Lesson Explainer: Solving a System of Two Equations Using a Matrix Inverse Mathematics

In this explainer, we will learn how to solve a system of two linear equations using the inverse of the matrix of coefficients.

We can solve a system of two linear equations, which are also called simultaneous equations, using the substitution or elimination methods, so it is fair to ask why we need to learn a different method to solve the same system. In fact, using a matrix inverse to solve a system of two linear equations is more involved than the previous two methods, which further justifies this question. We are studying this method as a model to understand the relationship between a system of linear equations and matrices. Since the system of two linear equations is the simplest model that relates a system of equations to matrices, it makes sense to start here.

The method we will learn in this explainer can be used for a system containing a larger number of linear equations and unknown variables, although we will not discuss larger systems here. While it is not too difficult to solve a system of two linear equations without using matrices, it is more challenging to do this when we have three or more equations involved. Understanding the relationship between a system of linear equations and matrices lets us organize the given system of equations into a concise matrix equation, which can be solved using a method analogous to what we will discuss here.

Before we discuss how to solve simultaneous equations using matrices, we need to understand how to solve a matrix equation. Recall the inverse matrix.

Definition: Inverse Matrices

Given a square matrix 𝐴, the inverse matrix 𝐴 is a square matrix of the same order satisfying 𝐴𝐴=𝐴𝐴=𝐼, where 𝐼 is the identity matrix of the same order. If such a matrix exists, we say that matrix 𝐴 is invertible.

Consider a matrix equation 𝐴𝑋=𝐡, where 𝐴 and 𝐡 are known 𝑛×𝑛 and π‘›Γ—π‘˜ matrices, respectively, and 𝑋 is an unknown π‘›Γ—π‘˜ matrix. We further assume that 𝐴 is an invertible matrix. We know that, in order to multiply a pair of matrices, the number of columns of the first matrix must equal the number of rows of the second matrix. We can see that the matrix multiplication 𝐴𝑋 is well defined.

Since matrix 𝐴 is invertible, there exists an 𝑛×𝑛 inverse matrix 𝐴. Multiplying from the left by 𝐴 on both sides of the equation 𝐴𝑋=𝐡, we obtain

𝐴𝐴𝑋=𝐴𝐡.(1)

On the left-hand side of the equation, we know that 𝐴𝐴=𝐼, where 𝐼 is the multiplicative identity. Hence, 𝐴𝐴𝑋=𝐼𝑋=𝑋.

Substituting this expression to the left-hand side of equation (1), we can write 𝑋=𝐴𝐡.

Both 𝐴 and 𝐡 are known matrices; hence, this gives the solution to the matrix equation 𝐴𝑋=𝐡.

How To: Solving Matrix Equations

Let 𝐴 be an invertible matrix and 𝐡 be a matrix such that the multiplication 𝐴𝐡 is defined. Matrix 𝑋 satisfying the equation 𝐴𝑋=𝐡 is given by 𝑋=𝐴𝐡.

This method gives us a way to solve any matrix equation of the form 𝐴𝑋=𝐡 if matrix 𝐴 is invertible. However, this method cannot be used when 𝐴 is not invertible. This could happen if 𝐴 is not a square matrix or if 𝐴 is square and det𝐴=0. In such cases, the matrix equation has either an infinite number of solutions or no solution. For a simple example, we can think of the case where 𝐴=𝑂, where 𝑂 is a zero matrix.

We know that 𝑂 is not invertible since det𝑂=0. The matrix equation 𝑂𝑋=𝐡 has no solution if 𝐡 is a nonzero matrix, since the multiplication of a zero matrix by any matrix results in a zero matrix. On the other hand, if 𝐡 is a zero matrix of the natural order resulting from this matrix multiplication, any matrix 𝑋 satisfies the equation 𝑂𝑋=𝑂. This means that this matrix equation has an infinite number of solutions.

In our first example, we will solve a matrix equation using the inverse matrix.

