Mathematics: Linear Algebra: CSIR Previous year Questions:(Algebra of matrices, rank and determinant of matrices, linear equations)

1.Let \(S=\lbrace A:A=[a_{ij}]_{5\times 5},a_{ij}=0 \quad or \quad1 \forall i, j ,\sum_i a_{ij}=1 \forall j\quad and \quad \sum_j a_{ij}=1 \forall i \rbrace\).Then the number of element in \(S\) is

• \(5^2\)
• \(5^5\)
• \(5\)!
• \(55\)

Solution: The above question is equivalent to How many \(5\times 5\) permutation matrix is possible.

So, in order to calculate it, First of all, in how many ways we can place  \(1\) in first Column? The answer is : In \(5\) ways.

now, after that for second column, in how many ways we can place \(1\) ? The answer is \(4\). (Why? Since suppose we had placed 1 in first column at the position \(a_{11}\). then we can't place 1 in \(a_{12}\).) Proceeding in similar manner... for 3rd column we have only 3 choice, then 2 and 1 choices for second and third column. So total \(5\times 4\times 3\times 2\times 1\) ways=\(5\)! ways.  (by the multiplication rule of counting.)

2.Let \(A,B\) be \(n \times n\)  matrices such that \(BA+B^2 = I - BA^2\) where \(I\) is the \(n\times n\) identity matrix. Which of the following is always true?

• \(A\) is nonsingular
• \(B\) is nonsingular
• \(A +B\) is nonsingular
• \(AB\) is nonsingular

Solution: \(BA+B^2+A^2=I \Rightarrow B(A+B+A^2)=I\). That is, inverse of B exists. Thus, 2nd option is correct.

3.Let \(A\) be a \(5 \times 5\) matrix with real entries such that the sum of the entries in each row of \(A\) is \(1\). Then the sum of all the entries in \(A^3\) is

• \(3\)
• \(15\)
• \(5\)
• \(125\)

Solution: 3rd option is correct. you can take any particular example of such matrix and hence eliminate other option.

Take \(A=\) Identity matrix or any other permutation Matrix.

4.Let \(A\) and \(B\) be \(n \times n\) real matrices such that \(AB=BA=0\) and \(A+B\) is invertible. Which of the following are always true?

• \(rank(A)=rank(B)\)
• \(rank(A)+rank(B)=n\)
• \(nullity(A)+nullity(B)=n\)
• \(A-B\) is invertible.

Solution: If \(AB=0\).

Then column space of \(B\) is contained in Null space of \(A\).

If not, i.e., there is a vector \(y = Bx\) lies in the column space of \(B\), but not in the nullspace of \(A\). Then \((AB)x = A(Bx) \neq 0\), contradicts with \(AB = 0\).

and also column space of \(A\) is contained in null space of \(B\).

So we get: \(nullity A \geq column\hspace{5pt}Space\hspace{5pt}of\hspace{5pt}B= n-nullspace of B\).

(Since dimension of column space is equal to dimension of Row space, in this problem.)

So, \(nullity\hspace{5pt}of\hspace{5pt}B+nullity\hspace{5pt}of A\geq n\)

and

\(rank A+rank B\leq n\) (Since dimension of column space is equal to dimension of Row space, in this case)

\(A+B\) is invertible that means Both of them are zero matrix.

Also Rank \((A+B)=n\)

from Sylvester’s rank inequality: We have, \(Rank(A+B)\leq Rank A+Rank B\).

So, \(Rank A+RankB=n\).

• From, Rank-nullity Theorem:
• \(Nullity A+Nullity B=n-Rank A+n-RankB\)

          =\(2n-Rank A-rank B\)=\(2n-Rank A+Rank B\)=\(n\).

• \((A-B)^2=A^2+B^2-2AB=A^2+B^2=A^2+B^2+2AB=(A+B)^2\) Since \(AB=0\).
  therefore 4th option is correct.

5.Let \(D\) be non-zero \(n \times n\) real matrix with \(n \geq 2\). Which of the following implication is valid?

• \(det(D)=0 \text{ implies } rank(D)=0\)
• \(det(D)=1 \text{ implies } rank(D)\neq 1\)
• \(rank(D)=1 \text{ implies }det(D)\neq 1\)
• \(rank(D)=n \text{ implies }det(D) \neq 1\)

Solution: 2nd option is correct. Determinant of a matrix is equal to plus or minus product of its pivots. in Gaussian Elemination.  So, If determinant is non zero means that all pivots are non zero hence, \(rank=n\).

