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Right ANSWER is (c) 3, -1, 3, 0

To explain: The two MATRICES \(\begin{bmatrix}a&b+c\\c+d&b\end{bmatrix}\)and\(\begin{bmatrix}3&2\\3&-1\end{bmatrix}\) are EQUAL matrices. COMPARING the two matrices, we get

a=3, b+c=2, c+d=3, b=-1

Solving the above equations, we get a=3, b=-1, c=3, d=0.

Right option is (c) \(\BEGIN{BMATRIX}\frac{-1}{4}&\frac{1}{4}&0\\ \frac{1}{40}&\frac{-1}{8}&\frac{1}{5}\\ \frac{3}{40}&0&\frac{-1}{10}\END{bmatrix}\)

For explanation: Consider the MATRIX A=\(\begin{bmatrix}1&2&4\\5&2&4\\3&6&2\end{bmatrix}\)

Using the elementary ROW operation, we write A=IA

\(\begin{bmatrix}1&2&4\\5&2&4\\3&6&2\end{bmatrix}\)=\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)A

Applying R1→R1-R2

\(R_1 \rightarrow \frac{R_1}{-4}\)

\(\begin{bmatrix}1&0&0\\5&2&4\\3&6&2\end{bmatrix}\)=\(\begin{bmatrix}\frac{-1}{4}&\frac{1}{4}&0\\0&1&0\\0&0&1\end{bmatrix}\)A

Applying R2→R2-5R1and R3→R3-3R1

\(\begin{bmatrix}1&0&0\\0&2&4\\0&6&2\end{bmatrix}\)=\(\begin{bmatrix}\frac{-1}{4}&\frac{1}{4}&0\\\frac{5}{4}&\frac{-1}{4}&0\\\frac{3}{4}&\frac{-3}{4}&1\end{bmatrix}\)A

Applying R2→R2-2R3and \(R_2 \rightarrow \frac{R_2}{-10}\)

\(\begin{bmatrix}1&0&0\\0&1&0\\0&6&2\end{bmatrix}\)=\(\begin{bmatrix}\frac{-1}{4}&\frac{1}{4}&0\\ \frac{1}{40}&\frac{-5}{40}&\frac{1}{5}\\\frac{3}{4}&\frac{-3}{4}&1\end{bmatrix}\)A

Applying R3→R3-6R2and \(R_2 \rightarrow \frac{R_2}{2}\)

\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)=\(\begin{bmatrix}\frac{-1}{4}&\frac{1}{4}&0\\ \frac{1}{40}&\frac{-1}{8}&\frac{1}{5}\\ \frac{3}{40}&0&\frac{-1}{10}\end{bmatrix}\)A

A^-1=\(\begin{bmatrix}\frac{-1}{4}&\frac{1}{4}&0\\ \frac{1}{40}&\frac{-1}{8}&\frac{1}{5}\\ \frac{3}{40}&0&\frac{-1}{10}\end{bmatrix}\).

The CORRECT answer is (a) A=\(\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}\)

The EXPLANATION: A matrix is called a scalar matrix if the elements ALONG the diagonal of the matrix are equal and are non-zero i.e. aij=k for i=j and aij=0 for i≠j.

Therefore, the matrix A=\(\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}\) is a scalar matrix.

Right OPTION is (B) A=A’

For EXPLANATION I would say: A MATRIX is A said to be a symmetric matrix if it is EQUAL to its transpose i.e. A=A’.

Correct OPTION is (d) A=\(\begin{bmatrix}1&-1\\-2&5\end{bmatrix}\)

The explanation: A=\(\begin{bmatrix}1&-2\\-1&5\end{bmatrix}\). To find the transpose of the MATRIX, interchange the ROWS with COLUMNS and columns with rows.

Hence, A’=\(\begin{bmatrix}1&-1\\-2&5\end{bmatrix}\).

The correct answer is (d) x=1, y=9

Explanation: Given that, 2\(\begin{bmatrix}5&x\\y-4&6\end{bmatrix}\)+\(\begin{bmatrix}-4&1\\3&2\end{bmatrix}\)=\(\begin{bmatrix}6&3\\10&14\end{bmatrix}\)

⇒\(\begin{bmatrix}2(5)-4&2x+1\\2(y-4)+3&2(6)+2\end{bmatrix}\)=\(\begin{bmatrix}6&3\\10&14\end{bmatrix}\)

Comparing the two MATRICES, 2x+1=3, 2y-8=10

Solving the two EQUATIONS we get, x=1, y=9.

