The following are all matrices:

(i). $A = \pmatrix {0 & \;\,\,4 \\ 1 & -1}$

(ii). $B = \pmatrix {\;50 & 25 & \;45 \\ 125 & 80 & \;60 \\ \;60 & 0 & \;75 \\ \;90 & 80 & 110}$.

(iii). $K = \pmatrix{\;\;\,1.2 & 3.8 & 0.2 & -0.2 & 3.1 \\ -2.3 & 2.1 & 3.6 & \;\;\,3.8 & 0.1}$

(iv). $M = \pmatrix{3 \\ 1}$.

The matrices in Example 1 have the following sizes,

(i). $A$ is a $2 \times 2$ matrix, i.e. 2 rows and 2 columns.

(ii). $B$ is a $4 \times 3$ matrix, i.e. 4 rows and 3 columns.

(iii). $K$ is a $2 \times 5$ matrix, i.e. 2 rows and 5 columns.

(iv). $M$ is a $2 \times 1$ matrix, i.e. 2 rows and 1 column.

For the matrix $A = \pmatrix{\;\,50 & \;\;\,25 & -45 \\ \,125 & \;\;\,80 & \;\;\,60 \\ \;\;60 & -50 & \;\;\,75 \\ \;\,90 & \,120 & \,110 }$

(i). element $a_{43} = 110$, as it is located at Row 4, Column 3.

(ii). element $a_{32} = -50$ , as it is located at Row 3, Column 2.

We can easily draw parallels between matrices and computer arrays as used in programming languages. For example, in C++ if $A$ is an array we would use the syntax $A[2][3]$ to index the element in the second row and third column of $A$, in Maple we would use the notation $A[2, 3]$ and in MATLAB we would write $A(2, 3)$.

Determine the values of $w, x, y$ and $z$ that guarantee the matrices $A$ and $B$ are equal,

$A = \pmatrix{4 & x \\ x+z & 0}$, $B = \pmatrix{y & 6 \\ 7 & w}$

Solution

We require $w = 0,\; x = 6,\; y = 4,\; x + z = 7 \implies z = 1$.

In each of the following carry out the specified addition.

(i). $\pmatrix{\;\;50 & \;\;25 & \;\;45 \\ 125 & \;\;80 & \;\;60 \\ \;\;60 & \;\;80 & \;\;75 \\ \;\;90 & 120 & 110} + \pmatrix {-50 & 35 & \;\;\,45 \\ -25 & 80 & -70 \\ -80 & 10 & \;\,-5 \\ \;\;\,90 & 20 & -135} = \pmatrix{\;\;\;0 & \;\;\,60 & \;\;\,90 \\ \,100 & \,160 & -10 \\ -20 & \;\;\,90 & \;\;\,70 \\ \,180 & \,140 & -25}$

(ii). $\pmatrix {2 & -1 \\3 & \;\;\,6} + \pmatrix{\;\;\,0 & 4 \\ -2 & 5} = \pmatrix{2 & \;\,3 \\ 1 & 11}$

(iii). $\pmatrix{\;\;\,0 & 4 \\ -2 & 5} + \pmatrix{2 & -1 \\ 3 & \;\;\,6} = \pmatrix{2 & \;\,3 \\ 1 & 11}$

In the following carry out the specified subtraction.

(i). $\pmatrix{3 & 1 & -2 \\ 4 & 0 & \;\;\,8} - \pmatrix{2 & -3 & -1 \\ 2 & -5 & \;\;\,1} = \pmatrix{1 & 4 & -1 \\ 2 & 5 & \;\;\,7}$

(ii). $\pmatrix{2 & -3 & -1 \\ 2 & -5 & \;\;\,1} - \pmatrix{3 & 1 & -2 \\ 4 & 0 & \;\;\,8} = \pmatrix{-1 & -4 & \;\;\,1 \\ -2 & -5 & -7}$

Simplify each of the following by performing the scalar multiplication.

(i). $2\pmatrix{0 & \;\;\,4 \\1 & -1} = \pmatrix{0 & \;\;\,8 \\ 2 & -2}$

(ii). $-0.1\pmatrix{-3 \\ -2 \\ \;\;\,0 \\ \;\;\,2} = \pmatrix{\;\;0.3 \\ \;\;0.2 \\ \;\;\;\;0 \\ -0.2}$

Let $A = \pmatrix{1 & 1 & 2 \\ 2 & 3 & 4}, B = \pmatrix{0 & -1 & \;\;\,0 \\ 1 & \;\;\,0 & -1}$ and $C = \pmatrix{\;\;\,1 & 0 \\ -9 & 4}$

If possible simplify each of the following:

(i). $2A + 3B$, (ii). $3B - 2A$(iii). $C - A$

Solution

(i). $2A + 3B = 2\pmatrix{1 & 1 & 2 \\ 2 & 3 & 4} + 3\pmatrix{0 & -1 & \;\;\,0 \\ 1 & \;\;\,0 & -1}$

$=\; \pmatrix{2 & 2 & 4 \\ 4 & 6 & 8} + \pmatrix{0 & -3 & \;\;\,0 \\ 3 & \;\;\,0 & -3}$

$=\; \pmatrix{2 & -1 & 4 \\ 7 & \;\;\,6 & 5}$

(ii). $3B - 2A = 3\pmatrix{0 & -1 & \;\;\,0 \\ 1 & \;\;\,0 & -1} - 2\pmatrix{1 & 1 & 2 \\ 2 & 3 & 4}$

$=\; \pmatrix{0 & -3 & \;\;\,0 \\ 3 & \;\;\,0 & -3} - \pmatrix{2 & 2 & 4 \\ 4 & 6 & 8}$

$=\; \pmatrix{-2 & -5 & \;\;-4 \\ -1 & -6 & -11}$

(iii). We are unable to calculate $C - A$ as the matrices have different sizes.

Here $C$ is a $2 \times 2$ matrix while $A$ is a $2 \times 3$ matrix.

Determine which of the following matrices are conformable for matrix multiplication

$A = \pmatrix{1 & \;\;\,2 & 3 \\ 4 & -5 & 6}, B = \pmatrix{1 & 2 \\ 3 & 4}, C = \pmatrix{1 & \;\;\,2 \\ 2 & \;\;\,4 \\ 3 & -1}$.

Solution

The matrix $A$ is $2 \times 3$.

The matrix $B$ is $2 \times 2$.

The matrix $C$ is $3 \times 2$.

(i). First consider the matrix product $AB$.

The “inner dimensions” are not equal and so we cannot perform the matrix multiplication. The number of columns in $A(3)$ does not equal the number of rows in $B( 2)$.

(ii). Now consider the matrix product $AC$.

The “inner dimensions” are equal and so we can perform the matrix multiplication and the product matrix will have size given by the “outer dimensions”, i.e. $2 \times 2$.

Exercise: Confirm the following: $BA, AC, CA, CB$ and $BB$ are valid multiplications; whereas we cannot calculate $AB, BC, AA$ or $CC$.

Let $A = \pmatrix{\;\;\,1 & 6 & -2 \\ \;\;\,3 & 0 & \;\;\,3 \\ -2 & 3 & -1}$ and $B = \pmatrix{1 & \;\;\,2 \\ 1 & -2 \\ 3 & \;\;\,0}$

The product matrix $AB$ can be calculated as $A$ has size $3 \times 3$ and $B$ has size $3 \times 2$, i.e. the number of columns of $A$ is the same as the number of rows of $B$.

Hence, the product matrix, $AB$ will have size $3 \times 2$.

For example, to obtain the element in Row 1/Column 1, i.e. position $(1, 1)$ of $AB$ we take the dot product of Row 1 of $A$ with Column 1 of $B$, i.e.

To obtain the element in Row 1/Column 2, i.e. position $(1, 2)$ we take the dot product of Row 1 of $A$ with Column 2 of $B$. In vector form we have,

$\pmatrix{1 & 6 & -2}. \pmatrix{\;\;\,2 \\ -2 \\ \;\;\,0} = 1 \times 2 + 6 \times (-2) + (-2) \times 0 = -10$.

This process can be continued to generate the 6 components of the $3 \times 2$ product matrix, $AB$.

$\pmatrix{\;\;\,1 & 6 & -2 \\ \;\;\,3 & 0 & \;\;\,3 \\ -2 & 3 & -1}\pmatrix{1 & \;\;\,2 \\ 1 & -2 \\3 & \;\;\,0} = \pmatrix{1 \times 1 + 6 \times 1 + (-2) \times 3 & 1 \times 2 + 6(-2) + (-2) \times 0 \\ 3 \times 1 + 0 \times 1 + 3 \times 3 & 3 \times 2 + 0 \times (-2) + 3 \times 0 \\ -2 \times 1 + 3 \times 1 + (-1) \times 3 & -2 \times 2 + 3 \times (-2) + (-1) \times 0}$

$= \pmatrix{\;\;\,1 & -10 \\ \;\;12 & \;\;\;\,6 \\ -2 & -10}$

Let $A = \pmatrix{0 & \;\;\;4 \\ 1 & -1}$ and $B = \pmatrix{1 & \;\;\;1 \\ 2 & -1}$. If possible calculate $AB$ and $BA$.

