This set of Numerical Methods Multiple Choice Questions & Answers (MCQs) focuses on “Matrix Inversion”.

1. The inverse of a matrix exists if and only if it is a non-singular matrix.

a) True

b) False

View Answer

Explanation: The inverse exists for a non-singular matrix because for a singular matrix, the value of determinant is zero, which in turn makes the value of inverse according to the following relation:

A

^{-1}= Adj (A)/|A|

2. Matrix inverse in for systems of equation is used to determine which of the following?

a) identified set of equation

b) variable set of equation

c) solution set of equations

d) constant set of equations

View Answer

Explanation: It is used to determine solution set of equations, by using different methods like Cramer’s rule, Gauss Jordan, etc.

3. Matrix which does not have an inverse by solving it, is classified as which of the following?

a) unidentified matrix

b) linear matrix

c) non-singular matrix

d) singular matrix

View Answer

Explanation: This is because singular matrices do not have inverse, because the value of their determinant is zero.

4. Multiplicative inverse matrix generalized form to solve is which of the following?

a) AA^{-1} = A^{-1}A = I

b) AA^{-2} = A-3A = I

c) 3AA^{4} = A^{-1}A = I

d) AA^{-1} = 2A^{-5}A = Q

View Answer

Explanation: This is because product of a matrix and its inverse is always an identity matrix.

5. Which of the following relation describes the solution of system of equation in form of AX = B?

a) A^{-1}AX = A^{-1}B

b) B^{-1}AX = A^{-1}B

c) AB^{-1}AX = AB^{-1}X

d) X^{-1}AB = A^{-1}BX

View Answer

Explanation: We have, A

^{-1}AX = A

^{-1}B which is same as X = A

^{-1}B, and this is the primary relation employed for finding inverse.

6. Equations are linearly dependent, in a system of linear equations, if and only if?

a) A^{-2} must exists

b) A^{-1} does not exist

c) A^{-3} does not exist

d) A^{-4} must exist

View Answer

Explanation: If the equations are linearly dependent, the determinant of the matrix would be zero and hence, no inverse exists for such a matrix.

7. Find inverse of the following matrix?

\(\begin{bmatrix}1&-1&1\\1&1&1\\1&2&4\end{bmatrix}\)

a) 1/6 \(\begin{bmatrix}1&-1&1\\1&1&1\\1&2&4\end{bmatrix}\)

b) 1/6 \(\begin{bmatrix}1&1&1\\5&-1&6\\1&6&4\end{bmatrix}\)

c) 1/6 \(\begin{bmatrix}1&-1&1\\8&1&6\\1&2&4\end{bmatrix}\)

d) 1/6 \(\begin{bmatrix}1&1&1\\0&-1&1\\1&3&4\end{bmatrix}\)

View Answer

Explanation: We have, the minor of the (1, 1) entry is \(\begin{vmatrix}1&1\\2&4\end{vmatrix}\)

.

The minor of the entry (1, 2) is\(\begin{vmatrix}1&1\\1&4\end{vmatrix}\)

.

Repeating this process, the matrix of minors is

\(\begin{pmatrix}2&3&1\\-6&3&3\\-2&0&2\end{pmatrix}\).

Next we negate every other entry, according to the pattern

\(\begin{pmatrix}+&-&+\\-&+&-\\+&-&+\end{pmatrix}\).

The matrix of minors becomes

\(\begin{pmatrix}2&-3&1\\6&3&-3\\-2&0&2\end{pmatrix}\).

Transposing, we get \(\begin{pmatrix}2&6&-2\\-3&3&0\\1&-3&2\end{pmatrix}\).

In order to divide by the determinant we must first compute it.

We can compute that det = 6.

Hence,

A

^{-1}= \(\frac{1}{6}\begin{pmatrix}2&6&-2\\-3&3&0\\1&-3&2\end{pmatrix}\).

8. Solve the following equation by matrix inversion.

\(\begin{pmatrix}2&3\\10&16\end{pmatrix}\) \(\begin{pmatrix}x\\y\end{pmatrix}\) = \(\begin{pmatrix}1\\2\end{pmatrix}\)

a) \(\begin{pmatrix}8\\-5\end{pmatrix}\)

b) \(\begin{pmatrix}5\\-3\end{pmatrix}\)

c) \(\begin{pmatrix}6\\-7\end{pmatrix}\)

d) \(\begin{pmatrix}9\\-5\end{pmatrix}\)

View Answer

Explanation: We observe that this matrix equation is in the standard form Ax=b, where

A = \(\begin{pmatrix}2&3\\10&16\end{pmatrix}\)

X = \(\begin{pmatrix}x\\y\end{pmatrix}\)

b = \(\begin{pmatrix}1\\2\end{pmatrix}\)

The determinant of A is 2(16) – 3(10) = 2.

Hence,

A

^{-1}= \(\begin{pmatrix}8&-3/2\\-5&1\end{pmatrix}\)

The solution to the linear system is given by x = A

^{-1}b.

X = \(\begin{pmatrix}8&-3/2\\-5&1\end{pmatrix}\)\(\begin{pmatrix}1\\2\end{pmatrix}\)

Computing the matrix multiplication,

X = \(\begin{pmatrix}x\\y\end{pmatrix}\) = \(\begin{pmatrix}8(1) -3/2(2)\\-5(1) +1(2)\end{pmatrix}\) =\(\begin{pmatrix}5\\-3\end{pmatrix}\).

9. Non square matrices do not have inverse.

a) True

b) False

View Answer

Explanation: Solving for an inverse needs an adjoint and a determinant which are found only for the square matrices.

10. What is the drawback of finding inverse by adjoint method?

a) It needs a lot of calculations

b) It gives incorrect answers

c) It assumes certain values

d) It is solved by approximating certain values

View Answer

Explanation: The main drawback is that it needs a lot of calculations and hence it is lengthy, so new faster methods are developed to remove this drawback.

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