The correct choice is (B) False

Easiest EXPLANATION: The given statement is false. Matrix ADDITION is COMMUTATIVE i.e. A+B=B+A. But matrix MULTIPLICATION is not commutative i.e.AB≠BA.

The correct CHOICE is (a) M=1/2 \(\BEGIN{bmatrix}9&11\\13&16\end{bmatrix}\), N=1/2 \(\begin{bmatrix}1&1\\1&0\end{bmatrix}\)

For explanation I would say: M+N = \(\begin{bmatrix}5&6\\7&8\end{bmatrix}\)-(1) and M-N = \(\begin{bmatrix}4&5\\6&8\end{bmatrix}\)-(2)

Adding equation (1) and equation (2), (M+N)+(M-N)=2M=\(\begin{bmatrix}5&6\\7&8\end{bmatrix}\)+\(\begin{bmatrix}4&5\\6&8\end{bmatrix}\)

M=1/2 \(\begin{bmatrix}9&11\\13&16\end{bmatrix}\).

Subtracting equation (1) and equation (2), (M+N)-(M-N)=2N=\(\begin{bmatrix}5&6\\7&8\end{bmatrix}\)–\(\begin{bmatrix}4&5\\6&8\end{bmatrix}\)

N=1/2 \(\begin{bmatrix}1&1\\1&0\end{bmatrix}\).

Right choice is (d) MINOR matrix

Easiest explanation: Minor matrix is not a TYPE of matrix. Scalar, diagonal, SYMMETRIC are VARIOUS type of MATRICES.

The CORRECT OPTION is (b) 3×2

To explain I WOULD say: The given MATRIX A=\(\begin{bmatrix}2&3\\1&9\\5&2\end{bmatrix}\) has 3 rows and 2 columns. Therefore, the order of the matrix is 3×2.

Correct ANSWER is (b) 2, 9, 7

The explanation: The elements following the condition i=j will have the same row number and COLUMN number. The elements are a11, a22, a33 which in the matrix A are 2, 3, 9 RESPECTIVELY.

Correct CHOICE is (b) \(\BEGIN{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\)

To explain: Consider matrix A=\(\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\), applying the elementary operation R1→2R1+3R2.

\(\begin{bmatrix}2(0)+3(0)&2(0)+3(0)&2(0)+3(0)\\0&0&0\\0&0&0\end{bmatrix}\)=\(\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\).

Therefore, the matrix A=\(\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\), remains same after applying the elementary operation.

Correct ANSWER is (c) C2→2+2C2

To explain: The column operation C2→2+2C2is incorrect. A non-zero number cannot be DIRECTLY added to any column or row in a matrix.

The correct answer is (b) B = \(\begin{bmatrix}4&2\\4&1\end{bmatrix}\)

Best EXPLANATION: Given that, A+B = \(\begin{bmatrix}6&7\\5&0\end{bmatrix}\)and A = \(\begin{bmatrix}2&5\\1&-1\end{bmatrix}\)

⇒B=(A+B)-A = \(\begin{bmatrix}6&7\\5&0\end{bmatrix}\)–\(\begin{bmatrix}2&5\\1&-1\end{bmatrix}\)

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

The correct choice is (c) 2(A+A’)

To explain: GIVEN that, A=\(\begin{bmatrix}a&B\\c&d\end{bmatrix}\)

⇒A’=\(\begin{bmatrix}a&c\\b&d\end{bmatrix}\)

LET B=A-A’=\(\begin{bmatrix}a&b\\c&d\end{bmatrix}\)–\(\begin{bmatrix}a&c\\b&d\end{bmatrix}\)=\(\begin{bmatrix}a-a&b-c\\c-b&d-d\end{bmatrix}\)=\(\begin{bmatrix}0&b-c\\c-b&0\end{bmatrix}\)

B’=\(\begin{bmatrix}0&c-b\\b-c&0\end{bmatrix}\)=B’

THUS, B=A-A’ is a SKEW – symmetric.

The correct OPTION is (a) A is a SKEW-symmetric matrix

To EXPLAIN: Given that, A=\(\BEGIN{bmatrix}1&0\\0&1\end{bmatrix}\)

∴A’=\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)

⇒-A’=\(\begin{bmatrix}-1&0\\0&-1\end{bmatrix}\)≠A. Hence, it is not a skew symmetric matrix.

Correct answer is (a) True

For explanation: The given statement is true. Every SQUARE MATRIX can be expressed as a SUM of sum of symmetric and skew-symmetric matrix.

