ex] 2x - 7y + 8z = -5 \\[3ex] -5x - 6y + 4z = 2 \end{cases} \\[10ex] R_2 = -2r_1 + r_2 \\[3ex] R_3 = 5r_1 + r_3 \\[5ex] \left[ \begin{array}{ccc|c} 1 & -5 & 2 & -2 \\[3ex] -2(1) + 2 & -2(-5) + -7 & -2(2) + 8 & -2(-2) + -5 \\[3ex] 5(1) + -5 & 5(-5) + -6 & 5(2) + 4 & 5(-2) + 2 \end{array} \right] \\[10ex] \left[ \begin{array}{ccc|c} 1 & -5 & 2 & -2 \\[3ex] -2 + 2 & 10 - 7 & -4 + 8 & 4 - 5 \\[3ex] 5 - 5 & -25 - 6 & 10 + 4 & -10 + 2 \end{array} \right] \\[10ex] \left[ \begin{array}{ccc|c} 1 & -5 & 2 & -2 \\[3ex] 0 & 3 & 4 & -1 \\[3ex] 0 & -31 & 14 & -8 \end{array} \right] \\[10ex] (d.)\;\;\begin{cases} 9x - 4y + z = -7 \\[3ex] 2x - 4y + 6z = -7 \\[3ex] -4x + y + 3z = 8 \end{cases} \\[10ex] R_1 = -4r_2 + r_1 \\[3ex] R_3 = 2r_2 + r_3 \\[5ex] \left[ \begin{array}{ccc|c} -4(2) + 9 & -4(-4) + -4 & -4(6) + 1 & -4(-7) + -7 \\[3ex] 2 & -4 & 6 & -7 \\[3ex] 2(2) + -4 & 2(-4) + 1 & 2(6) + 3 & 2(-7) + 8 \end{array} \right] \\[10ex] \left[ \begin{array}{ccc|c} -8 + 9 & 16 - 4 & -24 + 1 & 28 - 7 \\[3ex] 2 & -4 & 6 & -7 \\[3ex] 4 - 4 & -8 + 1 & 12 + 3 & -14 + 8 \end{array} \right] \\[10ex] \left[ \begin{array}{ccc|c} 1 & 12 & -23 & 21 \\[3ex] 2 & -4 & 6 & -7 \\[3ex] 0 & -7 & 15 & -6 \end{array} \right] $
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