Solution

In this case $A$ and $B$ are both $2 \times 2$ (square) matrices and so we can calculate $AB$ and $BA$. The result of both multiplications will be a $2 \times 2$ matrix. We have

$AB = \pmatrix{0 & \;\;\;4 \\ 1 & -1}\pmatrix{1 & \;\;\;1 \\ 2 & -1} = \pmatrix{\;\;\;8 & -4 \\ -1 & \;\;\;2}$

and

$BA = \pmatrix{1 & \;\;\;1 \\ 2 & -1}\pmatrix{0 & \;\;\;4 \\ 1 & -1} = \pmatrix{\;\;\;1 & 3 \\ -1 & 9}$

Note: This is an example of a very important result in matrix arithmetic. In general, for square matrices $A$ and $B$, we have that $AB \neq BA$.

If possible evaluate the following matrix products:

(i). $\pmatrix{\;\;\,1 & 2 & \;\;\,3 \\ \;\;\,2 & 0 & \;\;\,1 \\ -1 & 1 & -1}\pmatrix{\;\;\,1 & 2 \\ \;\;\,4 & 2 \\ -1 & 0}$.

(ii). $\pmatrix{a & b \\ b & a}\pmatrix{1 & 2 \\ 3 & 4}$.

(iii). $\pmatrix{1 & x \\ x & 2}\pmatrix{x & y \\ x & 1}$.

(iv). $\pmatrix{\;\;\,3 \\ -1 \\ \;\;\,2}\pmatrix{2 & 0 & -1 & 6}$

(v). $\pmatrix{2 & 0 & -1 & 6}\pmatrix{\;\;\,3 \\ -1 \\ \;\;\,2}$.

Solution

(i) $\pmatrix{\;\;\,1 & 2 & \;\;\,3 \\ \;\;\,2 & 0 & \;\;\,1 \\ -1 & 1 & -1}\pmatrix{\;\;\,1 & 2 \\ \;\;\,4 & 2 \\ -1 & 0}$.

The product can be formed as the first matrix has size $3 \times 3$ and the second matrix has size $3 \times 2$, i.e. the number of columns in the first matrix $(3)$ is the same as the number of rows $(3)$ in the second matrix. The product matrix will have size $3 \times 2$. Multiplying gives,

$\pmatrix{1 \times 1 + 2 \times 4 + 3 \times (-1) & 1 \times 2 + 2 \times 2 + 3 \times 0 \\ 2 \times 1 + 0 \times 4 + 1 \times (-1) & 2 \times 2 + 0 \times 2 + 1 \times 0 \\-1 \times 1 + 1 \times 4 + (-1) \times (-1) & -1 \times 2 + 1 \times 2 + (-1) \times 0} = \pmatrix{6 & 6 \\ 1 & 4 \\ 4 & 0}$

(ii). $\pmatrix{a+3b & 2a+4b \\ b+3a & 2b+4a}$ (Exercise: Check this answer)

(iii). $\pmatrix{x + x^2 & y + x \\ x^2 + 2x & xy + 2}$ (Exercise: Check this answer)

(iv). $\pmatrix{\;\;\,3 \\ -1 \\ \;\;\,2}\pmatrix{2 & 0 & -1 & 6}$

The product can be formed as the first matrix has size $3 \times 1$ and the second matrix has size $1 \times 4$, i.e. the number of columns in the first matrix $(1)$ is the same as the number of rows $(1)$ in the second matrix. The product matrix will have size $3 \times 4$. Multiplying gives,

$\pmatrix{\;\;\,3 \times 2 & \;\;\,3 \times 0 & \;\;\,3 \times (-1) & \;\;\,3 \times 6 \\ -1 \times 2 & -1 \times 0 & -1 \times (-1) & -1 \times 6 \\ \;\;\,2 \times 2 & \;\;\,2 \times 0 & \;\;\,2 \times (-1) & \;\;\,2 \times 6} = \pmatrix{\;\;\,6 & 0 & -3 & \;\;16 \\ -2 & 0 & \;\;\,1 & -6 \\ \;\;\,4 & 0 & -2 & \;\;12}$

(v). $\pmatrix{2 & 0 & -1 & 6}\pmatrix{\;\;\,3 \\ -1 \\ \;\;\,2}$

The product cannot be formed as the first matrix has size $1 \times 4$ and the second matrix has size $3 \times 1$, i.e. the number of columns in the first matrix $(4)$ is not the same as the number of rows $(3)$ in the second matrix.

Let $A = \pmatrix{1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8}$ and $B = \pmatrix{-40 & 16 & \;\;\,9 \\ \;\;\,13 & -5 & -3 \\ \;\;\;\;5 & -2 & -1}$. Calculate the matrix products $AB$ and $BA$.

Solution

(i). $AB = \pmatrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1}, BA = \pmatrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1}$.

Exercise: You should verify these calculations.