Example 1: Solving Equations of Matrices Using Their Inverses

Given that 𝐴=ο€Ό2βˆ’5βˆ’8βˆ’9,𝐴π‘₯𝑦=ο€Όβˆ’28, what is the value of 𝑦?

Answer

In this example, we are given a matrix equation. Matrix ο€»π‘₯𝑦 is an unknown matrix. If we find this matrix, we can find the value of 𝑦.

The example does not give us what matrix 𝐴 is, but it gives us the inverse of this matrix 𝐴. Recall that the inverse of a square matrix 𝐴, if it exists, is a matrix satisfying 𝐴𝐴=𝐴𝐴=𝐼, where 𝐼 is the identity matrix. We can multiply from the left by 𝐴 on both sides of the given equation to obtain 𝐴𝐴π‘₯𝑦=π΄ο€Όβˆ’28.

We know that 𝐴𝐴=𝐼, which is a multiplicative identity, so we can neglect the factor 𝐴𝐴 and fill in the provided expression for 𝐴 on the right-hand side to write ο€»π‘₯𝑦=ο€Ό2βˆ’5βˆ’8βˆ’9οˆο€Όβˆ’28.

Computing this matrix multiplication, we obtain ο€»π‘₯𝑦=ο€Ύ2Γ—(βˆ’2)+(βˆ’5)Γ—8(βˆ’8)Γ—(βˆ’2)+(βˆ’9)Γ—8=ο€Όβˆ’44βˆ’56.

This leads to the unknown matrix. We know that a pair of matrices are equal if each pair of corresponding entries in the matrices are equal. Hence, this leads to π‘₯=βˆ’44,𝑦=βˆ’56.

In particular, the example asks for the value of 𝑦, which is 𝑦=βˆ’56.

In the previous example, we solved a matrix equation when we were given the inverse matrix 𝐴. If we are not provided the expression for 𝐴, we can find the inverse matrix by using the following formula, as long as det𝐴≠0.

Formula: Inverse of a 2 Γ— 2 Matrix

Let 𝐴=ο€Όπ‘Žπ‘π‘π‘‘οˆ such that det𝐴≠0. Then, 𝐴=1π΄ο€»π‘‘βˆ’π‘βˆ’π‘π‘Žο‡,det where det𝐴=π‘Žπ‘‘βˆ’π‘π‘. If det𝐴=0, matrix 𝐴 is not invertible.

Let us consider an example where we solve a matrix equation by first finding the inverse of a 2Γ—2 matrix.

Example 2: Solving Equations of Matrices Using Their Inverses

Given that ο€Ό581βˆ’8οˆο€»π‘₯𝑦=ο€Όβˆ’431, determine the values of π‘₯ and 𝑦.

Answer

In this example, we are given a matrix equation. Matrix ο€»π‘₯𝑦 is an unknown matrix. If we find this matrix, we can find the values of π‘₯ and 𝑦.

We recall that, given matrices 𝐴 and 𝐡, an unknown matrix 𝑋 satisfying the equation 𝐴𝑋=𝐡 is given by 𝑋=𝐴𝐡, if the inverse matrix 𝐴 exists and the matrix multiplication 𝐴𝐡 can be defined. In our given example, this matrix 𝐴 corresponds to the 2Γ—2 matrix ο€Ό581βˆ’8. Hence, this can be written as

ο€»π‘₯𝑦=ο€Ό581βˆ’8οˆο€Όβˆ’431,(2)

if the inverse matrix exists and the matrix multiplication is well defined. Hence, we need to begin by finding the inverse of this matrix, if it exists.

We know that the inverse of a square matrix exists if its determinant is not equal to zero. Let us first compute the determinant of this matrix. We know that detο€Όπ‘Žπ‘π‘π‘‘οˆ=π‘Žπ‘‘βˆ’π‘π‘.