If A is a square matrix then it can be expressed as

A = \(\frac{1}{2}\)(A+A’)+\(\frac{1}{2}\)(A-A’), where (A+A’) is symmetric and (A-A’) is skew-symmetric.

The CORRECT ANSWER is (c) A+B = \(\begin{BMATRIX}1&7&3\\14&10&16\end{bmatrix}\)

Explanation: Given that, A = \(\begin{bmatrix}1&2&3\\9&10&11\end{bmatrix}\) and B = \(\begin{bmatrix}0&5&0\\5&0&5\end{bmatrix}\)

Then A+B = \(\begin{bmatrix}1+0&2+5&3+0\\9+5&10+0&11+5\end{bmatrix}\) = \(\begin{bmatrix}1&7&3\\14&10&16\end{bmatrix}\).

Correct choice is (C) A=\(\BEGIN{bmatrix}4&5&6 \\5&2&6\\6&4&7\\7&7&1\end{bmatrix}\)

Explanation: The MATRIX A=\(\begin{bmatrix}4&5&6\\5&2&6\\6&4&7\\7&7&1\end{bmatrix}\) is a 3×4 matrix as it as 3 ROWS and 4 columns.

The correct OPTION is (B) 2×2

To explain: The number of ROWS (m) and the number of columns (n) in the given matrix A=\(\begin{bmatrix}3&5\\7&9\end{bmatrix}\) is 2. Therefore, the ORDER of the matrix is 2×2(m×n).

Right option is (a) \(\begin{bmatrix}2&5&4\\16&19&24\\7&2&1\end{bmatrix}\)

The best EXPLANATION: Consider A=\(\begin{bmatrix}2&5&4\\5&2&6\\7&2&1\end{bmatrix}\), after APPLYING R2→2R2+3R1

⇒\(\begin{bmatrix}2&5&4\\2(5)+3(2)&2(2)+3(5)&2(6)+3(4)\\7&2&1\end{bmatrix}\)=\(\begin{bmatrix}2&5&4\\16&19&24\\7&2&1\end{bmatrix}\).

Right choice is (b) False

To elaborate: The given statement is false. A MATRIX is non-invertible if it has all ZEROES in ONE or more rows on L.H.S. This is because after APPLYING all the elementary operations on the matrix, we should get an IDENTITY matrix on the L.H.S. to obtain an inverse of the given matrix, which is not possible if we obtain all zeroes in one or more rows.

The CORRECT option is (C) AB = \(\begin{bmatrix}7&9\\15&23\end{bmatrix}\)

For EXPLANATION I would SAY: Given that, A = \(\begin{bmatrix}1&2\\3&4\end{bmatrix}\) and B = \(\begin{bmatrix}1&5\\3&2\end{bmatrix}\)

Then, AB = \(\begin{bmatrix}1&2\\3&4\end{bmatrix}\)\(\begin{bmatrix}1&5\\3&2\end{bmatrix}\)

=\(\begin{bmatrix}1×1+2×3&1×5+2×2\\3×1+4×3&3×5+4×2\end{bmatrix}\)=\(\begin{bmatrix}7&9\\15&23\end{bmatrix}\).

Right ANSWER is (d) COLUMN matrix

To explain: The given matrix A = \(\begin{bmatrix}4\\12\\36\end{bmatrix}\) is of the ORDER 3×1. The matrix has only one column (n=1). HENCE, it is a column matrix.

Right ANSWER is (d) (3,1)

Explanation: The POSSIBLE ORDERS in which the matrix with 6 ELEMENTS can be formed are 2×3, 3×2, 1×6, 6×1. THEREFORE, the possible orders pairs are (2,3), (3,2), (1,6), (6,1). Thus, (3,1) is not possible.

The correct CHOICE is (a) mn

Best EXPLANATION: The NUMBER of ELEMENTS for a matrix with the ORDER m×n is equal to mn, where m is the number of rows and n is the number of columns in the matrix.

The correct OPTION is (C) \(\begin{bmatrix}\frac{2}{15}&-\frac{1}{15}\\-\frac{1}{15}&\frac{8}{15}\end{bmatrix}\)

Easy explanation: Consider the matrix A=\(\begin{bmatrix}8&1\\1&2\end{bmatrix}\)

Using the elementary row operation, we write A=IA

Applying R2→8R2-R1and R2→R2/15, we get

\(\begin{bmatrix}8&1\\0&1\end{bmatrix}\)=\(\begin{bmatrix}1&0\\-\frac{1}{15}&\frac{8}{15}\end{bmatrix}\)A

Applying R1→R1-R2and R1→R1/8, we get

\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)=\(\begin{bmatrix}\frac{2}{15}&-\frac{1}{15}\\-\frac{1}{15}&\frac{8}{15}\end{bmatrix}\)A

A^-1=\(\begin{bmatrix}\frac{2}{15}&-\frac{1}{15}\\-\frac{1}{15}&\frac{8}{15}\end{bmatrix}\).