Unlike Example 11 we find that in this case matrix multiplication is commutative and $AB = BA$. This particular square matrix, with 1’s on the main diagonal (top left to bottom right) and 0’s everywhere else, is known as the identity matrix – see Sections 4.2 and 5 for further discussions.

Determine the values of $x$ and $y$ that satisfy the following matrix equation.

$\pmatrix{1 & \;\;\,3 \\ x & -2}\pmatrix{2 & 4 \\ 3 & y} = \pmatrix{11 & \;\;\,1 \\ \;\;\,6 & 26}$

Solution

Expanding the left-hand-side (LHS) gives,

$\pmatrix{1 & \;\;\,3 \\ x & -2}\pmatrix{2 & 4 \\ 3 & y} = \pmatrix{11 & 4 + 3y \\ 2x - 6 & 4x - 2y} = \pmatrix{11 & \;\;\,1 \\ \;\;\,6 & 26}$

We therefore have three equations. The first two of these are equations in one variable which can easily be solved for $x$ and $y$. The third equation can be used to check our answers.

$ {\array{2x -6 = 6 \\ 4 + 3y = 1 \\ 4x - 2y = 26}}\Bigg\} \implies x = 6,\; y = -1 $.

Let $A = \pmatrix{1 & -2 & 7 \\ 6 & \;\;\,0 & 5}$. Write down the matrix $A^T$.

Solution

The matrix $A^T$ is obtained by interchanging rows and columns of the matrix $A$. Hence,

$A^T = \pmatrix{\;\;\,1 & 6 \\ -2 & 0 \\ \;\;\,7 & 5}$.

Note that Row 1 of $A$ is Column 1 of $A^T$ and Row 2 of $A$ is Column 2 of $A^T$.

Alternatively, Column 1 of $A$ is Row 1 of $A^T$, etc.

(i). The $2 \times 2$ identity matrix is, $I = \pmatrix{1 & 0 \\ 0 & 1}$.

(ii). We met the $3 \times 3$ identity matrix, $I = \pmatrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1}$ in Example 13.

If $A = \pmatrix{\;\;\,1 & 9 \\ -2 & 3}$ then

$\pmatrix{\;\;\,1 & 9 \\ -2 & 3} + \pmatrix{0 & 0 \\ 0 & 0} = \pmatrix{\;\;\,1 & 9 \\ -2 & 3}$

The matrix $A$ is unchanged by addition of the zero matrix.

The following matrices are all examples of diagonal matrices:

(i). $\pmatrix{6 & \;\;\,0 \\ 0 & -1} $ (ii). $\pmatrix{2 & \;\;\,0 & 0 \\ 0 & -5 & 0 \\0 & \;\;\,0 & 3}$(iii). $\pmatrix{4 & 0 & \;\;\,0 \\ 0 & 0 & \;\;\,0 \\0 & 0 & -1}$.

The identity matrix is a special case of a diagonal matrix where all the diagonal entries are equal to 1.

The matrices below are all examples of upper triangular matrices :

(i). $\pmatrix{6 & \;\;2 \\ 0 & -1}$ (ii). $\pmatrix{2 & -5 & 4 \\ 0 & \;\;9 & 6 \\ 0 & \;\;0 & 3}$(iii). $\pmatrix{4 & 2 & \;\;0 \\ 0 & 0 & \;\;3 \\ 0 & 0 & -1}$.

The matrices below are all examples of lower triangular matrices :

(i). $\pmatrix{6 & \;\;0 \\ 2 & -1}$ (ii). $\pmatrix{2 & \;\;0 & 0 \\ 6 & \;\;9 & 0 \\ 4 & -5 & 3}$(iii). $\pmatrix{4 & 0 & 0 \\ 3 & 1 & 0 \\ 2 & 0 & 0}$.

The following matrices are all symmetric,

(i). $\pmatrix{\;\;\,3 & -1 \\ -1 & \;\;\,3}$ (ii). $\pmatrix{2 & \;\;\,0 & 1 \\ 0 & -6 & 3 \\ 1 & \;\;\,3 & 5}$(iii). $\pmatrix{3 & \;\;\,2 & \;\;\,1 \\ 2 & -9 & -3 \\ 1 & -3 & \;\;\,4}$

Calculate the determinant of each of the following matrices. Hence, identify which matrices are invertible and for each invertible matrix calculate its inverse.

(i). $A = \pmatrix{\;\;\,2 & -1 \\ -2 & \;\;\,1}$ (ii). $B = \pmatrix{7 & 3 \\ 2 & 1}$

(iii). $C = \pmatrix{0 & 1 \\ 0 & 1}$ (iv). $M = \pmatrix{\;\;\,2 & \;\;\,0 \\ -2 & -3}$

Solution

(i). $det(A) = 2 \times 1 - (-1) \times (-2) = 2 - 2 = 0$. No Inverse.

