In the previous section,
we saw the basics of invertible matrices. In this
section, we will see the two theorems related to invertible matrices. Later in this section, we will see the method to find the inverse of a given matrix.
Theorem 1:
There will be only one inverse matrix.
• The proof can be written in 8 steps:
1. Let A be a square matrix. Let us assume that, B and C are inverse matrices of A.
2. Since B is an inverse of A, we can write: AB = BA = I
3. Since C is an inverse of A, we can write: AC = CA = I
4. Any matrix multiplied by I will give the same matrix. So we have: B = BI
5. But from (3), we have: I = AC
• So the result in (4) becomes: B = BI = B(AC)
6. Recall the associative law: X(YZ) = (XY)Z (Details here)
• Applying the law, the result in (5) becomes:
B = BI = B(AC) = (BA)C
7. But from (2), we have: BA = I
• So the result in (6) becomes:
B = BI = B(AC) = (BA)C = IC = C
8. Thus we get B = C
That means, there can be only one inverse.
Theorem 2
If A and B are invertible matrices of the same order, then (AB)-1 = B-1A-1.
• Proof can be written in steps:
1. We know that (X)(X)-1 = I
So we can write: (AB)(AB)-1 = I
2. Pre multiplying both sides by A-1, we get:
A-1(AB)(AB)-1 = A-1I
⇒ A-1(AB)(AB)-1 = A-1
3. Applying associative law, we get:
(A-1A)B(AB)-1 = A-1.
⇒ IB(AB)-1 = A-1.
⇒ B(AB)-1 = A-1.
4. Pre multiplying both sides by B-1, we get:
B-1B(AB)-1 = B-1A-1.
⇒ I(AB)-1 = B-1A-1.
⇒ (AB)-1 = B-1A-1.
Inverse of a Matrix by elementary operations
First we will see how elementary operations can be applied to a matrix equation. It can be written in 3 steps:
1. Consider the equation: X = AB, where X, A and B are square matrices of the same order.
2. We can perform elementary operations on this equation.
• But while performing the operations, the equality should be maintained.
3. For maintaining the equality, we follow a procedure:
• Whatever elementary row operation is performed on the L.H.S, the same operation must be performed on the first matrix A on the R.H.S.
• Whatever elementary column operation is performed on the L.H.S, the same operation must be performed on the second matrix B on the R.H.S.
The above steps give us a method to find the inverse of a matrix. It can be written in 3 steps:
1. Consider the equation A = IA
• We can perform a series of elementary row operations on this equation, till we reach the form: I = BA
• So our aim is to transform A on the L.H.S to I
• Based on step (3), we can say that, A on the L.H.S, and I on the R.H.S will get transformed during the process.
♦ A in the L.H.S will become I.
♦ I in the R.H.S will become B.
• Once we get I = BA, it implies that, B is the inverse of A.
2. There is an alternate method to find the inverse.
• Consider the equation A = AI
• We can perform a series of elementary column operations on this equation, till we reach the form: I = AB
• So here also, our aim is to transform A on the L.H.S to I
• Based on step (3), we can say that, A on the L.H.S, and I on the R.H.S will get transformed during the process.
♦ A in the L.H.S will become I.
♦ I in the R.H.S will become B.
• Once we get I = AB, it implies that, B is the inverse of A.
3. Let us compare the two methods:
(i) Use of elementary operations:
♦ In (1), we use elementary row operations.
♦ In (2), we use elementary column operations.
(ii) Initial equation:
♦ In (1), the initial equation is A = IA
♦ In (2), the initial equation is A = AI
(iii) Final equation:
♦ In (1), the final equation is I = BA
♦ In (2), the final equation is I = AB
(iv) In both (1) and (2),
♦ A on the L.H.S is transformed into I.
♦ I on the R.H.S is transformed into B.
Let us see a solved example:
Solved Example 19.16
By using elementary operations, find the inverse of the matrix A = $\left[\begin{array}{r}
1 &{ 2 } \\
2 &{ -1 } \\
\end{array}\right]
$
Solution:
Method 1: Using elementary row operations.
• We can write a general procedure:
1. We are given a 2×2 matrix.
• So the diagonal elements are: a11 and a22. These elements must be ‘1’ in the final answer.
2. First we change a11 to 1.
• Then we change the remaining elements in that column, to zeros. We do that step by step.
3. Next, we change a22 to 1.
• Then we change the remaining elements in that column, to zeros. We do that step by step.
