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

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