Applying this formula to the given 2Γ—2 matrix, detο€Ό581βˆ’8=5Γ—(βˆ’8)βˆ’8Γ—1=βˆ’48.

Since the determinant is nonzero, we can proceed to find its inverse. We recall the formula for the inverse of a 2Γ—2 matrix: 𝐴=1π΄ο€»π‘‘βˆ’π‘βˆ’π‘π‘Žο‡.det

Hence, using the determinant of the matrix we found earlier, ο€Ό581βˆ’8=1βˆ’48ο€Όβˆ’8βˆ’8βˆ’15.

Substituting this expression into equation (2), ο€»π‘₯𝑦=1βˆ’48ο€Όβˆ’8βˆ’8βˆ’15οˆο€Όβˆ’431.

We know that, in order to multiply a pair of matrices, the number of columns of the first matrix must equal the number of rows of the second matrix. Computing this matrix multiplication, we obtain ο€»π‘₯𝑦=1βˆ’48ο€Ύ(βˆ’8)Γ—(βˆ’43)+(βˆ’8)Γ—1(βˆ’1)Γ—(βˆ’43)+5Γ—1=1βˆ’48ο€Ό33648.

Finally, computing the scalar multiplication, ο€»π‘₯𝑦=βŽ›βŽœβŽœβŽ336βˆ’4848βˆ’48⎞⎟⎟⎠=ο€Όβˆ’7βˆ’1.

This leads to the unknown matrix. We know that a pair of matrices are equal if each pair of corresponding entries in the matrices are equal. Hence, this leads to π‘₯=βˆ’7,𝑦=βˆ’1.

So far, we have considered a few examples where we solved matrix equations using the inverse matrix. Let us turn our attention to the system of two linear equations.

How To: Representing Systems of Two Equations as Matrix Equations

Consider a system of equations given by π‘Žπ‘₯+𝑏𝑦=𝑒,𝑐π‘₯+𝑑𝑦=𝑓, for some known constants π‘Ž, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓. We can write this system of two equations as one matrix equation ο€Όπ‘Žπ‘π‘π‘‘οˆο€»π‘₯𝑦=ο€Όπ‘’π‘“οˆ.

If we carry out the matrix multiplication on the left-hand side of the matrix equation, this is the same as ο€Ύπ‘Žπ‘₯+𝑏𝑦𝑐π‘₯+π‘‘π‘¦οŠ=ο€Όπ‘’π‘“οˆ.

Equating the corresponding entries of the matrices on both sides of this equation leads back to the system of two linear equations. Hence, this matrix equation is equivalent to the system of two linear equations. Since we can write the system of equations as a matrix equation, we can solve this system using the matrix inverse.

We can see that the coefficients of π‘₯ and 𝑦 in the system of equations became the 2Γ—2 matrix in the matrix equation. This is called the coefficient matrix, because its entries come from the coefficients of the simultaneous equations. When writing down the coefficient matrix, we need to be careful with the order of the entries. Since the variable matrix has π‘₯ as its first entry, the coefficients of π‘₯ go in the first column. Hence, the same coefficient matrix is used even if the first equation in the system is written as 𝑏𝑦+π‘Žπ‘₯=𝑒. Rather than following the order of the coefficients written in the given equation, we need to consider which variable it is the coefficient of.

We also note that the column matrix on the right-hand side of the matrix equation contains the constant terms from the right-hand sides of the simultaneous equations. The order of these constants must be consistent with the coefficient matrix. Since the coefficients of the first equation, π‘Žπ‘₯+𝑏𝑦=𝑒, are written in the first row of the coefficient matrix, the constant 𝑒 from this equation must also appear in the first row of this matrix equation.

Just like we discussed when solving matrix equations, this also means that the system of equations has either no solution or an infinite number of solutions when the coefficient matrix is not invertible.

In the next example, we will write a pair of simultaneous equations into a matrix equation and then solve the matrix equation using the matrix inverse.