(ii). $det(B) = 7 \times 1 - 3 \times 2 = 7 - 6 = 1 \neq 0$. Matrix has an inverse.

$B^{-1} = \ {\Large\frac{1}{1}}\pmatrix{\;\;\,1 & -3 \\ -2 & \;\;\,7} =\pmatrix{\;\;\,1 & -3 \\ -2 & \;\;\,7}$

Exercise: Check that $BB^{-1} = B^{-1}B = I$ .

(iii). $det(C) = 0 \times 1 - 1 \times 0 = 0 - 0 = 0$. No inverse.

(iv). $det(M) = 2 \times (-3) - 0 \times (-2) = -6 \neq 0$.

$M^{-1} = -{\Large\frac{1}{6}}\pmatrix{-3 & 0 \\ \;\;\,2 & 2}$

$= {\Large\frac{1}{6}}\pmatrix{\;\;\,3 & \;\;\,0 \\ -2 & -2}$

Exercise: Check that $MM^{-1} = M^{-1}M = I$.

Calculate the determinant of the matrix, $A = \pmatrix{3 & \;\;\,2 & -1 \\ 1 & \;\;\,6 & \;\;\,3 \\ 2 & -4 & \;\;\,0}$ using cofactor expansion.

Solution

We choose to expand along Row 1 and first calculate the minor for each entry :

the minor of entry $a_{11} = 3$ is:

the minor of entry $a_{12} = 2$ is:

the minor of entry $a_{13} = -1$ is:

Now calculate the cofactors :

the cofactor of entry $a_{11}$ is: $C_{11} = (-1)^{1 + 1} \times M_{11} = (-1)^2 \times 12 = 12$.

the cofactor of entry $a_{12}$ is: $C_{12} = (-1)^{1 + 2} \times M_{12} = (-1)^3 \times (-6) = 6$.

the cofactor of entry $a_{13}$ is: $C_{13} = (-1)^{1 + 3} \times M_{13} = (-1)^4 \times (-16) = -16$.

Finally, multiply each entry in Row 1 by its cofactor and sum the values. Hence,

$$\text{det}(A) = \sum_{j = 1}^{3}a_{1j}C_{1j}$$

$= a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}$

$= 3 \times 12 + 2 \times 6 + (-1) \times (-16)$

$= 36 + 12 + 16$

$=$ $64$.

Notes

(i). The value of the determinant will be the same regardless of which row or column we expand on.

(ii). When calculating the determinant of a matrix we expand along the row or column containing the most zeros in order to minimise the arithmetic.

Calculate the determinant of the matrix, $A = \pmatrix{3 & \;\;\,2 & -1 \\1 & \;\;\,6 & \;\;\,3 \\ 2 & -4 & \;\;\,0}$ by expanding on Row 1. Repeat the calculation by expanding on Column 2.

Solution

Expand along the first row of $A$ calculating the minors. Multiply each minor by the corresponding entry from the matrix $A$, augmented by the appropriate sign from the first row of the sign array, i.e. $(+ \;\; - \;\; +)$ and sum the results.

$\text{det}(A) = + 3. \bigg |\matrix{\;\;\,6 & 3 \\ -4 & 0}\bigg | - 2. \bigg |\matrix{1 & 3 \\ 2 & 0} \bigg | + (-1). \bigg |\matrix{1 & \;\;\,6 \\ 2 & -4}\bigg |$

$= 3.(6 \times 0 - 3 \times (-4)) - 2.(1 \times 0 - 3 \times 2) + (-1).(1 \times(-4) - 6 \times 2)$

$= 36 + 12 + 16$

$=$ $64$.

Now expand along the second column of $A$ using the appropriate signs from the second column of the sign array, i.e. $(- \;\; + \;\; -)$ :

$\text{det}(A) = -2. \bigg | \matrix{1 & 3 \\ 2 & 0} \bigg | + 6. \bigg | \matrix{3 & -1 \\ 2 & \;\;\, 0} \bigg | - (-4). \bigg | \matrix{3 & -1 \\ 1 & \;\;\, 3} \bigg |$

$= - 2.(1 \times 0 - 3 \times 2) + 6.(3 \times 0 - (-1) \times 2) + 4.(3 \times 3 - (-1) \times 1)$

$= 12 + 12 + 40$

$=$ $64$ as obtained earlier.

Calculate the determinant of the matrix, $A = \pmatrix{3 & \;\;\,2 & -1 \\ 1 & \;\;\,6 & \;\;\, 3 \\ 2 & -4 & \;\;\,0}$ using the Rule of Sarrus.

