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Matrices and Determinants questions for IPMAT, with solutions

10 free Matrices and Determinants practice questions for IPMAT Indore, Rohtak and JIPMAT, each with the answer and a full worked solution. No sign-in. Try each one first, then open the solution. The lessons for this chapter are in Matrices and Determinants on Learn.

1. Question 1Type the answerWritten by IPMATwayDeterminants

   The value of the determinant $\begin{vmatrix}1&2&3\\4&5&6\\7&8&9\end{vmatrix}$ is

   Show the answer and solution

   Answer: 0

   Subtract row 1 from row 2 and row 2 from row 3: $R_2-R_1=(3,3,3)$ and $R_3-R_2=(3,3,3)$. Row operations of this type do not change the determinant. Now two rows are equal, and a determinant with two equal rows is $0$. So the value is $0$. (Direct expansion: $1(45-48)-2(36-42)+3(32-35)=-3+12-9=0$.)

2. Question 2Type the answerWritten by IPMATwayDeterminants

   If $\begin{vmatrix}a&b&c\\p&q&r\\x&y&z\end{vmatrix}=6$, then $\begin{vmatrix}2a&2b&2c\\3p&3q&3r\\x&y&z\end{vmatrix}$ equals

   Show the answer and solution

   Answer: 36

   A common factor of a single row can be taken out of a determinant. Row 1 has the common factor $2$ and row 2 has the common factor $3$. So the new determinant is $2\cdot3\cdot\begin{vmatrix}a&b&c\\p&q&r\\x&y&z\end{vmatrix}=6\cdot6=36$.

3. Question 3Multiple choiceWritten by IPMATwayMatrices

   If $A=\begin{bmatrix}1&2\\3&4\end{bmatrix}$ and $B=\begin{bmatrix}0&1\\1&0\end{bmatrix}$, then $AB$ equals

   1. A.$\begin{bmatrix}1&3\\2&4\end{bmatrix}$
   2. B.$\begin{bmatrix}3&4\\1&2\end{bmatrix}$
   3. C.$\begin{bmatrix}2&1\\4&3\end{bmatrix}$
   4. D.$\begin{bmatrix}0&2\\3&0\end{bmatrix}$
   Show the answer and solution

   Answer: C. $\begin{bmatrix}2&1\\4&3\end{bmatrix}$

   Multiply row by column. Row 1 of $A$ is $(1,2)$: with column 1 of $B$, $(0,1)$, it gives $2$; with column 2, $(1,0)$, it gives $1$. Row 2 of $A$ is $(3,4)$: it gives $4$ and $3$. So $AB=\begin{bmatrix}2&1\\4&3\end{bmatrix}$; multiplying by $B$ on the right swaps the columns of $A$.

4. Question 4Type the answerWritten by IPMATwayMatrices

   If $A$ is a $2\times3$ matrix and $B$ is a $3\times4$ matrix, the number of entries of $AB$ is

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   Answer: 8

   An $m\times n$ matrix times an $n\times p$ matrix is an $m\times p$ matrix. Here $A$ is $2\times3$ and $B$ is $3\times4$, so $AB$ is $2\times4$. The number of entries is $2\cdot4=8$.

5. Question 5Type the answerWritten by IPMATwayMatrices

   If $A=\begin{bmatrix}2&0\\0&3\end{bmatrix}$, then the sum of all the entries of $A^{4}$ is

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   Answer: 97

   For a diagonal matrix, each power just raises the diagonal entries: $A^4=\begin{bmatrix}2^4&0\\0&3^4\end{bmatrix}=\begin{bmatrix}16&0\\0&81\end{bmatrix}$. The off-diagonal entries stay $0$. So the sum of all entries is $16+81=97$.

6. Question 6Type the answerWritten by IPMATwayMatrices

   The number of entries that can be chosen independently in a symmetric matrix of order $4$ is

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   Answer: 10

   A symmetric matrix has $a_{ij}=a_{ji}$, so it is fixed by its entries on and above the diagonal. A $4\times4$ matrix has $4$ diagonal entries and $\frac{16-4}{2}=6$ entries above the diagonal. So $4+6=10$ entries can be chosen independently.

7. Question 7Type the answerWritten by IPMATwayDeterminants

   If $A=\begin{bmatrix}2&1\\1&2\end{bmatrix}$, then $\det(A^{2}-4A)$ equals

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   Answer: 9

   Factor: $A^2-4A=A(A-4I)$, so $\det(A^2-4A)=\det A\cdot\det(A-4I)$. $\det A=2\cdot2-1\cdot1=3$. $A-4I=\begin{bmatrix}-2&1\\1&-2\end{bmatrix}$, with determinant $4-1=3$. So $\det(A^2-4A)=3\cdot3=9$.

8. Question 8Type the answerWritten by IPMATwayDeterminants

   If $A$ is a $3\times3$ matrix with $A(\operatorname{adj}A)=10I$, then $|\operatorname{adj}A|$ equals

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   Answer: 100

   For any square matrix, $A(\operatorname{adj}A)=|A|I$. Comparing with $10I$ gives $|A|=10$. For order $n$, $|\operatorname{adj}A|=|A|^{n-1}$; here $n=3$. So $|\operatorname{adj}A|=10^2=100$.

9. Question 9Type the answerWritten by IPMATwayDeterminants

   If $x+y+z=6$, $x-y+z=2$ and $2x+y-z=1$, then $z$ equals

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   Answer: 3

   Subtract the second equation from the first: $(x+y+z)-(x-y+z)=6-2$, so $2y=4$ and $y=2$. Then the first equation gives $x+z=4$ and the third gives $2x-z=1-2=-1$. Adding these: $3x=3$, so $x=1$ and $z=3$. Check: $1+2+3=6$, $1-2+3=2$, $2+2-3=1$.

10. Question 10Multiple choiceWritten by IPMATwayMatrices

    Given below are two statements about an invertible matrix $A$.

    Statement I: $(A^{T})^{-1}=(A^{-1})^{T}$.

    Statement II: $(2A)^{-1}=2A^{-1}$.

    Choose the correct answer.

    1. A.Both Statement I and Statement II are false
    2. B.Statement I is true but Statement II is false
    3. C.Both Statement I and Statement II are true
    4. D.Statement I is false but Statement II is true
    Show the answer and solution

    Answer: B. Statement I is true but Statement II is false

    Statement I: transpose $AA^{-1}=I$ to get $(A^{-1})^TA^T=I$, so $(A^{-1})^T$ is the inverse of $A^T$ - true. Statement II: $(2A)\left(\frac{1}{2}A^{-1}\right)=I$, so $(2A)^{-1}=\frac{1}{2}A^{-1}$, not $2A^{-1}$ - false. So I is true and II is false.