Example 3: Solving a Pair of Simultaneous Equations Using Matrices

Consider the simultaneous equations 4π‘₯βˆ’2𝑦=0,3𝑦+5π‘₯=βˆ’11.

  1. Express the given simultaneous equations as a matrix equation.
  2. Write down the inverse of the coefficient matrix.
  3. Multiply through by the inverse, on the left-hand side, to solve the matrix equation.

Answer

Part 1

Recall that a pair of simultaneous equations given by π‘Žπ‘₯+𝑏𝑦=𝑒,𝑐π‘₯+𝑑𝑦=𝑓 can be written as the matrix equation ο€Όπ‘Žπ‘π‘π‘‘οˆο€»π‘₯𝑦=ο€Όπ‘’π‘“οˆ.

Here, it is important to note that the coefficient matrix ο€Όπ‘Žπ‘π‘π‘‘οˆ is the coefficient of the simultaneous equations in the order given in the variable matrix ο€»π‘₯𝑦. This means that the first column of the coefficient matrix contains the coefficients of variable π‘₯, while the second column contains the coefficients of variable 𝑦. In particular, we should first notice that 5π‘₯ and 3𝑦 are written in the opposite order. We can rearrange this pair of simultaneous equations to say 4π‘₯βˆ’2𝑦=0,5π‘₯+3𝑦=βˆ’11.

Then, we can write ο€Ό4βˆ’253οˆο€»π‘₯𝑦=ο€Ό0βˆ’11.

Part 2

In this part, we need to find the inverse of the coefficient matrix. In the previous part, we found the coefficient matrix ο€Ό4βˆ’253. We know that the inverse of a square matrix exists if its determinant is not equal to zero. Let us first compute the determinant of this matrix. We know that detο€Όπ‘Žπ‘π‘π‘‘οˆ=π‘Žπ‘‘βˆ’π‘π‘.

Applying this formula to the coefficient matrix, detο€Ό4βˆ’253=4Γ—3βˆ’(βˆ’2)Γ—5=22.

Since the determinant is nonzero, we can proceed to find its inverse. We recall the formula for the inverse of a 2Γ—2 matrix: 𝐴=1π΄ο€»π‘‘βˆ’π‘βˆ’π‘π‘Žο‡.det

Hence, using the determinant of the matrix we found earlier, ο€Ό4βˆ’253=122ο€Ό32βˆ’54.

Part 3

In this part, we need to multiply through by the inverse on the left-hand side and solve the matrix equation. We begin with the matrix equation ο€Ό4βˆ’253οˆο€»π‘₯𝑦=ο€Ό0βˆ’11.

Multiplying both sides of the equation by the inverse of the coefficient matrix, we have ο€Ό4βˆ’253οˆο€Ό4βˆ’253οˆο€»π‘₯𝑦=ο€Ό4βˆ’253οˆο€Ό0βˆ’11.

On the left-hand side of this equation, the inverse of the coefficient matrix is multiplied by the coefficient matrix. Recall that, for any invertible matrix 𝐴, we have 𝐴𝐴=𝐼, where 𝐼 is the identity matrix. This means ο€Ό4βˆ’253οˆο€Ό4βˆ’253=𝐼.

Since 𝐼 is the identity matrix, which is multiplied by the variable matrix, we can neglect this term. This leads to ο€»π‘₯𝑦=ο€Ό4βˆ’253οˆο€Ό0βˆ’11.

We can now substitute the inverse matrix from the previous part: ο€»π‘₯𝑦=122ο€Ό32βˆ’54οˆο€Ό0βˆ’11.

Computing this matrix multiplication, we obtain ο€»π‘₯𝑦=122ο€Ύ3Γ—0+2Γ—(βˆ’11)βˆ’5Γ—0+4Γ—(βˆ’11)=122ο€Όβˆ’22βˆ’44.

Finally, computing the scalar multiplication, ο€»π‘₯𝑦=βŽ›βŽœβŽœβŽβˆ’2222βˆ’4422⎞⎟⎟⎠=ο€Όβˆ’1βˆ’2.

