We're asked to determine if the following
three by three matrices are invertible or nonsingular.
A matrix is invertible if it has an inverse matrix.
Looking at the notes below,
matrix A is invertible or nonsingular
if matrix A is an N by N matrix,
so the matrix must be a square matrix,
such that eight times some matrix B
equals B times A, which equals I, the identity matrix,
where again I sub N is the identity matrix,
and B is the inverse of matrix A.
And we write B equals A inverse using this notation here.
The easiest way to determine
if a square matrix is invertible
is by using the determinants.
If matrix A is invertible or nonsingular,
then the determinant of matrix A does not equal zero.
We'll find the determinant of matrix A and matrix C
to determine if it is invertible, or nonsingular.
For a quick review, there are two main methods
to find a determinant.
The diagonal method which is shown here,
we'll find one determinant using this method,
and also, the method of expansion by minors,
often called the method of cofactors.
We'll find the second determinant
using this method here for review.
So going back to our question,
let's first find the determinant of matrix A
using the diagonal method.
To apply the diagonal method to find the determinant,
we first copy down the elements from matrix A,
and then we copy down column one and column two again
to the right of column three.
Column one is one, three, negative one,
and column two is negative two, five, three.
There are three diagonals from the upper left
to the lower right
and three diagonals from the upper right to the lower left.
From the upper right to the lower left,
here's one, two,
three diagonals.
And then from the upper right to the lower left
we also have three diagonals.
One, two, and three.
So to find the value of the determinant of matrix A,
we want to find the sum of the products of the diagonals
from the upper left to the lower right, or the blue ones,
and then subtract the sum of the products
of the diagonals from the upper right to the lower left,
or the green diagonals.
So looking at the diagonals from the upper left
to the lower right, we have one times five
times negative four, that's negative 20,
plus negative two times two times negative one.
That's positive four, plus three times three times three,
that's 27, and then we have minus.
Now we want to find the sum of the products
of the diagonals from the upper right to the lower left,
or the green diagonals.
We have three times five times negative one,
that's negative 15, plus one times two times three,
that's six, plus negative two times three
times negative four, that's positive 24.
Now simplifying, here we have negative 20,
plus four, plus 27.
That's 11, minus,
here we have negative 15 plus six plus 24.
That's positive 15.
11 minus 15 equals negative four,
which does not equal zero,
which means matrix A is invertible, or nonsingular.
Meaning matrix A does have an inverse
where A times A inverse, or A inverse times A
would be equal to the three by three identity matrix.
Notice in this question we're not asked
to find the inverse matrix.
So we just say matrix A is invertible.
Now I'll find the determinant of matrix C,
but for this determinant, let's use
the expansion by minors method,
or the method of cofactors given here.
You may want to pause this,
but if I use this first row for the expansion by minors,
we can use this formula here to evaluate the determinant
of the three by three matrix.
So the determinant of matrix C is going to be equal to
going across the first row we'd have one,
and then we're going to multiply this
by a two by two determinant,
where the elements in this determinant are found
by eliminating the row and column of element one.
So we eliminate row one, column one,
leaving us with these four elements.
So we have seven, negative 10, 16, negative 21,
and then we have minus, again going across row one,
the next element is two.
And we'll have times a two by two determinant.
Where to find the elements in this determinant
we eliminate the row and the column of this element here,
so we eliminate row one, column two.
So we have three, negative 10, seven, negative 21.
And then we have plus, the last element
in row one is negative one,
so we'll have plus, and then negative one times
a two by two determinant formed by eliminating
row one, and now column three.
So we have three, seven, seven, 16.
Now to evaluate these two by two determinant,
we'll find this product minus this product.
Here we have one times, we have seven times negative 21,
which is equal to negative 147,
minus negative 10 times 16.
That's minus negative 160, which equals positive 13.
So here we have one times 13, minus two times
here we have three times negative 21.
That's negative 63, minus negative 10 times seven.
That's negative 70.
This simplifies to positive seven,
so we have minus two times seven.
Let's write this as just minus one times
here we have three times 16, that's 48,
minus seven times seven, that's 49,
which equals negative one.
Simplifying one last time here,
we have 13 minus 14, plus one, which equals zero.
So here the determinant of matrix A is equal to zero,
which means matrix C is not invertible,
meaning matrix C does not have an inverse.
So we'll say matrix C
is not invertible.
We can also say that matrix C is singular
because if a matrix is invertible,
we can say it's nonsingular.
I hope you found this helpful.