Which matrices have a determinant, and what kind of object is it?

Only square matrices have a determinant. It is a single number , which may be positive, negative or zero, worked out from the elements of the matrix.

A LevelFurther MathsEdexcelCore PureFlashcardsMatricesProperties of Matrices

Properties of Matrices (Edexcel A Level Further Maths: Core Pure): Flashcards

Exam code: 9FM0

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FrontIntroduction to Matrices
Define the order of a matrix.

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Define the order of a matrix.
The order of a matrix is its number of rows followed by its number of columns, written $m \times n$ .
So a matrix with 2 rows and 3 columns has order $2 \times 3$ , and the rows are always quoted first.
In the notation $a_{i j}$ for the elements of a matrix, what do $i$ and $j$ refer to?
$i$ is the row number and $j$ is the column number, so $a_{i j}$ is the element sitting in row $i$ and column $j$ .
In $(\begin{matrix} 5 & - 3 & 7 \\ - 1 & 2 & 4 \end{matrix})$ the element $a_{23}$ is $4$ .
Complete the descriptions of these two special matrices:
A zero matrix has every element equal to $______$ and an identity matrix is square with $______$ all along its leading diagonal and zeros everywhere else.
The completed descriptions are: a zero matrix has every element equal to $0$ , and an identity matrix is square with $1$ all along its leading diagonal and zeros everywhere else.
For $2 \times 2$ matrices these are $(\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix})$ and $(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})$ .
How is the transpose $A^{T}$ formed from the matrix $A$ itself?
Interchange the rows and the columns, so that row 1 of $A$ becomes column 1 of $A^{T}$ .
An $m \times n$ matrix therefore transposes into an $n \times m$ matrix.
True or False?
For any two matrices of the same order, $A + B = B + A$ .
True.
Addition works element by element, and adding two numbers gives the same answer either way round.
The two matrices must be of the same order for the sum to exist at all.
When can two matrices be added, and what is the order of the result?
They can be added only when they have the same order, and then corresponding elements are added.
The answer has that same order, so adding two $3 \times 2$ matrices gives another $3 \times 2$ matrix.
When does the product $A B$ exist, and what is its order?
It exists only when the number of columns of $A$ equals the number of rows of $B$ .
If $A$ is $m \times n$ and $B$ is $n \times p$ then $A B$ has order $m \times p$ .
How is each individual element of a matrix product worked out?
The element in row $i$ , column $j$ of $A B$ comes from row $i$ of $A$ paired with column $j$ of $B$ .
Multiply the entries in pairs as you move along the row and down the column, then add all those products together.
$A$ has order $2 \times 3$ and $B$ has order $3 \times 2$ . Why can $A B$ and $B A$ not possibly be equal?
$A B$ is $2 \times 2$ while $B A$ is $3 \times 3$ , so they are not even the same size.
Even for two square matrices of the same order the two products are usually different, because matrix multiplication is not commutative.
Why is ${(A + B)}^{2}$ not equal to $A^{2} + 2 A B + B^{2}$ in general?
Multiplying out gives $A^{2} + A B + B A + B^{2}$ , and the two middle terms cannot be collected into $2 A B$ .
The safe route is to add the matrices inside the bracket first and then multiply that single matrix by itself.
For $A = [\begin{matrix} a & b \\ c & d \end{matrix}]$ complete the determinant:
$\det A =______-______$
The completed determinant is:
$\det A = a d - b c$
It is the product along the leading diagonal minus the product along the other diagonal.
Which matrices have a determinant, and what kind of object is it?
Only square matrices have a determinant.
It is a single number, which may be positive, negative or zero, worked out from the elements of the matrix.
Define the minor of an element of a $3 \times 3$ matrix.
The minor of an element is the determinant of the $2 \times 2$ matrix that is left when you cross out the row and the column containing that element.
So the minor of $a$ in $[\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}]$ is $e i - f h$ .
How do you use minors to find the determinant of a $3 \times 3$ matrix?
Choose any one row or column, multiply each of its elements by its own minor, and combine the products using the matrix of signs $[\begin{matrix} + & - & + \\ - & + & - \\ + & - & + \end{matrix}]$ .
The top row is the most convenient choice, because its signs are always $+, -, +$ .
Why is it worth choosing a row or column that contains a zero?