Hence, the solution to the matrix equation is ο€»π‘₯𝑦=ο€Όβˆ’1βˆ’2.

In the previous example, we solved the matrix equation corresponding to a given pair of simultaneous equations. While we did not explicitly verify this, it can be shown that the values we found for π‘₯ and 𝑦 satisfy the given simultaneous equations. In the next problem, we will solve a pair of simultaneous equations by using matrices.

Example 4: Solving a System of Two Equations Using Matrices

Use matrices to solve the system βˆ’π‘₯+5𝑦=8,βˆ’3π‘₯+𝑦=8.

Answer

In this example, we need to solve the system of two linear equations by using matrices. We know that we can write a system of two linear equations into an equivalent matrix equation. Let us recall this process. Given the system of equations π‘Žπ‘₯+𝑏𝑦=𝑒,𝑐π‘₯+𝑑𝑦=𝑓, we can write an equivalent matrix equation ο€Όπ‘Žπ‘π‘π‘‘οˆο€»π‘₯𝑦=ο€Όπ‘’π‘“οˆ.

Here, it is important to note that the coefficient matrix ο€Όπ‘Žπ‘π‘π‘‘οˆ is the coefficient of the simultaneous equations in the order given in the variable matrix ο€»π‘₯𝑦. This means that the first column of the coefficient matrix contains the coefficients of variable π‘₯, while the second column contains the coefficients of variable 𝑦.

We note that the variable π‘₯ in the first equation is only accompanied by a negative sign. This indicates that the coefficient of π‘₯ in this equation is βˆ’1. Also, the variable 𝑦 in the second equation does not display any coefficients meaning that its coefficient is equal to 1. We can rewrite this pair of simultaneous equations with this information: βˆ’1π‘₯+5𝑦=8,βˆ’3π‘₯+1𝑦=8.

Then, we can write the matrix equation

ο€Όβˆ’15βˆ’31οˆο€»π‘₯𝑦=ο€Ό88.(3)

We can solve this matrix equation by multiplying from the left the inverse of the coefficient matrix ο€Όβˆ’15βˆ’31 if it exists. We know that the inverse of a square matrix exists if its determinant is not equal to zero. Let us first compute the determinant of this matrix. We know that detο€Όπ‘Žπ‘π‘π‘‘οˆ=π‘Žπ‘‘βˆ’π‘π‘.

Applying this formula to the coefficient matrix, detο€Όβˆ’15βˆ’31=(βˆ’1)Γ—1βˆ’5Γ—(βˆ’3)=14.

Since the determinant is nonzero, we can proceed to find its inverse. We recall the formula for the inverse of a 2Γ—2 matrix: 𝐴=1π΄ο€»π‘‘βˆ’π‘βˆ’π‘π‘Žο‡.det

Hence, using the determinant of the matrix we found earlier, ο€Όβˆ’15βˆ’31=114ο€Ό1βˆ’53βˆ’1.

Recall that, for any invertible matrix 𝐴, we have 𝐴𝐴=𝐼, where 𝐼 is the identity matrix. This means that we will be able to remove the coefficient matrix from the left-hand side of equation (3) by multiplying from the left the inverse of the coefficient matrix. Multiplying both sides of the equation by the inverse of the coefficient matrix, we have ο€»π‘₯𝑦=114ο€Ό1βˆ’53βˆ’1οˆο€Ό88.

We know that, in order to multiply a pair of matrices, the number of columns of the first matrix must equal the number of rows of the second matrix. We can see that the matrix multiplication on the right-hand side of the equation above is well defined. Computing this matrix multiplication, we obtain ο€»π‘₯𝑦=114ο€Ύ1Γ—8+(βˆ’5)Γ—83Γ—8+(βˆ’1)Γ—8=114ο€Όβˆ’3216.