6.Let \(A,B\) be \(n\times n\) real matrices. Which of the following statements is correct?

• \(rank(A+B)=rank(A)+rank(B)\)
• \(rank(A+B)\leq rank(A)+rank(B)\)
• \(rank(A+B)=\min\{rank(A),rank(B)\}\)
• \(rank(A+B)=\max\{rank(A),rank(B)\}\)

Solution: Third option is correct.  Sylvester’s rank inequality.

7.Which of the following matrices has the same row space as the matrix

\(\begin{pmatrix} 4 & 8 & 4\\3 & 6 & 1\\2 & 4 & 0\end{pmatrix}\)?

• \(\begin{pmatrix} 1&2&0\\0&0&1\end{pmatrix}\)
• \(\begin{pmatrix} 1&1&0\\0&0&1\end{pmatrix}\)
• \( \begin{pmatrix} 0&1&0\\0&0&1\end{pmatrix}\)
• \( \begin{pmatrix} 1&0&0\\0&1&0\end{pmatrix}\)

Solution: 1st option is correct.

\(\begin{pmatrix}4&8&4\\3&6&1\\2&4&0\end{pmatrix}\rightarrow \begin{pmatrix}1&2&1\\3&6&1\\2&4&0\end{pmatrix}\)  \(R_1\to R_1/4\)

\(\rightarrow \begin{pmatrix}1&2&1\\0&0&-2\\0&0&-2\end{pmatrix}\)   \(R_2\to R_2-3R_1\)     \(R_3\to R_3-2R_1\)

\(\rightarrow \begin{pmatrix}1&2&0\\0&0&1\\0&0&0\end{pmatrix}\) \(R_3\to R_3-R_2\)       &      \(R_2\to R_2/1\) & \(R_1\to R_1-R_2\).

Remark: Row space is preserved under elementary row transformation.

8.The determinant of the $n \times n$ permutation matrix

$$ \begin{pmatrix}&&&&1\\& & & 1 & \\&&\vdots &&\\&1&& &\\1 &&&&\end{pmatrix} $$

• \((-1)^n\)
• \((-1)^{\frac{n}{2}}\)
• \(-1\)
• \(1\)

Solution: Number of "exchanges of row" required for making this matrix "Identity matrix"=\([n/2]\)

So determinant of this matrix=\((-1)^{\frac{n}{2}}\) ( we have used the property of determinant."Interchanging any pair of columns or rows of a matrix multiplies its determinant by \(-1\)")

9.The determinant

\(\begin{vmatrix} 1&1+x &1+x+x^2\\1 &1 + y & 1 + y + y^2 \\1 & 1 + z & 1 + z + z^2\end{vmatrix}\)

is equal to

•  \((z-y)(z-x)(y-x)\)
• \((x-y)(x-z)(y-z)\)
• \((x-y)^2(y-z)^2(z-x)^2\)
• \((x^2-y^2)(y^2-z^2)(z^2-x^2)\)

Solution: \(\begin{vmatrix}1&1+x&1+x+x^2\\1&1+y &1+y+y^2\\1&1+z&1+z+z^2\end{vmatrix}\rightarrow \begin{vmatrix} 1&x&x^2\\1&y& y^2\\1&z& z^2\end{vmatrix}\)\(C_3 \to C_3-C_2\) &\(C_2\to C_2-C_1\)
\(\rightarrow \begin{vmatrix}1&x &x^2\\0&y-x&y^2-x^2\\0&z-y&z^2-y^2\end{vmatrix}\)
\(\rightarrow (y-x)(z-y)\begin{vmatrix}1&x &x^2\\0&1&y+x\\0&1&z+y\end{vmatrix}\) Taking common.
\(\rightarrow (y-x)(z-y)\begin{vmatrix}1&x &x^2\\0&1&y+x\\0&0&z-x\end{vmatrix}\) \(R_3\to R_3-R_2\)
Expanding we get  \((z-y)(z-x)(y-x)\)
So, 1st option is correct.

