Inverse matrix questions and solutions are given here to help students learn how to find the inverse of different matrices using different formulas and techniques. As we know, matrices are one of the most scoring concepts for students. Finding the inverse matrix is simple for 2×2 matrices. However, we can easily find the inverse matrix for 3×3 and 4×4 matrices using some simple rules. In this article, you will learn how to find the inverse of a given matrix using a suitable method.
What is an inverse matrix?
Suppose A is a non-singular square matrix of order n×n, and there is a matrix B of the same order, such that AB = BA = I, then B is called the inverse matrix of A, and I is the identity matrix.
Inverse matrix formula for 2×2 matrix
\(\begin{array}{l}A=\begin{bmatrix}a & b \\c & d \\\end{bmatrix}\end{array} \)
|A| = ad – bc
\(\begin{array}{l}adjA=\begin{bmatrix}d & -b \\-c & a \\\end{bmatrix}\end{array} \)
Therefore, A-1 = adjA/|A|
That means,
\(\begin{array}{l}A^{-1}=\frac{1}{ad-bc}\begin{bmatrix}d & -b \\-c & a \\\end{bmatrix}\end{array} \)
Inverse matrix formula for 3×3 or n×n matrix
Step 1: Find the determinant of the given matrix, say A.
Step 2: Find the cofactor matrix Cij = (-1)i+j det (Mij), where Mij is the (i,j)th minor matrix after removing the ith row and the jth column.
Step 3: Find the transpose of the cofactor matrix to get the adj A.
Step 4: A-1 = adj A/det(A)
Learn more about the inverse matrix here.
Inverse Matrix Questions and Answers
1. Find the inverse of the matrix \(\begin{array}{l}A=\begin{bmatrix}2 & 3 \\1 & 2 \\\end{bmatrix}\end{array} \)
.
Solution:
Given,
\(\begin{array}{l}A=\begin{bmatrix}2 & 3 \\1 & 2 \\\end{bmatrix}\end{array} \)
Let us find the determinant of A.
\(\begin{array}{l}|A|=\begin{vmatrix}2 & 3 \\1 & 2 \\\end{vmatrix}=2\times2-3\times 1=4-3=1\end{array} \)
Here, |A| ≠0, so the inverse of A exists.
\(\begin{array}{l}adjA=\begin{bmatrix}2 & -3 \\-1 & 2 \\\end{bmatrix}\end{array} \)
Now, A-1 = adjA/|A|
Therefore,
\(\begin{array}{l}A^{-1}=\frac{1}{1}\begin{bmatrix}2 &-3 \\-1 & 2 \\\end{bmatrix}=\begin{bmatrix}2 & -3 \\-1 & 2 \\\end{bmatrix}\end{array} \)
2. What is the inverse of the matrix \(\begin{array}{l}A=\begin{bmatrix}1 & 2 \\-3 & 0 \\\end{bmatrix}\end{array} \)
?
Solution:
Given,
\(\begin{array}{l}A=\begin{bmatrix}1 & 2 \\-3 & 0 \\\end{bmatrix}\end{array} \)
Let us calculate the determinant of A.
\(\begin{array}{l}|A|=\begin{vmatrix}1 & 2 \\-3 & 0 \\\end{vmatrix}=1\times 0-2\times(-3)=6\end{array} \)
Here, |A| ≠0, so the inverse of A exists.
Now,
\(\begin{array}{l}adjA=\begin{bmatrix}0 & -2 \\3 &1 \\\end{bmatrix}\end{array} \)
As we know, A-1 = adjA/|A|
Hence,
\(\begin{array}{l}A^{-1}=\frac{1}{6}\begin{bmatrix}0 &-2 \\3 &1 \\\end{bmatrix}=\begin{bmatrix}0 &-\frac{1}{3} \\\frac{1}{2} & \frac{1}{6} \\\end{bmatrix}\end{array} \)
3. If \(\begin{array}{l}A=\begin{bmatrix}cos\ \theta & sin\ \theta \\-sin\ \theta & cos\ \theta \\\end{bmatrix}\end{array} \)
and A-1 = AT, find the value of θ.