Solution

Rewrite the first two columns of the matrix to the right of it :

$\matrix{3 & \;\;\,2 & -1 & 3 & \;\;\,2 \\ 1 & \;\;\,6 & \;\;\,3 & 1 & \;\;\,6 \\ 2 & -4 &\;\;\,0 & 2 & -4}$

Calculate the determinant as described above :

$\text{det}(A) = 3 \times 6 \times 0 + 2 \times 3 \times 2 + (-1) \times 1 \times (-4)$

$-\, 2 \times 1 \times 0 - 3 \times 3 \times (-4) - (-1) \times 6 \times 2$

$= 0 + 12 + 4 - 0 - (-36) - (-12)$

$=$ $64$.

Note that, as expected, this is the same answer as we obtained using the previous method.

Now watch the following:

📹 Rule of Sarrus

Aside: Geometrically, the absolute value of the determinant of a $3 \times 3$ matrix is the volume of a parallelepiped whose edges are the vectors $\boldsymbol{u} = (a_1, \,a_2,\,a_3)$, $\boldsymbol{v} = (b_1,\,b_2,\,b_3)$ and $\boldsymbol{w} = (c_1, \,c_2,\,c_3)$.

Determine the inverse of the matrix, $A = \pmatrix{3 & \;\;\, 2 & -1 \\ 1 & \;\;\, 6 & \;\;\, 3 \\ 2 & -4 & \;\;\, 0}$ , if it exists.

Solution

Although we have performed some of the calculations for this example previously for completeness we present a full solution.

STEP 1: Calculate the determinant of $A$

Expanding along the third row the determinant of $A$ is :

$\text{det}(A) = 2. \bigg | \matrix{2 & -1 \\ 6 & \;\;\, 3} \bigg | - (- 4) \bigg | \matrix{3 & -1 \\ 1 & \;\;\, 3} \bigg | + 0. \bigg | \matrix{3 & 2 \\ 1 & 6} \bigg | = 2 \times 12 + 4 \times 10 = 64$.

Since $\text{det}(A) \neq 0,\; A^{-1}$ exists.

STEP 2: Calculate the matrix of minors

The minor of entry $a_{ij}$, denoted by $M_{ij}$, is obtained as follows:

remove the $i^{\text{th}}$ row.

remove the $j^{\text{th}}$ column.

the minor $M_{ij}$ is the determinant of the remaining $2 \times 2$ submatrix.

The minor of entry $a_{11}$ is:

The minor of entry $a_{12}$ is:

The minor of entry $a_{13}$ is:

The minor of entry $a_{21}$ is:

The minor of entry $a_{22}$ is:

The minor of entry $a_{23}$ is:

The minor of entry $a_{31}$ is:

The minor of entry $a_{32}$ is:

The minor of entry $a_{33}$ is:

Hence, the matrix of minor is : $\pmatrix{\;\,12 & -6 & -16 \\-4 & \;\;\,2 & -16 \\ \;\,12 & \;\,10 & \;\;\, 16}$.

STEP 3: Calculate the cofactor matrix

The cofactor of entry $a_{ij}$, denoted by $C_{ij}$, is defined as $C_{ij} = (-1)^{i + j}M_{ij}$.

To obtain the cofactor matrix $cof(A)$ we use a shortcut and simply change signs of the elements of the matrix of minors, calculated in Step 2, using the sign matrix :

$\pmatrix{+ & - & + \\ - & + & - \\ + & - & +}$

Hence, we obtain the cofactor matrix :

$cof(A) = \pmatrix{12 & \;\;\,6 & -16 \\\;\,4 & \;\;\,2 & -16 \\ 12 & -10 & \;\;\, 16}$

STEP 4: Calculate the adjoint matrix

The adjoint of $A$ is defined to be the transpose of the cofactor matrix, i.e.

$adj(A) = \pmatrix{\;\;\,12 & \;\,4 & \;\;\,12 \\ \;\;\;6 & \;\,2 & -10 \\ -16 & 16 & \;\;\,16}$

STEP 5: Calculate the inverse matrix

The inverse of the matrix $A$ is calculated as :

$$A^{-1} = \frac{1}{\text{det}(A)}adj(A)$$

so that $A^{-1} = {\Large\frac{1}{64}} \pmatrix{\;\;\,12 & \;\,4 & \;\;12 \\ \;\;\;6 & \;\,2 & -10 \\-16 & 16 & \;\;16}$ which simplifies to give $A^{-1} = {\Large\frac{1}{32}} \pmatrix{\;\;\;6 & 2 & \;\;\;6 \\ \;\;\;3 & 1 & -5 \\ -8 & 8 & \;\;\;8}$ .