Finally, computing the scalar multiplication, ο€»π‘₯𝑦=βŽ›βŽœβŽœβŽβˆ’32141614⎞⎟⎟⎠=βŽ›βŽœβŽœβŽβˆ’16787⎞⎟⎟⎠.

This gives us the solution of the matrix equation ο€»π‘₯𝑦=βŽ›βŽœβŽœβŽβˆ’16787⎞⎟⎟⎠.

We know that a pair of matrices are equal if each pair of corresponding entries are equal. Hence, we obtain the solution to the given system of equations π‘₯=βˆ’167,𝑦=87.

In our final example, we will solve a real-world problem involving a system of two equations using matrices.

Example 5: Solving Two Equations with Two Unknowns Using Matrices

The length of a rectangle is 6 cm more than twice its width, and twice its length is 39 cm more than its width. Given this, use matrices to determine the perimeter of the rectangle.

Answer

In this example, we need to find the perimeter of the rectangle whose length and width are related according to the given description. We know that the perimeter of a rectangle is given by twice the sum of its length and width. Let us begin by denoting the length and width of the rectangle by unknown constants π‘₯ and 𝑦 respectively. Then, perimeter=2(π‘₯+𝑦).

Let us begin by writing down the relationship between π‘₯ and 𝑦 in the form of equations. First, we are given that the length of a rectangle is 6 cm more than twice its width. This can be written as π‘₯=2𝑦+6.

Second, we are given that twice the length is 39 cm more than its width. This can be written as 2π‘₯=39+𝑦.

Let us rearrange these equations so that the left-hand sides of the equations contain the variables and the right-hand sides contain the constants: π‘₯βˆ’2𝑦=6,2π‘₯βˆ’π‘¦=39.

Now, we will solve this system of equations by using matrices. Recall that the system of equations π‘Žπ‘₯+𝑏𝑦=𝑒,𝑐π‘₯+𝑑𝑦=𝑓 can be written as the matrix equation ο€Όπ‘Žπ‘π‘π‘‘οˆο€»π‘₯𝑦=ο€Όπ‘’π‘“οˆ.

Hence, we can write our system of equations as

ο€Ό1βˆ’22βˆ’1οˆο€»π‘₯𝑦=ο€Ό639.(4)

We can solve this matrix equation by multiplying from the left the inverse of the coefficient matrix ο€Ό1βˆ’22βˆ’1 if it exists. We know that the inverse of a square matrix exists if its determinant is not equal to zero. Let us first compute the determinant of this matrix. We know that detο€Όπ‘Žπ‘π‘π‘‘οˆ=π‘Žπ‘‘βˆ’π‘π‘.

Applying this formula to the coefficient matrix, detο€Ό1βˆ’22βˆ’1=1Γ—(βˆ’1)βˆ’(βˆ’2)Γ—2=3.

Since the determinant is nonzero, we can proceed to find its inverse. We recall the formula for the inverse of a 2Γ—2 matrix: 𝐴=1π΄ο€»π‘‘βˆ’π‘βˆ’π‘π‘Žο‡.det

Hence, using the determinant of the matrix we found earlier, ο€Ό1βˆ’22βˆ’1=13ο€Όβˆ’12βˆ’21.

Recall that, for any invertible matrix 𝐴, we have 𝐴𝐴=𝐼, where 𝐼 is the identity matrix. This means that we will be able to remove the coefficient matrix from the left-hand side of equation (4) by multiplying from the left the inverse of the coefficient matrix. Multiplying both sides of the equation by the inverse of the coefficient matrix, we have ο€»π‘₯𝑦=13ο€Όβˆ’12βˆ’21οˆο€Ό639.

We know that, in order to multiply a pair of matrices, the number of columns of the first matrix must equal the number of rows of the second matrix. We can see that the matrix multiplication on the right-hand side of the equation above is well defined. Computing this matrix multiplication, we obtain ο€»π‘₯𝑦=13ο€Όβˆ’1Γ—6+2Γ—39βˆ’2Γ—6+1Γ—39=13ο€Ό7227.