Solution:
Given,
\(\begin{array}{l}A=\begin{bmatrix}cos\ \theta & sin\ \theta \\-sin\ \theta & cos\ \theta \\\end{bmatrix}\end{array} \)
Also, given that,
A-1 = AT
⇒ AA-1 = AAT
⇒ I = AAT
Now,
\(\begin{array}{l}A^{T}=\begin{bmatrix}cos\ \theta & -sin\ \theta \\sin\ \theta & cos\ \theta \\\end{bmatrix}\end{array} \)
\(\begin{array}{l}AA^{T}=I\\\begin{bmatrix}cos\ \theta & sin\ \theta \\-sin\ \theta & cos\ \theta \\\end{bmatrix}.\begin{bmatrix}cos\ \theta & -sin\ \theta \\sin\ \theta & cos\ \theta \\\end{bmatrix}=\begin{bmatrix}1 0 \\0 1 \\\end{bmatrix}\\\begin{bmatrix}cos^2\theta+sin^2\theta & -cos\theta sin\theta+cos\theta sin\theta \\-cos\theta sin\theta+cos\theta sin\theta & cos^2\theta+sin^2\theta \\\end{bmatrix}=\begin{bmatrix}1 & 0 \\0 & 1 \\\end{bmatrix}\end{array} \)
From the above,
cos2θ + sin2θ = 1
This is one of the trigonometric identities and is true for all real values of θ.
4. Calculate the inverse of the matrix \(\begin{array}{l}A=\begin{bmatrix}2 & 4 & -6 \\7 & 3 & 5 \\1 & -2 & 4 \\\end{bmatrix}\end{array} \)
.
Solution:
Given,
\(\begin{array}{l}A=\begin{bmatrix}2 & 4 & -6 \\7 & 3 & 5 \\1 & -2 & 4 \\\end{bmatrix}\end{array} \)
First, find the determinant of matrix A.
\(\begin{array}{l}A=\begin{vmatrix}2 & 4 & -6 \\7 & 3 & 5 \\1 & -2 & 4 \\\end{vmatrix}\end{array} \)
= 2(12 + 10) – 4(28 – 5) – 6(-14 – 3)
= 2(22) – 4(23) – 6(-17)
= 44 – 92 + 102
= 54 ≠0
Thus, the inverse matrix exists.
Thus, the minor matrix of A
\(\begin{array}{l}=\begin{bmatrix}22 & 23 &-17 \\4 & 14 & -8 \\38 & 52 & -22 \\\end{bmatrix}\end{array} \)
Cofactor matrix of A
\(\begin{array}{l}=\begin{bmatrix}22 & -23 & -17 \\-4 & 14 & 8 \\38 & -52 & -22 \\\end{bmatrix}\end{array} \)
Also, adjA
\(\begin{array}{l}=\begin{bmatrix}22 & -4 & 38 \\-23 & 14 & -52 \\-17 & 8 & -22 \\\end{bmatrix}\end{array} \)
Therefore, A-1 = adjA/|A|
\(\begin{array}{l}A^{-1}=\frac{1}{54}\begin{bmatrix}22 & -4 & 38 \\-23 & 14 & -52 \\-17 & 8 & -22 \\\end{bmatrix}=\begin{bmatrix}
\frac{11}{27} & -\frac{2}{27} & \frac{19}{27} \\
-\frac{23}{54} & \frac{7}{27} &-\frac{26}{27} \\
-\frac{17}{54} & \frac{4}{27} & -\frac{11}{27} \\
\end{bmatrix}
\end{array} \)
5. If \(\begin{array}{l}A=\begin{bmatrix}2 & 1 \\7 & 2 \\\end{bmatrix}\end{array} \)
, show that (A-1)-1 = A.
Solution:
Given,
\(\begin{array}{l}A=\begin{bmatrix}2 & 1 \\7 & 2 \\\end{bmatrix}\end{array} \)
Now,
\(\begin{array}{l}|A|=\begin{vmatrix}2 & 1 \\7 & 2 \\\end{vmatrix}=4-7=-3\end{array} \)
Here, matrix A is non-singular.
\(\begin{array}{l}adjA=\begin{bmatrix}2 & -1 \\-7 & 2 \\\end{bmatrix}\end{array} \)
Thus,
\(\begin{array}{l}A^{-1}=\frac{adjA}{|A|}=\frac{1}{-3}\begin{bmatrix}2 & -1 \\-7 & 2 \\\end{bmatrix}=\begin{bmatrix}-\frac{2}{3} & \frac{1}{3} \\\frac{7}{3} & -\frac{2}{3}\\ \end{bmatrix}\end{array} \)
Let A-1 = B
So,
\(\begin{array}{l}B=\begin{bmatrix}-\frac{2}{3} & \frac{1}{3} \\\frac{7}{3} & -\frac{2}{3}\\ \end{bmatrix}\end{array} \)
And
\(\begin{array}{l}|B|=\begin{vmatrix}-\frac{2}{3} & \frac{1}{3} \\\frac{7}{3} & -\frac{2}{3}\\ \end{vmatrix}\end{array} \)
= (-â…”)(-â…”) – (â…“)(7/3)
= (4/9) – (7/9)
= (4 – 7)/9
= -3/9
= -â…“
Now,
\(\begin{array}{l}adjB=\begin{bmatrix}-\frac{2}{3} -\frac{1}{3} \\-\frac{7}{3} -\frac{2}{3} \\\end{bmatrix}\end{array} \)
Also,
\(\begin{array}{l}B^{-1}=\frac{adjB}{|B|}=\frac{1}{-\frac{1}{3}}\begin{bmatrix}-\frac{2}{3} &-\frac{1}{3} \\-\frac{7}{3} &-\frac{2}{3} \\\end{bmatrix}=-3\begin{bmatrix}-\frac{2}{3} &-\frac{1}{3} \\-\frac{7}{3} &-\frac{2}{3} \\\end{bmatrix}=\begin{bmatrix}2 & 1 \\7 & 2 \\\end{bmatrix}\end{array} \)
That means B-1 = (A-1)-1 = A
6. Find x, y, z if \(\begin{array}{l}A=\begin{bmatrix}0 & 2y & z \\x & y & -z \\x & -y & z \\\end{bmatrix}\end{array} \)
satisfies AT = A-1.