Any element equal to zero contributes nothing to the sum, so its minor never has to be worked out at all.
With one zero in the chosen line only two $2 \times 2$ determinants are needed instead of three.
For $n \times n$ matrices complete these two determinant properties:
$\det (k A) =______\det (A) and \det (A B) =______$
The completed properties are:
$\det (k A) = k^{n} \det (A) and \det (A B) = \det (A) \det (B)$
The index $n$ appears because every one of the $n$ rows gets multiplied by $k$ , not just one of them.
True or False?
For two square matrices of the same order, $A B$ and $B A$ always have the same determinant.
True.
Both determinants come out as $\det (A) \times \det (B)$ , and multiplying two numbers gives the same answer either way round.
So the determinants agree even though the two product matrices usually do not.
What are the determinants of an identity matrix and of a zero matrix?
The determinant of an identity matrix is $1$ and the determinant of a zero matrix is $0$ .
Both hold whatever the size, so $\det (I) = 1$ for a $3 \times 3$ identity just as it does for a $2 \times 2$ one.
Define a singular matrix.
A matrix is singular when its determinant is equal to $0$ .
A square matrix whose determinant is not zero is described as non-singular.
Given that $\det (A) = 5$ what is the determinant of the inverse matrix $A^{- 1}$ equal to?
It is $\frac{1}{5}$ , because $\det (A^{- 1}) = \frac{1}{\det (A)}$ .
That follows from applying $\det (A B) = \det (A) \det (B)$ to the statement $A A^{- 1} = I$ .
Define the inverse $A^{- 1}$ of a square matrix.
The inverse $A^{- 1}$ is the matrix for which $A A^{- 1} = A^{- 1} A = I$ .
Multiplying by it undoes multiplying by $A$ , which makes it the matrix equivalent of a reciprocal.
Which square matrices have an inverse, and how can you tell?
A square matrix has an inverse exactly when its determinant is not zero.
A singular matrix, one whose determinant is $0$ , has no inverse at all, so the determinant is always the first thing to check.
True or False?
The inverse of a product is the product of the inverses, so ${(A B)}^{- 1}$ equals $A^{- 1} B^{- 1}$ in that order.
False.
The order reverses, giving ${(A B)}^{- 1} = B^{- 1} A^{- 1}$ .
Written that way the product with $A B$ cancels from the inside outwards, leaving the identity.
To get $B$ out of the equation $A B = C$ which side do you multiply by $A^{- 1}$ on?
Multiply on the left, giving $A^{- 1} A B = A^{- 1} C$ and so $B = A^{- 1} C$ .
The side matters because matrix multiplication is not commutative: for $B A = C$ you would multiply on the right instead, giving $B = C A^{- 1}$ .
For $A = [\begin{matrix} a & b \\ c & d \end{matrix}]$ complete the inverse:
$A^{- 1} = \frac{1}{\det A} [\begin{matrix} ______ & ______ \\ - c & a \end{matrix}]$
The completed inverse is:
$A^{- 1} = \frac{1}{\det A} [\begin{matrix} d & - b \\ - c & a \end{matrix}]$
Swap the two entries on the leading diagonal, change the sign of the other two, then divide by the determinant.
Why does the $2 \times 2$ inverse involve dividing by the determinant?
Multiplying $[\begin{matrix} a & b \\ c & d \end{matrix}]$ by $[\begin{matrix} d & - b \\ - c & a \end{matrix}]$ gives $[\begin{matrix} a d - b c & 0 \\ 0 & a d - b c \end{matrix}]$ , which is the determinant times the identity.
Dividing by the determinant turns that into $I$ exactly, and it is impossible when the determinant is $0$ .
Two matrices satisfy $M N = k I$ for some non-zero constant $k$ . What is the inverse of $M$ in terms of $N$ ?
Dividing both sides by $k$ gives $M (\frac{1}{k} N) = I$ , so $M^{- 1} = \frac{1}{k} N$ .
The two matrices are almost inverses of each other already, and the scalar $\frac{1}{k}$ is all that is needed to finish the job.
Why is the determinant the first thing to work out when inverting a $3 \times 3$ matrix?
It is the number you divide by at the very end, so it is needed whatever else happens.
It also settles immediately whether there is an inverse to find at all, before a single minor has been calculated.
Define the matrix of cofactors of a $3 \times 3$ matrix.
It is the matrix of minors with the alternating matrix of signs applied to it, so each minor either keeps its value or has its sign reversed according to its position.
The cofactors are what the inverse is built from, once they have been transposed.
Once the matrix of cofactors $C$ has been found, how is $A^{- 1}$ formed from it?
Transpose $C$ to get the adjugate $C^{T}$ , then divide by the determinant, so that $A^{- 1} = \frac{1}{\det A} C^{T}$ .
It is usually tidier to leave the answer as a scalar multiple of a whole-number matrix than to write out nine separate fractions.

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