Finally, computing the scalar multiplication, ο€»π‘₯𝑦=βŽ›βŽœβŽœβŽ723273⎞⎟⎟⎠=ο€Ό249.

We know that a pair of matrices are equal if each pair of corresponding entries are equal. Hence, we obtain the solution to the given system of equations π‘₯=24,𝑦=9.

This leads to the perimeter of the rectangle: 2(π‘₯+𝑦)=2(24+9)=66.cm

Lastly, we note that matrices and their multiplicative inverses can be applied in the field of cryptography. In this context, the aim is to encode a message before it is transmitted, to prevent it from being understood if it is intercepted during transmission. The message can then be decoded once it reaches its desired destination.

As an illustration, we might want to transmit the message β€œhelp”. We start by representing each letter as a number, using the following simple alphabet table.

a: 1
b: 2
c: 3
d: 4
e: 5
f: 6
g: 7
h: 8
i: 9
j: 10
k: 11
l: 12
m: 13
n: 14
o: 15
p: 16
q: 17
r: 18
s: 19
t: 20
u: 21
v: 22
w: 23
x: 24
y: 25
z: 26

Then, taking our message two letters at a time, we write each two-letter section as a 2Γ—1 matrix: help=ο€Ό85,=ο€Ό1216. Next, to encode our message, we choose an invertible 2Γ—2 matrix as our coding matrix 𝐢 and multiply each of the above 2Γ—1 matrices on the left by 𝐢. For instance, choosing 𝐢=ο€Ό1245, our message encodes as ο€Ό1245οˆο€Ό85=ο€Ό1857,ο€Ό1245οˆο€Ό1216=ο€Ό44128.

Therefore, we transmit the message ο€Ό1857οˆο€Ό44128.

Now, recall that for a matrix 𝐢=ο€Όπ‘Žπ‘π‘π‘‘οˆ with det𝐢≠0, its multiplicative inverse is given by 𝐢=1πΆο€»π‘‘βˆ’π‘βˆ’π‘π‘Žο‡,det where det𝐢=π‘Žπ‘‘βˆ’π‘π‘. In this case, we have det𝐢=(1Γ—5)βˆ’(2Γ—4)=βˆ’3, so 𝐢=1βˆ’3ο€Ό5βˆ’2βˆ’41=βŽ›βŽœβŽœβŽβˆ’532343βˆ’13⎞⎟⎟⎠. Once the message ο€Ό1857οˆο€Ό44128 has been received, to decode it we must multiply each of these 2Γ—1 matrices on the left by 𝐢. This will β€œundo” the effect of 𝐢 and give us back the original pair of 2Γ—1 matrices ο€Ό85οˆο€Ό1216. We can then read off the message β€œhelp” from these four numbers by looking up the corresponding letters in the alphabet table.

Let us finish by recapping a few important concepts from the explainer.

Key Points

  • Let 𝐴 be an invertible matrix and 𝐡 be a matrix such that the multiplication 𝐴𝐡 is defined. Matrix 𝑋 satisfying the equation 𝐴𝑋=𝐡 is given by 𝑋=𝐴𝐡.
  • If matrix 𝐴 is not invertible, the matrix equation 𝐴𝑋=𝐡 has either an infinite number of solutions or no solution.
  • Consider a system of equations given by π‘Žπ‘₯+𝑏𝑦=𝑒,𝑐π‘₯+𝑑𝑦=𝑓 for some known constants π‘Ž, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓. We can write this system of two equations as one matrix equation ο€Όπ‘Žπ‘π‘π‘‘οˆο€»π‘₯𝑦=ο€Όπ‘’π‘“οˆ. This equation can then be solved using the matrix inverse: ο€»π‘₯𝑦=ο€Όπ‘Žπ‘π‘π‘‘οˆο€Όπ‘’π‘“οˆ.

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