Solution:
Given,
\(\begin{array}{l}A=\begin{bmatrix}0 & 2y & z \\x & y & -z \\x & -y & z \\\end{bmatrix}\end{array} \)
AT = A-1
⇒ AAT = AA-1
⇒ AAT = I {since A-1A = AA-1 = I}
Now,
\(\begin{array}{l}A^{T}=\begin{bmatrix}0 & x & x \\2y & y & -y \\z & -z & z \\\end{bmatrix}\end{array} \)
\(\begin{array}{l}AA^{T}=I\\\begin{bmatrix}0 & 2y & z \\x & y & -z \\x & -y & z \\\end{bmatrix}\begin{bmatrix}0 & x & x \\2y & y & -y \\z & -z & z \\\end{bmatrix} =\begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \\\end{bmatrix}\end{array} \)
By performing multiplication on the LHS, we get:
By equating the corresponding elements, we have:
4y2 + z2 = 1 …(1)
x2 + y2 + z2 = 1 …(2)
2y2 – z2 = 0 …(3)
Adding equations (1) and (3), we get:
4y2 + z2 + 2y2 – z2 = 1 + 0
6y2 = 1
y2 = 1/6
⇒ y = ±1/√6
Substituting the value of y in equation (3), we get:
z2 = 2y2
z2 = 2(1/6)
z2 = 1/3
⇒ z = ±1/√3
Substituting the values of y and z in equation (2), we get:
x2 = 1 – y2 – z2
x2 = 1 – (1/6) – (1/3)
x2 = (6 – 1 – 2)/6
x2 = 3/6
x2 = 1/2
⇒ x = ±1/√2
Therefore, x = ±1/√2, y = ±1/√6 and z = ±1/√3.
7. Find the value of x for which the matrix \(\begin{array}{l}A=\begin{bmatrix}2 & 0 & 10 \\ 0 & x+7 & -3 \\0 & 4 & x \\\end{bmatrix}\end{array} \)
is invertible.
Solution:
\(\begin{array}{l}A=\begin{bmatrix}2 & 0 & 10 \\ 0 & x+7 & -3 \\0 & 4 & x \\\end{bmatrix}\end{array} \)
Let us find the determinant of the given matrix.
\(\begin{array}{l}|A|=\begin{vmatrix}2 & 0 & 10 \\ 0 & x+7 & -3 \\0 & 4 & x \\\end{vmatrix}\end{array} \)
= 2[(x + 7)x – (-3)(4)] – 0 + 10(0 – 0)
= 2(x2 + 7x + 12)
We know that a matrix is invertible if and only if its determinant is not equal to 0.
Let |A| = 0
2(x2 + 7x + 12) = 0
⇒ x2 + 7x + 12 = 0
⇒ x2 + 3x + 4x + 12 = 0
⇒ x(x + 3) + 4(x + 3) = 0
⇒ (x + 3)(x + 4) = 0
⇒ x + 3 = 0, x + 4 = 0
⇒ x = -3, x = -4
Thus, for x = -3 and -4, the given matrix is invertible.
| Inverse matrix by elementary transformation
We can perform various operations on rows and columns to find the inverse of a given matrix.
To find the inverse of matrix A, we can apply a sequence of row operations on A = IA until we get an identity matrix on the LHS .i.e I = BA. Hence, the matrix B on the RHS is the inverse of matrix A.
To find the inverse of matrix A, we can apply a sequence of column operations on A = AI until we get an identity matrix on the LHS .i.e I = AB. Hence, the matrix B on the RHS is the inverse of matrix A.
We can also find the inverse matrix by writing the augmented matrix of the form [A | I] and applying row operations until we get [I | B], where B is the inverse of A.