It is straightforward to check that $AA^{-1} = A^{-1}A = I$. We have,

$AA^{-1} = {\Large\frac{1}{32}}\pmatrix{3 & \;\;\;2 & -1 \\ 1 & \;\;\;6 & \;\;\;3 \\ 2 & -4 & \;\;\;0}\pmatrix{\;\;\;6 & 2 & \;\;\;6 \\ \;\;\;3 & 1 & -5 \\ -8 & 8 & \;\;\;8} = {\Large\frac{1}{32}}\pmatrix{32 & \;\;\,0 & \;\;\,0 \\ \;\;0 & 32 & \;\;0 \\ \;\;\,0 & \;\;\,0 & 32} = \pmatrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1} = I$ as required.

Determine the inverse of the matrix $A = \pmatrix{3 & -1 & 6 \\ 9 & -5 & 2 \\ 0 & \;\;\,4 & 7}$ if it exists.

Solution

STEP 1: Calculate the determinant of $A$

$\text{det}(A) = 3. \bigg | \matrix{-5 & 2 \\ \;\;\,4 & 7} \bigg | - (-1). \bigg | \matrix{9 & 2 \\0 & 7} \bigg | + 6. \bigg | \matrix{9 & -5 \\ 0 & \;\;\,4} \bigg | = 3 \times (-43) + 63 + 6 \times 36 = 150.$

Since $\text{det}(A) \neq 0,\; A^{-1}$ exists.

STEP 2: Calculate the matrix of minors

The minor of entry $a_{11}$ is:

The minor of entry $a_{12}$ is:

The minor of entry $a_{13}$ is:

The minor of entry $a_{21}$ is:

The minor of entry $a_{22}$ is:

The minor of entry $a_{23}$ is:

The minor of entry $a_{31}$ is:

The minor of entry $a_{32}$ is:

The minor of entry $a_{33}$ is:

Hence, the matrix of minors is: $\pmatrix{-43 & \;\;63 & \;36 \\ -31 & \;\;21 & \;12 \\ \;\;\,28 & -48 & -6}$

STEP 3: Calculate the cofactor matrix

The cofactor of entry $a_{ij}$, denoted by $C_{ij}$, is defined as $C_{ij} = (-1)^{i+j}M_{ij}$.

To obtain the cofactor matrix $cof(A)$ we simply change signs of the elements of the matrix of minors in Step 2 using the sign matrix :

$\pmatrix{+ & - & + \\ - & + & - \\ + & - & +}$

Hence, we obtain the cofactor matrix :

$cof(A) = \pmatrix{-43 & -63 & \;\;36 \\ \;\;\,31 & \;\;\,21 & -12 \\ \;\;\,28 & \;\;\,48 & \;-6}$.

STEP 4: Calculate the adjoint matrix

The adjoint of $A$ is determined as the transpose of the cofactor matrix, i.e.

$adj(A) = \pmatrix{-43 & \;\;\;31 & \;28 \\ -63 & \;\;\;21 & \;48 \\ \;\;\;36 & -12 & -6}$

STEP 5: Calculate the inverse matrix

The inverse of the matrix $A$ is calculated as

$A^{-1} = {\Large\frac{1}{\text{det}(A)}}adj(A)$

so that $A^{-1} = {\Large\frac{1}{150}}\pmatrix{-43 & \;\;\;31 & \;28 \\ -63 & \;\;\;21 & \;48 \\ \;\;\;36 & -12 & -6}$.

Check the inverse is correct as follows :

$AA^{-1} = {\Large\frac{1}{150}} \pmatrix{3 & -1 & 6 \\ 9 & -5 & 2 \\ 0 & \;\;\;4 & 7} \pmatrix{-43 & \;\;\;31 & \;28 \\ -63 & \;\;\;21 & \;48 \\ \;\;\;36 & -12 & -6} = {\Large\frac{1}{150}}\pmatrix{150 & 0 & 0 \\ 0 & 150 & 0 \\ 0 & 0 & 150} = \pmatrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1} = I$ as required.

Calculate the determinant of the matrix, $A = \pmatrix{1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9}$ and find $A^{-1}$ if it exists.

Solution

As an alternative we use the Rule of Sarrus to calculate the determinant. Writing

$\matrix{1 & 2 & 3 & 1 & 2 \\ 4 & 5 & 6 & 4 & 5 \\ 7 & 8 & 9 & 7 & 8}$

$\text{det}(A) = 1 \times 5 \times 9 + 2 \times 6 \times 7 + 3 \times 4 \times 8 - 2 \times 4 \times 9 - 1 \times 6 \times 8 - 3 \times 5 \times 7$

$= 45 + 84 + 96 - 72 - 48 - 105 = 225 - 225 = 0$.

As $\text{det}(A) = 0$ the matrix $A$ is not invertible.

Aside: Geometrically the result that $\det(A) = 0$ means that the vectors $(1, 2, 3)$, $(4, 5, 6)$, and $(7, 8, 9)$ are coplanar, i.e. they lie in the same plane.