The correct ANSWER is (a) True

Explanation: The GIVEN statement is true.

(AB) (AB)^-1=I (Using the formula AA^-1=I)

MULTIPLYING both SIDES by A^-1, we get

A^-1 (AB) (AB)^-1=A^-1 I

(A^-1 A)B(AB)^-1=A^-1

IB(AB^-1)=A^-1

B(AB^-1)=A^-1

⇒B^-1 B(AB^-1)=B^-1 A^-1

(AB^-1)=B^-1 A^-1

Correct answer is (d) Ri→1+kRi

The explanation is: The ELEMENTARY operation Ri→1+kRiis incorrect, the valid elementary operations on matrices are as follows.

i) INTERCHANGING any two rows and columns

ii) The multiplication of the elements of any ROW or column by a non-zero number.

iii) The addition to the elements of any row or column, the CORRESPONDING elements of any other row and column multiplied by any non-zero number.

Right OPTION is (B) False

To EXPLAIN: GIVEN that, A=\(\begin{bmatrix}0&1&1\\1&0&-1\\-1&1&0\end{bmatrix}\)

⇒A’=\(\begin{bmatrix}0&1&-1\\1&0&1\\1&-1&0\end{bmatrix}\). ∴A ≠ A’. Hence, it is not symmetric.

The CORRECT CHOICE is (d) \(\begin{bmatrix}3&-7\\-4&-5\end{bmatrix}\)

To elaborate: Given that, A = \(\begin{bmatrix}3&4\\1&2\end{bmatrix}\)and B = \(\begin{bmatrix}1&5\\2&3\end{bmatrix}\)

⇒2A=2\(\begin{bmatrix}3&4\\1&2\end{bmatrix}\)=\(\begin{bmatrix}6&8\\2&4\end{bmatrix}\) and 3B=3\(\begin{bmatrix}1&5\\2&3\end{bmatrix}\)=\(\begin{bmatrix}3&15\\6&9\end{bmatrix}\)

∴2A-3B = \(\begin{bmatrix}6&8\\2&4\end{bmatrix}\)–\(\begin{bmatrix}3&15\\6&9\end{bmatrix}\)=\(\begin{bmatrix}3&-7\\-4&-5\end{bmatrix}\).

Correct CHOICE is (C) A=\(\begin{bmatrix}4&0&0\\0&5&0\\0&0&9\end{bmatrix}\)

Explanation: The matrix is said to be a diagonal matrix if the elements ALONG the diagonal of the matrix are non – zero.

i.e. aij=0 for i≠j andaij≠0 for i=j.

Therefore, the matrix A=\(\begin{bmatrix}4&0&0\\0&5&0\\0&0&9\end{bmatrix}\) is a diagonal matrix.

The correct option is (c) Horizontal matrix

The explanation: The matrix in which number of ROWS is SMALLER than the number of columns is called is called a horizontal matrix. In the given matrix A=\(\BEGIN{bmatrix}4&6&9\\12&11&10\end{bmatrix}\), m=3 and n=2 i.e.

3<2. HENCE, it is a horizontal matrix.

The CORRECT option is (b) \(\begin{BMATRIX}\frac{-1}{13}&\frac{3}{13}\\ \frac{5}{13}&\frac{-2}{13}\end{bmatrix}\)

The explanation: Consider the matrix A=\(\begin{bmatrix}2&3\\5&1\end{bmatrix}\)

Using ELEMENTARY row operation, we write A=IA.

\(\begin{bmatrix}2&3\\5&1\end{bmatrix}\)=\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)A

\(\begin{bmatrix}-13&0\\5&1\end{bmatrix}\)=\(\begin{bmatrix}1&-3\\0&1\end{bmatrix}\)A(Applying R1→R1-3R2)

\(\begin{bmatrix}1&0\\5&1\end{bmatrix}\)=\(\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\0&1\end{bmatrix}\)A(Applying \(R_1 \rightarrow -\frac{R_1}{13}\))

\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)=\(\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\ \frac{5}{13}&\frac{-2}{13}\end{bmatrix}\)A(Applying R2→R2-5R1)

\(A^{-1}=\begin{bmatrix}\frac{-1}{13}&\frac{3}{13}\\ \frac{5}{13}&\frac{-2}{13}\end{bmatrix}\).