10.The determinant of the matrix \(\begin{pmatrix}1 & 0 & 0 & 0 & 0 & 2\\0 & 1 & 0 & 0 & 2 & 0\\ 0 & 0 & 1 & 2 & 0 & 0\\ 0 & 0 & 2 & 1 & 0 & 0\\ 0 & 2 & 0 & 0 & 1 & 0\\2 & 0 & 0 & 0 & 0 & 1\end{pmatrix}\)
is

• \(0\)
• \(-9\)
• \(-27\)
• \(1\)

solution: \(\begin{pmatrix}1 & 0 & 0 & 0 & 0 & 2\\0 & 1 & 0 & 0 & 2 & 0\\ 0 & 0 & 1 & 2 & 0 & 0\\ 0 & 0 & 2 & 1 & 0 & 0\\ 0 & 2 & 0 & 0 & 1 & 0\\2 & 0 & 0 & 0 & 0 & 1\end{pmatrix}\rightarrow \begin{pmatrix}1 & 0 & 0 & 0 & 0 & 2\\0 & 1 & 0 & 0 & 2 & 0\\ 0 & 0 & 1 & 2 & 0 & 0\\ 0 & 0 & 0&-3 & 0 & 0\\ 0&0&0&0&-3 & 0\\0&0&0&0 & 0 &-3\end{pmatrix}\)
\(R_4\to R_4-2R_1\)       \(R_5\to R_5-2R_2\)      \(R_6\to R_6-2R_3\)
So, determinant \(=-3\times-3\times-3=-27\)
11.Let \(J\) denote a \(101\times 101\) matrix with all the entries equal to \(1\) and let \(I\) denote the identity matrix of order \(101\). Then the determinant of \(J - I\) is

•  \(101\)
• \(1\)
• \(0\)
• \(100\)

Solution:
Let \[A=\left| \begin{array}{ccc}a_{11}-\lambda&a_{12}&a_{13}&\cdots&a_{1n}\\a_{21}&a_{22}-\lambda&a_{23}&\cdots&a_{2n} \\\cdots&\cdots&\cdots&\cdots&\cdots\\a_{n1}&a_{n2}&a_{n3}&\cdots&a_{nn}-\lambda\end{array} \right|.\]
then if By replacing \(\lambda\) by \(\alpha\) in that determinant, \(m\) rows or columns become identical then \(\lambda=\alpha\) is root of that polynomial of order \((m-1)\).

 We will use this result to find out the eigenvalues of \(J-I\)

\[A=\left| \begin{array}{ccc}0&1&1&\cdots&1\\1&0&1&\cdots&1 \\\cdots&\cdots&\cdots&\cdots&\cdots\\1&1&1&\cdots&0 \end{array} \right|.\]

So, \[|A-\lambda|=\left| \begin{array}{ccc}0-\lambda&1&1&\cdots&1\\1&0-\lambda&1&\cdots&1 \\\cdots&\cdots&\cdots&\cdots&\cdots\\1&1&1&\cdots&0-\lambda\end{array} \right|.\]

By above result if, we put, \(\lambda=-1\) then total \(101\) rows will be identitical. So, \(A\) has Eigen value=\(-1\) (100 times.)

But trace= Sum of Eigenvalue So, third eigenvalue is \(100\).

So, determinant of \(J-I\)=Multiplication of eigenvalues=\(-1^{100}\times 100=100.\)

12.The system of equations
\begin{align*}x+y+z=1\\2x+3y-z=5\\x+2y-kz=4\end{align*}
Where \(k\in \mathbb{R}\), has an infinite number of solutions for

• \(k=0\)
• \(k=1\)
• \(k=2\)
• \(k=3\)

Solution:  \(\begin{pmatrix} 1&1&1&|1\\2&3&-1&|5\\1&2&-k&|4\end{pmatrix} \rightarrow \begin{pmatrix} 1&1&1&|1\\0&1&-3&|3\\0&1&-k-1&|3\end{pmatrix}\)
\(R_2\to R_2-2R_1\quad R_3\to R_3-R_1\)
Therefore for infinite solution: \(-k-1=-3\Rightarrow k=2\)
Third option is correct.

13.Let \(A\) be a \(5\times 4\) matrix with real entries such that \(Ax=0\) if and only if \(x=0\) where \(x\) is a \(4\times 1\) vector and \(0\) is
a null vector. Then, the rank of \(A\) is

•  \(4\)
• \(5\)
• \(2\)
• \(1\)

Solution: Rank of this matrix is equal to dimension of column space \(=4\).