(i). Solve the system of equations

$x + 2y = - 1$

$4x - 3y = 18$

using the inverse matrix method.

Solution

First write the equations in the matrix form, $A\boldsymbol{x} = \boldsymbol{b}$ :

$\pmatrix{1 & \;\;\,2 \\ 4 & -3} \pmatrix{x \\ y} = \pmatrix{-1 \\ \;18} \\ \;\;\;\;\;\;\; A \;\;\;\;\;\;\;\;\;\, \boldsymbol{x} \;\;\;=\;\;\;\;\; \boldsymbol{b}$

Calculate $A^{-1}$ if it exists.

$\text{det}(A) = 1 \times (-3) - 2 \times 4 = - 11 \neq 0$ and so the matrix is invertible.

Hence,

$A^{-1} = - {\Large\frac{1}{11}}\pmatrix{-3 & -2 \\ -4 & \;\;\;1} = {\Large\frac{1}{11}} \pmatrix{3 & \;\;\;2 \\ 4 & -1}$.

Now solve the system by forming, $\boldsymbol{x} = A^{-1} \boldsymbol{b}$:

$\pmatrix{x \\ y} = {\Large\frac{1}{11}} \pmatrix{3 & \;\;\;2 \\ 4 & -1}\pmatrix{-1 \\ \;18} = {\Large\frac{1}{11}} \pmatrix{\;\;\,33 \\ -22} = \pmatrix{\;\;\,3 \\ -2} \\ \\ \;\;\boldsymbol{x} \;\;\; = \;\;\;\;\;\;\;\;\;\;\;A^{-1}\;\;\;\;\;\;\;\; \boldsymbol{b}$

Hence, $x = 3$ and $y = -2$.

(ii). Solve the system of equations

$4x + 2y = 10$

$3x - 7y = -1$

using the inverse matrix method.

Solution

First write the equations in the matrix form, $A \boldsymbol{x} = \boldsymbol{b}$ :

$\pmatrix{4 & \;\;\;2 \\ 3 & -7} \pmatrix{x \\ y} = \pmatrix{\;10 \\ -1}$.

Calculate $A^{-1}$ if it exists.

$\det(A) = -28 - 6 = - 34 \neq 0$ and so the matrix is invertible.

Hence,

$A^{-1} = - {\Large\frac{1}{34}}\pmatrix{-7 & -2 \\ -3 & \;\;\;4} = {\Large\frac{1}{34}} \pmatrix{7 & \;\;\;2 \\ 3 & -4}$.

Now solve the system by forming, $\boldsymbol{x} = A^{-1} \boldsymbol{b}$:

$\pmatrix{x \\ y} = {\Large\frac{1}{34}} \pmatrix{7 & \;\;\;2 \\ 3 & -4} \pmatrix{\;10 \\ -1} = {\Large\frac{1}{34}} \pmatrix{68 \\ 34} = \pmatrix{2 \\ 1}$.

Hence, $x = 2$ and $y = 1$.

(iii). Solve the system of equations given by

$3x + 2y - \;\;z = \;\;\;3$

$\;\;x + 6y + 3z = \;\;29$.

$2x - 4y \;\;\;\;\;\;\;\;= -18$

using the inverse matrix method.

Solution

In matrix form, $A \boldsymbol{x} = \boldsymbol{b}$, we have

$\pmatrix{3 &\;\;\; 2 & -1 \\ 1 & \;\;\;6 & \;\;\;3 \\ 2 & -4 & \;\;\;0} \pmatrix{x \\ y \\ z} = \pmatrix{\;\;\;3 \\ \;\;29 \\ -18}$.

We found the inverse of $A$ in Example 26, $A^{-1} = {\Large\frac{1}{32}} \pmatrix{\;\;\;6 & 2 & \;\;\;6 \\ \;\;\;3 & 1 & -5 \\ -8 & 8 & \;\;\; 8}$.

Now solve the system by forming, $\boldsymbol{x} = A^{-1} \boldsymbol{b}$ :

$\pmatrix{x \\ y \\ z} = {\Large\frac{1}{32}} \pmatrix{\;\;\; 6 & 2 & \;\;\;6 \\ \;\;\;3 & 1 & -5 \\ -8 & 8 & \;\;\;8} \pmatrix{\;\;\;\,3 \\ \;\;29 \\ -18} = {\Large\frac{1}{32}} \pmatrix{\;\;\;\,3 \\ \;\;29 \\ -18} = {\Large\frac{1}{32}} \pmatrix{-32 \\ \;128 \\ \;\;64} = \pmatrix{-1 \\ \;\;\; 4 \\ \;\;\; 2}$.

Hence, $x = -1$,  $y = 4$  and  $z = 2$.