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{A} & {~=~} &{IA} \\
{~\color{magenta} 2 } &{\implies} &{\left[\begin{array}{r} 1&{2}\\ 2&{-1}\\ \end{array}\right] } & {~=~} &{\left[\begin{array}{r} 1&{0}\\ 0&{1}\\ \end{array}\right]A } \\
{~\color{magenta} 3 } &{R_2 \rightarrow {R_2 \;–\; 2 R_1}} &{\left[\begin{array}{r} 1&{2}\\ 0&{-5}\\ \end{array}\right] } & {~=~} &{\left[\begin{array}{r} 1&{0}\\ -2&{1}\\ \end{array}\right] A} \\
{~\color{magenta} 4 } &{R_2 \rightarrow {-\frac{1}{5} R_2}} &{\left[\begin{array}{r} 1&{2}\\ 0&{1}\\ \end{array}\right] } & {~=~} &{\left[\begin{array}{r} 1&{0}\\ \frac{2}{5}&{-\frac{1}{5}}\\ \end{array}\right] A} \\
{~\color{magenta} 5 } &{R_1 \rightarrow {R_1 \,–\, 2 R_2}} &{\left[\begin{array}{r} 1&{0}\\ 0&{1}\\ \end{array}\right] } & {~=~} &{\left[\begin{array}{r} \frac{1}{5}&{\frac{2}{5}}\\ \frac{2}{5}&{-\frac{1}{5}}\\ \end{array}\right] A} \\
{~\color{magenta} 6 } &{\implies} &{I} & {~=~} &{\left[\begin{array}{r} \frac{1}{5}&{\frac{2}{5}}\\ \frac{2}{5}&{-\frac{1}{5}}\\ \end{array}\right] A} \\
{~\color{magenta} 7 } &{~\text{But}} &{I} & {~=~} &{A^{-1} A} \\
{~\color{magenta} 8 } &{\implies} &{A^{-1}} & {~=~} &{\left[\begin{array}{r} \frac{1}{5}&{\frac{2}{5}}\\ \frac{2}{5}&{-\frac{1}{5}}\\ \end{array}\right]} \\
\end{array}$
◼ Remarks:
• a11 is already ‘1’.
♦ The remaining element in that column is a21.
♦ We must change it to zero. This is achieved in (3).
• a22 is changed to ‘1’ in (4).
♦ The remaining element in that column is a12.
♦ We must change it to zero. This is achieved in (5).
Method 2: Using elementary column operations.
• We can write a general procedure:
1. We are given a 2×2 matrix.
• So the diagonal elements are: a11 and a22. These elements must be ‘1’ in the final answer.
2. First we change a11 to 1.
• Then we change the remaining elements in that row, to zeros. We do that step by step.
3. Next, we change a22 to 1.
• Then we change the remaining elements in that row, to zeros. We do that step by step.
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{A} & {~=~} &{AI} \\
{~\color{magenta} 2 } &{\implies} &{\left[\begin{array}{r} 1&{2}\\ 2&{-1}\\ \end{array}\right] } & {~=~} &{A \left[\begin{array}{r} 1&{0}\\ 0&{1}\\ \end{array}\right]} \\
{~\color{magenta} 3 } &{C_2 \rightarrow {C_2 \;–\; 2 C_1}} &{\left[\begin{array}{r} 1&{0}\\ 2&{-5}\\ \end{array}\right] } & {~=~} &{A \left[\begin{array}{r} 1&{-2}\\ 0&{1}\\ \end{array}\right] } \\
{~\color{magenta} 4 } &{C_2 \rightarrow {-\frac{1}{5} C_2}} &{\left[\begin{array}{r} 1&{0}\\ 2&{1}\\ \end{array}\right] } & {~=~} &{A \left[\begin{array}{r} 1&{\frac{2}{5}}\\ 0&{-\frac{1}{5}}\\ \end{array}\right] } \\
{~\color{magenta} 5 } &{C_1 \rightarrow {C_1 \,–\, 2 C_2}} &{\left[\begin{array}{r} 1&{0}\\ 0&{1}\\ \end{array}\right] } & {~=~} &{A \left[\begin{array}{r} \frac{1}{5}&{\frac{2}{5}}\\ \frac{2}{5}&{-\frac{1}{5}}\\ \end{array}\right] } \\
{~\color{magenta} 6 } &{\implies} &{I} & {~=~} &{A \left[\begin{array}{r} \frac{1}{5}&{\frac{2}{5}}\\ \frac{2}{5}&{-\frac{1}{5}}\\ \end{array}\right] } \\
{~\color{magenta} 7 } &{~\text{But}} &{I} & {~=~} &{A A^{-1} } \\
{~\color{magenta} 8 } &{\implies} &{A^{-1}} & {~=~} &{\left[\begin{array}{r} \frac{1}{5}&{\frac{2}{5}}\\ \frac{2}{5}&{-\frac{1}{5}}\\ \end{array}\right]} \\
\end{array}$
◼ Remarks:
• a11 is already ‘1’.
♦ The remaining element in that column is a12.
♦ We must change it to zero. This is achieved in (3).
• a22 is changed to ‘1’ in (4).
♦ The remaining element in that row is a21.
♦ We must change it to zero. This is achieved in (5).
In the next section, we will see a few more solved examples.
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