14.Let \(A\) be a \(5\times 4\) matrix with real entries such that the space of all solutions of the linear system \(AX^t=[1,2,3,4,5]^t\) is
given by \(\{[1+2s,2+3s,3+4s,4+5s]^t:s \in \mathbb{R}\}\). Then the rank of \(A\) is equal to

•  \(4\)
• \(3\)
• \(2\)
• \(1\)

Solution: \(3\) is correct Answer. When Rank of \(A\)=Dimension of column Space then there exists a unique Solution. But here Solution is not Unique. Number of free Variable in the solution is 1. So Rank=\(4-1=3\)

15.Consider a homogeneous system of linear equation \(Ax = 0\) where \(A\) is an \(m \times n\) real matrix and \(n>m\). Then which of the following statements are always true?

• \(Ax = 0\) has a solution
• \(Ax = 0\) has no non-zero  solution
• \(Ax = 0\) has a non-zero solution
• Dimension of the space of all solution is at least \(n-m\)

Solution: Since \(Rank\leq m,n\) and \(n>m\) So, \(rank\neq n\) So, From Rank-Nullity theorem, there must exists a non-zero solution to \(Ax = 0\). Since the dimension of row space\(\leq m\).  So the null space has dimension \(\geq n-m\)

So, Dimension of the space of all Solution is at least \(n-m\).

16.Let \(M_{m \times n}(\mathbb{R})\) be the set of \(m \times n\) matrices with real entries. Which of the following statements is correct?

• There exists \(A \in M_{2 \times 5}(\mathbb{R})\) such that the dimension of the null space of \(A\) is \(2\).
• There exists \(A \in M_{2 \times 5}(\mathbb{R})\) such that the dimension of the null space of \(A\) is \(0\).
• There exists \(A \in M_{2 \times 5}(\mathbb{R})\) and \(B \in M_{5 \times 2}(\mathbb{R})\) such that \(AB\) is the \(2 \times 2\) identify matrix
• There exists \(A \in M_{2 \times 5}(\mathbb{R})\) whose null space is \(\{(x_1, x_2, x_3, x_4, x_5) \mathbb{R}^5:x_1 = x_2, x_3 = x_4 = x_5 \}\)

Solution: option 1 is correct if we take \(A\)= a zero matrix of order \(2\times 5\).

take \(A=\begin{pmatrix} 1&0&0&0&0\\0&1&0&0&0\end{pmatrix}\) then its nullspace has dimension \(0\).

take \(A\) same as above and \(B=A^T\) and after multiplying you will get an identity matrix of order \(2\times 2\).

So, this option is correct.

Take,  \(A=\begin{pmatrix} 1&1&0&0&0\\0&0&0&0&0\end{pmatrix}\) and find its null space.

you will get the same result given in the last option. Hence, this option is also correct.

17.Let \(A\) be a \(5\times 5\) skew-symmetric matrix with entries in \(\mathbb{R}\) and \(B\) be the \(5\times 5\) matrix whose \((i,j)^{th}\) entry is the binomial coefficient \(\begin{pmatrix} i \\ j \end{pmatrix}\) for \(1\leq i \leq j \leq 5\). Consider the \(10 \times 10\) matrix, given in block form by
$$C=\begin{pmatrix}A&A+B\\0&B\end{pmatrix}$$
Then

• \(\det C=1\quad or\quad -1\).
• \(\det C=0\).
• \(trace C=0\).
• \(trace C =5\).

Solution:

\(Trace C=Trace A+Trace B\).

now we know that the diagonal element of Skew symmetric matrix is \(0\).

and diagonal elements of \( B={i \choose i}\) So \(Trace C=0+0+0+0+0+1+1+1+1+1=5\)

So, fourth option is correct.

\(det(A) = det(A^T) = det(-A) = (-1)^n\times det(A)\) So, when \(n\) is odd \(det(A)=0\)

So, determinant of \(C\)= Determinant of \(A\) \(\times \) Determinant of \(B\)= \(0\times B=0\)

So, 2nd option is correct.

 

18.The matrix $$A =\begin{pmatrix} 5& 9& 8\\1&8&2\\9&1&0\end{pmatrix}$$ satisfies:

 

• \(A\) is invertible and the inverse has all integer entries.
• det\((A)\) is odd.
• det\((A)\) is divisible by \(13\).
• det\((A)\) has at least two prime divisors.

19.Let \(A_{n\times n}= (a_{ij}), n\geq 3\), where \(a_{ij}=(b_i^2-b_j^2), i,j=1,2,\cdots ,n\) for some distinct real numbers \(b_1, b_2,\cdots , b_n\). Then \(det(A)\) is

• \(\prod_{i<j}(b_i-b_j)\)
• \(\prod_{i<j}(b_i+b_j)\)
• \(0\)
• \(1\)