Read more: Inverse matrix using elementary operations |
8. Find the inverse of \(\begin{array}{l}A=\begin{bmatrix}1 & 2 & 3 \\2 & 4 & 5 \\3 & 5 & 6 \\\end{bmatrix}\end{array} \)
using row operations.
Solution:
Given,
\(\begin{array}{l}A=\begin{bmatrix}1 & 2 & 3 \\2 & 4 & 5 \\3 & 5 & 6 \\\end{bmatrix}\end{array} \)
Let us write the augmented matrix [A | I ] such that I is a square matrix of the order same as A.
R3 → R3 – 3R1
R2 → R2 – 2R1
Now, interchange R2 and R3.
R2 → (-1).R2 and R3 → (-1).R3
R2 → R2 – 3R3
R1 → R1 – 2R2
R1 → R1 – 3R3
This is of the form [ I | B].
Here, B is the inverse of A.
Therefore,
\(\begin{array}{l}B=A^{-1}=\begin{bmatrix}1 & -3 & 2 \\-3 & 3 & -1 \\2 & -1 & 0 \\\end{bmatrix}\end{array} \)
9. Determine the formula for the inverse of matrix \(\begin{array}{l}A=\begin{bmatrix}p & 0 & 0 & 0 \\0 & q & 0 & 0 \\0 & 0 & r & 0 \\0 & 0 & 0 & s \\\end{bmatrix}\end{array} \)
, where p, q, r, s ≠0.
Solution:
Given,
\(\begin{array}{l}A=\begin{bmatrix}p & 0 & 0 & 0 \\0 & q & 0 & 0 \\0 & 0 & r & 0 \\0 & 0 & 0 & s \\\end{bmatrix}\end{array} \)
Let us write the augmented matrix [A | I ].
\(\begin{array}{l}[A|I]=\begin{bmatrix}p & 0 & 0 & 0 | 1 & 0 & 0 & 0\\0 & q & 0 & 0 | 0 & 1 & 0 & 0\\0 & 0 & r & 0 | 0 & 0 & 1 & 0\\0 & 0 & 0 & s | 0 & 0 & 0 & 1\\\end{bmatrix}\end{array} \)
R1 → (1/p) R1
R2 → (1/q) R2
R3 → (1/r) R3
R4 → (1/s) R4
\(\begin{array}{l}=\begin{bmatrix}1 & 0 & 0 & 0 | \frac{1}{p} & 0 & 0 & 0\\0 & 1 & 0 & 0 | 0 & \frac{1}{q} & 0 & 0\\0 & 0 & 1 & 0 | 0 & 0 & \frac{1}{r} & 0\\0 & 0 & 0 & 1 | 0 & 0 & 0 \frac{1}{s}\\\end{bmatrix}\end{array} \)
Hence, the inverse of A is:
\(\begin{array}{l}A^{-1}=\begin{bmatrix} \frac{1}{p} & 0 & 0 & 0\\ 0 & \frac{1}{q} & 0 & 0\\ 0 & 0 & \frac{1}{r} & 0\\0 & 0 & 0 & \frac{1}{s}\\\end{bmatrix}\end{array} \)
10. If A is 3 × 3 invertible matrix, then show that for any scalar k (non-zero), kA is invertible and (kA)–1 = (1/k)A –1.
Solution:
Consider (kA) [(1/k) A-1]
= [k (1/k)] (A A-1)
= 1. (AA-1)
= I {since AA-1 = A-1A = I}
That means kA is the inverse of (1/k)A-1.
Therefore, (kA)-1 = (1/k) A-1
Practice Questions on Inverse Matrix
- Find the inverse of the matrix
\(\begin{array}{l}A=\begin{bmatrix}3 & 5 \\-2 & 4 \\\end{bmatrix}\end{array} \)
.
- If
\(\begin{array}{l}A=\begin{bmatrix}2 & 1 & 1\\1 & 2 & 1\\1 & 1 & 2\\\end{bmatrix}\end{array} \)
, find A-1 exists.
- Using elementary row operations, find the inverse of the matrix
\(\begin{array}{l}\begin{bmatrix}-4 & 8 & 4\\-1 & 2 & 1\\-3 & 6 & 3\\\end{bmatrix}\end{array} \)
.
- Calculate the inverse matrix for
\(\begin{array}{l}B=\begin{bmatrix}4+3i & -i \\i &4-3i \\\end{bmatrix}\end{array} \)
.
- What is the inverse of the matrix
\(\begin{array}{l}A=\begin{bmatrix}1 & 0 & 1 & 2 \\-1 & 1 & 2 & 0 \\-2 & 0 & 1 & 2 \\0 & 0 & 0 & 1 \\\end{bmatrix}\end{array} \)
?