Correct choice is (a) SYMMETRIC MATRIX

To explain I would say: GIVEN that, A=\(\begin{bmatrix}1&2\\2&1\end{bmatrix}\)

⇒ A’=\(\begin{bmatrix}1&2\\2&1\end{bmatrix}\)

i.e.A=A’. HENCE, it is a symmetric matrix.

The correct answer is (a) Square matrix

The best EXPLANATION: A square matrix is a matrix in which the NUMBER of rows(m) is EQUAL to the number of columns(N). THEREFORE, the matrix which follows the condition m=n is a square matrix.

Correct ANSWER is (a) \(\begin{BMATRIX}1&5\\0&8\end{bmatrix}\)

Best explanation: CONSIDER the MATRIX A=\(\begin{bmatrix}1&5\\0&8\end{bmatrix}\)

Using the elementary column operations, we write A=AI

\(\begin{bmatrix}1&5\\0&8\end{bmatrix}\)=A\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)

Applying C2→C2-5C1

\(\begin{bmatrix}1&0\\0&8\end{bmatrix}\)=A\(\begin{bmatrix}1&-5\\0&1\end{bmatrix}\)

Applying C2→C2/8

\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)=A\(\begin{bmatrix}1&-\frac{5}{8}\\0&\frac{1}{8}\end{bmatrix}\)

A^-1=\(\begin{bmatrix}1&-\frac{5}{8}\\0&\frac{1}{8}\end{bmatrix}\).

Correct answer is (d) skew-symmetric MATRIX

To EXPLAIN: The given matrix A=\(\begin{bmatrix}0&1&-1\\-1&0&1\\1&-1&0\end{bmatrix}\) is skew symmetric.

⇒A’=\(\begin{bmatrix}0&-1&1\\1&0&-1\\-1&1&0\end{bmatrix}\)=A

∴A=-A’. HENCE, it is a skew-symmetric matrix.

The correct CHOICE is (C) A=\(\begin{bmatrix}2&3\\3&4\end{bmatrix}\)

Easy explanation: a11=1+1=2, a12=1+2=3, a21=2+1=3, a22=2+2=4

∴ A=\(\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22} \end{bmatrix}=\begin{bmatrix}2&3\\3&4\end{bmatrix}\).

The correct answer is (a) (AB)^-1=A^-1 B^-1

Easiest EXPLANATION: (AB)^-1=A^-1 B^-1 is incorrect. The correct formula is (AB)^-1=B^-1 A^-1. B^-1 A^-1 ≠ A^-1 B^-1 as MATRIX multiplication is not COMMUTATIVE.

Right choice is (b) A=\(\begin{BMATRIX}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\)

The explanation: The MATRIX A=\(\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\)=A’=-A’.HENCE, a null matrix is both symmetric and skew-symmetric.

The correct choice is (B) False

For explanation I would say: The given statement is false. In the matrix A=\(\begin{BMATRIX}1&5&9\\4&8&6\end{bmatrix}\) the number of rows(m) is 3 and the number of columns(n) is 2. Therefore, the order of the matrix is 3×2.

Correct option is (d) R1→R1+2R2

To elaborate: The GIVEN matrix is A=\(\begin{bmatrix}8&5\\2&8\end{bmatrix}\)

APPLYING the elementary operation R1→R1+2R2, we GET

\(\begin{bmatrix}8+2(2)&5+2(8)\\2&8\end{bmatrix}\)=\(\begin{bmatrix}12&21\\2&8\end{bmatrix}\).

Correct CHOICE is (b) \(\begin{BMATRIX}20&8\\-4&2\\12&-4\end{bmatrix}\)

The BEST I can EXPLAIN: Given matrix A=\(\begin{bmatrix}5&8\\-1&2\\3&-4\end{bmatrix}\)

Applying the column operation, C1→4C1 we get

\(\begin{bmatrix}4(5)&8\\4(-1)&2\\4(3)&-4\end{bmatrix}\)=\(\begin{bmatrix}20&8\\-4&2\\12&-4\end{bmatrix}\)

Right answer is (a) TRUE

Easiest explanation: The given STATEMENT is true. The POSSIBLE orders in which the MATRIX can be formed are 3×3,9×1,1×9. Therefore, the ORDERED pairs are (3,3),(1,9),(9,1).

The correct ANSWER is (a) A=IA

To ELABORATE: A MATRIX is said to be skew-symmetric if it is equal to the NEGATIVE of its transpose i.e. A=-A’.